Displaying 1-10 of 108 results found.
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1, 2, 1, 5, 4, 1, 12, 14, 6, 1, 30, 44, 27, 8, 1, 74, 133, 104, 44, 10, 1, 185, 388, 369, 200, 65, 12, 1, 460, 1110, 1236, 814, 340, 90, 14, 1, 1150, 3120, 3980, 3072, 1560, 532, 119, 16, 1, 2868, 8666, 12432, 10984, 6542, 2715, 784, 152, 18, 1, 7170, 23816, 37938, 37688, 25695, 12516, 4403, 1104, 189, 20, 1
FORMULA
T(n,k) = (2*k*sum(j=0..n+k, binomial(j,-n-3*k+2*j)*binomial(n+k,j)))/(n+k). - Vladimir Kruchinin, Oct 12 2011 T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + Sum_{i>=0} T(n-1,k+1+i)*(-2)^i. - Philippe Deléham, Feb 23 2012
EXAMPLE
The triangle starts
1;
2, 1;
5, 4, 1;
12, 14, 6, 1;
30, 44, 27, 8, 1;
MAPLE
A053121 := proc(n, k) if n< k or type(n-k, 'odd') then 0; else (k+1)*binomial(n+1, (n-k)/2)/(n+1) ; end if; end proc:
A038207 := proc(i, j) if j> i then 0; else binomial(i, j)*2^(i-j) ; end if; end proc:
MATHEMATICA
a53121[n_, k_] /; n < k || OddQ[n - k] = 0;
a53121[n_, k_] := (k + 1) Binomial[n + 1, (n - k)/2]/(n + 1);
a38207[n_, k_] := Sum[Binomial[n, i] Binomial[i, k], {i, 0, n}];
T[n_, k_] := Sum[a53121[n, j] a38207[j, k], {j, 0, n}];
PROG
(Maxima)
T(n, k):=(2*k*sum(binomial(j, -n-3*k+2*j)*binomial(n+k, j), j, 0, n+k))/(n+k); /* Vladimir Kruchinin, Oct 12 2011 */
EXTENSIONS
Edited, definition rewritten, program and more terms added by R. J. Mathar, Mar 25 2010
Triangle read by rows, square of A053121.
+20
3
1, 0, 1, 2, 0, 1, 0, 4, 0, 1, 7, 0, 6, 0, 1, 0, 18, 0, 8, 0, 1, 29, 0, 33, 0, 10, 0, 1, 0, 86, 0, 52, 0, 12, 0, 1, 131, 0, 179, 0, 75, 0, 14, 0, 1, 0, 427, 0, 316, 0, 102, 0, 16, 0, 1, 625, 0, 972, 0, 505, 0, 133, 0, 18, 0, 1
COMMENTS
Left border = aerated version of A007852.
EXAMPLE
First few rows of the triangle:
1;
0, 1;
2, 0, 1;
0, 4, 0, 1;
7, 0, 6, 0, 1;
0, 18, 0, 8, 0, 1;
29, 0, 33, 0, 10, 0, 1;
0, 86, 0, 52, 0, 12, 0, 1;
131, 0, 179, 0, 75, 0, 14, 0, 1;
0, 427, 0, 316, 0, 102, 0, 16, 0, 1;
625, 0, 972, 0, 505, 0, 133, 0, 18, 0, 1;
...
CROSSREFS
This triangle is the square of A053121.
1, 0, 1, 1, 0, 2, 0, 2, 0, 3, 2, 0, 6, 0, 6, 0, 5, 0, 12, 0, 10, 5, 0, 18, 0, 30, 0, 20, 0, 14, 0, 42, 0, 60, 0, 35, 14, 0, 56, 0, 120, 0, 140, 0, 70, 0, 42, 0, 144, 0, 270, 0, 280, 0, 126
COMMENTS
Row sums = A145974: (1, 1, 3, 5, 14, 27, 73,...).
FORMULA
infinite lower triangular matrix with A001405 as the main diagonal and the rest zeros.
EXAMPLE
First few rows of the triangle =
1;
0, 1;
1, 0, 2;
0, 2, 0, 3;
2, 0, 6, 0, 6;
0, 5, 0, 12, 0, 10;
5, 0, 18, 0, 30, 0, 20;
0, 14, 0, 42, 0, 60, 0, 35;
14, 0, 56, 0, 120, 0, 140, 0, 70;
0, 42, 0, 144, 0, 270, 0, 280, 0, 126;
42, 0, 180, 0, 450, 0, 700, 0, 630, 0, 252;
...
Example: Row 4 = (2, 0, 6, 0, 6) = termwise products of (2, 0, 3, 0, 1) and (1, 1, 2, 3, 6).
Row maximum of Catalan triangle with zeros ( A053121), i.e., maximum value of (m+1)*binomial(n+1,(n-m)/2)/(n+1) for given n with m same parity as n.
+20
2
1, 1, 1, 2, 3, 5, 9, 14, 28, 48, 90, 165, 297, 572, 1001, 2002, 3640, 7072, 13260, 25194, 48450, 90440, 177650, 326876, 653752, 1225785, 2414425, 4601610, 8947575, 17298645, 33266625, 65132550, 124062000, 245642760, 463991880, 927983760
COMMENTS
There are two maximum values when n is of the form k^2 + 2k - 1 (i.e., 2 less than a square, A008865 offset) in which case m = k +/- 1. In general m is the integer with the same parity as n closest to sqrt(n+2) - 1. The largest difference between adjacent binomial coefficients on n-th row of Pascal's triangle. - Vladimir Reshetnikov, Sep 16 2019
FORMULA
a(n) = (m+1)*binomial(n+1, (n-m)/2)/(n+1) where m = floor(sqrt(n+2) - (1 + (-1)^floor(n + sqrt(n+2) - 1))/2). a(n) seems to be slightly less than 2^n/n.
1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 4, 3, 4, 1, 1, 4, 8, 4, 5, 1, 1, 9, 8, 13, 5, 6, 1, 1, 9, 22, 13, 19, 6, 7, 1, 1, 23, 22, 41, 19, 26, 7, 8, 1, 1, 23, 64, 41, 67, 26, 34, 8, 9, 1, 1, 65, 64, 131, 67, 101, 34, 43, 9, 10, 1, 1
COMMENTS
Row sums = A036256: (1, 2, 4, 7, 13, 23, 43,...).
EXAMPLE
1;
0, 1;
1, 0, 1;
0, 2, 0, 1;
...
Taking partial sums from the top, we get A145972: 1;
1, 1;
2, 1, 1;
2, 3, 1, 1;
4, 3, 4, 1, 1;
4, 8, 4, 5, 1, 1;
9, 8, 13, 5, 6, 1, 1;
9, 22, 13, 19, 6, 7, 1, 1;
23, 22, 41, 19, 26, 7, 8, 1, 1;
23, 64, 41, 67, 26, 34, 8, 9, 1, 1;
...
Riordan transform of the Fibonacci sequence with the Riordan matrix A053121.
+20
1
0, 1, 1, 4, 6, 18, 32, 85, 165, 411, 839, 2013, 4237, 9933, 21317, 49236, 107014, 244750, 536500, 1218888, 2687362, 6077644, 13453606, 30329434, 67326816, 151439158, 336842092, 756452890, 1684953360, 3779590010, 8427441240
COMMENTS
The Riordan matrix A053121 of the Bell type R = (c(x^2), x*c(x^2)), with the o.g.f. c of A000108 (Catalan), is the inverse of the Riordan matrix A049310 (Chebyshev S). The Riordan transform of a sequence {a_n}, n >= 0 to a sequence {b_n}, n >= 0 is b = R a
in matrix notation, with the lower triangular Riordan matrix (N x N, with arbitrary large N).
FORMULA
G.f.: c(x^2)*A(x*c(x^2)) with the g.f. c of A000108 (Catalan) and A of A000045 (Fibonacci). a(n) = Sum_{m=0..n} R(n, m)*F(n), with R(n, m) = A053121(n, m) and F(m) = A000045(m), n >= 0.
a(n) = Sum_{k=0..n} binomial(n, k)*CT(n, k) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*CT(n, k), where CT(n, k) is the Catalan triangle A053121.
+20
1
1, 1, 2, 7, 21, 66, 216, 715, 2395, 8101, 27598, 94568, 325612, 1125632, 3904512, 13583195, 47373255, 165585883, 579907758, 2034443127, 7148313381, 25151582046, 88607951512, 312518438532, 1103393962996, 3899415207676, 13792683831176, 48825746365672, 172971084083752
FORMULA
a(n) = Sum_{j=0..n} even(n + j)*binomial(n, j)*binomial(n + 1, (n - j)/2)*(j + 1)/(n + 1), where even(k) = 1 if k is even and otherwise 0.
Recurrence: 2*(n+1)*(2*n + 1)*(13*n^3 - 17*n^2 - 12*n + 10)*a(n) = (143*n^5 - 44*n^4 - 525*n^3 + 320*n^2 + 94*n - 60)*a(n-1) + 4*(n-1)*(26*n^4 + 5*n^3 - 108*n^2 + 7*n + 22)*a(n-2) + 16*(n-2)*(n-1)*(13*n^3 + 22*n^2 - 7*n - 6)*a(n-3).
a(n) ~ sqrt((247 - 9131*13^(2/3)/(1788163 + 409728*sqrt(78))^(1/3) + (13*(1788163 + 409728*sqrt(78)))^(1/3))/13) * ((11 + (1/3)*(172017 - 16848*sqrt(78))^(1/3) + (6371 + 624*sqrt(78))^(1/3))^n / (sqrt(Pi*n) * 2^(2*n + 3) * 3^(n + 1/2))). (End)
MAPLE
a := n -> add((if n + j mod 2 = 1 then 0 else binomial(n, j)*binomial(n + 1, (n - j)/2)*(j + 1)/(n + 1) fi), j = 0..n): seq(a(n), n = 0..28);
MATHEMATICA
Table[Sum[(1 + (-1)^(n+j))/2 * Binomial[n, j] * Binomial[n+1, (n-j)/2] * (j+1)/(n+1), {j, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Apr 21 2024 *)
1, 1, 1, 2, 4, 9, 22, 56, 152, 428, 1257, 3827, 12023, 38975, 129801, 443833, 1554849, 5572334, 20409121, 76283610, 290737945, 1128730320, 4460079331, 17924115817, 73207479057, 303691230784, 1278796858227, 5462983288161, 23664467118878, 103895077228910
FORMULA
a(n) = Sum_{k=0..n-1} A053121(n-1,k)*a(k) for n>0 with a(0)=1.
T(n, k) = P(n-k, k) where P(n, x) = Sum_{k=0..n} A053121(n, k)*x^k. Triangle read by rows, for 0 <= k <= n.
+20
0
1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 2, 3, 5, 3, 1, 0, 6, 12, 10, 4, 1, 5, 10, 30, 33, 7, 5, 1, 0, 20, 74, 110, 72, 26, 6, 1, 14, 35, 185, 366, 306, 135, 37, 7, 1, 0, 70, 460, 1220, 1300, 702, 228, 50, 8, 1, 42, 126, 1150, 4065, 5525, 3650, 1406, 357, 65, 9, 1
EXAMPLE
Triangle starts:
[0] [ 1]
[1] [ 0, 1]
[2] [ 1, 1, 1]
[3] [ 0, 2, 2, 1]
[4] [ 2, 3, 5, 3, 1]
[5] [ 0, 6, 12, 10, 4, 1]
[6] [ 5, 10, 30, 33, 17, 5, 1]
[7] [ 0, 20, 74, 110, 72, 26, 6, 1]
[8] [14, 35, 185, 366, 306, 135, 37, 7, 1]
[9] [ 0, 70, 460, 1220, 1300, 702, 228, 50, 8, 1]
MAPLE
A053121 := (n, k, x) -> irem(n+k+1, 2)*x^k*(k+1)*binomial(n+1, (n-k)/2)/(n+1):
P := (n, x) -> add( A053121(n, k, x), k=0..n): seq(seq(P(n-k, k), k=0..n), n=0..10);
Catalan numbers: C(n) = binomial(2n,n)/(n+1) = (2n)!/(n!(n+1)!).
(Formerly M1459 N0577)
+10
4070
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152, 69533550916004, 263747951750360, 1002242216651368, 3814986502092304
COMMENTS
These were formerly sometimes called Segner numbers.
A very large number of combinatorial interpretations are known - see references, esp. R. P. Stanley, "Catalan Numbers", Cambridge University Press, 2015. This is probably the longest entry in the OEIS, and rightly so.
The solution to Schröder's first problem: number of ways to insert n pairs of parentheses in a word of n+1 letters. E.g., for n=2 there are 2 ways: ((ab)c) or (a(bc)); for n=3 there are 5 ways: ((ab)(cd)), (((ab)c)d), ((a(bc))d), (a((bc)d)), (a(b(cd))).
Consider all the binomial(2n,n) paths on squared paper that (i) start at (0, 0), (ii) end at (2n, 0) and (iii) at each step, either make a (+1,+1) step or a (+1,-1) step. Then the number of such paths that never go below the x-axis (Dyck paths) is C(n). [Chung-Feller]
Number of noncrossing partitions of the n-set. For example, of the 15 set partitions of the 4-set, only [{13},{24}] is crossing, so there are a(4)=14 noncrossing partitions of 4 elements. - Joerg Arndt, Jul 11 2011 a(n-1) is the number of ways of expressing an n-cycle (123...n) in the symmetric group S_n as a product of n-1 transpositions (u_1,v_1)*(u_2,v_2)*...*(u_{n-1},v_{n-1}) where u_i<v_i and u_k <= u_j for k < j; see example. If the condition is dropped, one obtains A000272. - Joerg Arndt and Greg Stevenson, Jul 11 2011 a(n) is the number of ordered rooted trees with n nodes, not including the root. See the Conway-Guy reference where these rooted ordered trees are called plane bushes. See also the Bergeron et al. reference, Example 4, p. 167. - Wolfdieter Lang, Aug 07 2007 As shown in the paper from Beineke and Pippert (1971), a(n-2)=D(n) is the number of labeled dissections of a disk, related to the number R(n)= A001761(n-2) of labeled planar 2-trees having n vertices and rooted at a given exterior edge, by the formula D(n)=R(n)/(n-2)!. - M. F. Hasler, Feb 22 2012 Shifts one place left when convolved with itself.
For n >= 1, a(n) is also the number of rooted bicolored unicellular maps of genus 0 on n edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 15 2001
Number of ways of joining 2n points on a circle to form n nonintersecting chords. (If no such restriction imposed, then the number of ways of forming n chords is given by (2n-1)!! = (2n)!/(n!*2^n) = A001147(n).) Arises in Schubert calculus - see Sottile reference.
Inverse Euler transform of sequence is A022553. With interpolated zeros, the inverse binomial transform of the Motzkin numbers A001006. - Paul Barry, Jul 18 2003 The Hankel transforms of this sequence or of this sequence with the first term omitted give A000012 = 1, 1, 1, 1, 1, 1, ...; example: Det([1, 1, 2, 5; 1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132]) = 1 and Det([1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132; 14, 42, 132, 429]) = 1. - Philippe Deléham, Mar 04 2004 a(n) equals the sum of squares of terms in row n of triangle A053121, which is formed from successive self-convolutions of the Catalan sequence. - Paul D. Hanna, Apr 23 2005 Also coefficients of the Mandelbrot polynomial M iterated an infinite number of times. Examples: M(0) = 0 = 0*c^0 = [0], M(1) = c = c^1 + 0*c^0 = [1 0], M(2) = c^2 + c = c^2 + c^1 + 0*c^0 = [1 1 0], M(3) = (c^2 + c)^2 + c = [0 1 1 2 1], ... ... M(5) = [0 1 1 2 5 14 26 44 69 94 114 116 94 60 28 8 1], ... - Donald D. Cross (cosinekitty(AT)hotmail.com), Feb 04 2005
The multiplicity with which a prime p divides C_n can be determined by first expressing n+1 in base p. For p=2, the multiplicity is the number of 1 digits minus 1. For p an odd prime, count all digits greater than (p+1)/2; also count digits equal to (p+1)/2 unless final; and count digits equal to (p-1)/2 if not final and the next digit is counted. For example, n=62, n+1 = 223_5, so C_62 is not divisible by 5. n=63, n+1 = 224_5, so 5^3 | C_63. - Franklin T. Adams-Watters, Feb 08 2006 Koshy and Salmassi give an elementary proof that the only prime Catalan numbers are a(2) = 2 and a(3) = 5. Is the only semiprime Catalan number a(4) = 14? - Jonathan Vos Post, Mar 06 2006 The answer is yes. Using the formula C_n = binomial(2n,n)/(n+1), it is immediately clear that C_n can have no prime factor greater than 2n. For n >= 7, C_n > (2n)^2, so it cannot be a semiprime. Given that the Catalan numbers grow exponentially, the above consideration implies that the number of prime divisors of C_n, counted with multiplicity, must grow without limit. The number of distinct prime divisors must also grow without limit, but this is more difficult. Any prime between n+1 and 2n (exclusive) must divide C_n. That the number of such primes grows without limit follows from the prime number theorem. - Franklin T. Adams-Watters, Apr 14 2006 The number of ways to place n indistinguishable balls in n numbered boxes B1,...,Bn such that at most a total of k balls are placed in boxes B1,...,Bk for k=1,...,n. For example, a(3)=5 since there are 5 ways to distribute 3 balls among 3 boxes such that (i) box 1 gets at most 1 ball and (ii) box 1 and box 2 together get at most 2 balls:(O)(O)(O), (O)()(OO), ()(OO)(O), ()(O)(OO), ()()(OOO). - Dennis P. Walsh, Dec 04 2006 a(n) is also the order of the semigroup of order-decreasing and order-preserving full transformations (of an n-element chain) - now known as the Catalan monoid. - Abdullahi Umar, Aug 25 2008 a(n) is the number of trivial representations in the direct product of 2n spinor (the smallest) representations of the group SU(2) (A(1)). - Rutger Boels (boels(AT)nbi.dk), Aug 26 2008
Limit_{n->oo} a(n)/a(n-1) = 4. - Francesco Antoni (francesco_antoni(AT)yahoo.com), Nov 24 2008
C(n) is the degree of the Grassmannian G(1,n+1): the set of lines in (n+1)-dimensional projective space, or the set of planes through the origin in (n+2)-dimensional affine space. The Grassmannian is considered a subset of N-dimensional projective space, N = binomial(n+2,2) - 1. If we choose 2n general (n-1)-planes in projective (n+1)-space, then there are C(n) lines that meet all of them. - Benji Fisher (benji(AT)FisherFam.org), Mar 05 2009
Starting with offset 1 = A068875: (1, 2, 4, 10, 18, 84, ...) convolved with Fine numbers, A000957: (1, 0, 1, 2, 6, 18, ...). a(6) = 132 = (1, 2, 4, 10, 28, 84) dot (18, 6, 2, 1, 0, 1) = (18 + 12 + 8 + 10 + 0 + 84) = 132. - Gary W. Adamson, May 01 2009 Convolved with A032443: (1, 3, 11, 42, 163, ...) = powers of 4, A000302: (1, 4, 16, ...). - Gary W. Adamson, May 15 2009 Sum_{k>=1} C(k-1)/2^(2k-1) = 1. The k-th term in the summation is the probability that a random walk on the integers (beginning at the origin) will arrive at positive one (for the first time) in exactly (2k-1) steps. - Geoffrey Critzer, Sep 12 2009 C(p+q)-C(p)*C(q) = Sum_{i=0..p-1, j=0..q-1} C(i)*C(j)*C(p+q-i-j-1). - Groux Roland, Nov 13 2009 Leonhard Euler used the formula C(n) = Product_{i=3..n} (4*i-10)/(i-1) in his 'Betrachtungen, auf wie vielerley Arten ein gegebenes polygonum durch Diagonallinien in triangula zerschnitten werden könne' and computes by recursion C(n+2) for n = 1..8. (Berlin, 4th September 1751, in a letter to Goldbach.) - Peter Luschny, Mar 13 2010 a(n) is also the number of quivers in the mutation class of type B_n or of type C_n. - Christian Stump, Nov 02 2010 Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1) equals the number of ways to choose 0 or more balls of each color while satisfying the following conditions: 1. No two colors are chosen the same positive number of times. 2. For any two colors (c, d) that are chosen at least once, color c is chosen more times than color d iff color c appears more times in the original set than color d. If the second requirement is lifted, the number of acceptable ways equals A000110(n+1). See related comments for A016098, A085082. (End) Deutsch and Sagan prove the Catalan number C_n is odd if and only if n = 2^a - 1 for some nonnegative integer a. Lin proves for every odd Catalan number C_n, we have C_n == 1 (mod 4). - Jonathan Vos Post, Dec 09 2010 a(n) is the number of functions f:{1,2,...,n}->{1,2,...,n} such that f(1)=1 and for all n >= 1 f(n+1) <= f(n)+1. For a nice bijection between this set of functions and the set of length 2n Dyck words, see page 333 of the Fxtbook (see link below). - Geoffrey Critzer, Dec 16 2010 Postnikov (2005) defines "generalized Catalan numbers" associated with buildings (e.g., Catalan numbers of Type B, see A000984). - N. J. A. Sloane, Dec 10 2011 Number of permutations in S(n) for which length equals depth. - Bridget Tenner, Feb 22 2012 a(n) is the number of binary sequences of length 2n+1 in which the number of ones first exceed the number of zeros at entry 2n+1. See the example below in the example section. - Dennis P. Walsh, Apr 11 2012 Number of binary necklaces of length 2*n+1 containing n 1's (or, by symmetry, 0's). All these are Lyndon words and their representatives (as cyclic maxima) are the binary Dyck words. - Joerg Arndt, Nov 12 2012 Number of sequences consisting of n 'x' letters and n 'y' letters such that (counting from the left) the 'x' count >= 'y' count. For example, for n=3 we have xxxyyy, xxyxyy, xxyyxy, xyxxyy and xyxyxy. - Jon Perry, Nov 16 2012 a(n) is the number of Motzkin paths of length n-1 in which the (1,0)-steps come in 2 colors. Example: a(4)=14 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 8 paths of shape HHH, 2 paths of shape UHD, 2 paths of shape UDH, and 2 paths of shape HUD. - José Luis Ramírez Ramírez, Jan 16 2013 If p is an odd prime, then (-1)^((p-1)/2)*a((p-1)/2) mod p = 2. - Gary Detlefs, Feb 20 2013 Conjecture: For any positive integer n, the polynomial Sum_{k=0..n} a(k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013 For 0 < p < 1, define f(p) = Sum_{n>=0} a(n)*(p*(1-p))^n, then f(p) = min{1/p, 1/(1-p)}, so f(p) reaches its maximum value 2 at p = 0.5, and p*f(p) is constant 1 for 0.5 <= p < 1. - Bob Selcoe, Nov 16 2013 [Corrected by Jianing Song, May 21 2021] No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013 Prime p divides a((p+1)/2) for p > 3. See A120303(n) = Largest prime factor of Catalan number. Reciprocal Catalan Constant C = 1 + 4*sqrt(3)*Pi/27 = 1.80613.. = A121839. Log(Phi) = (125*C - 55) / (24*sqrt(5)), where C = Sum_{k>=1} (-1)^(k+1)*1/a(k). See A002390 = Decimal expansion of natural logarithm of golden ratio. 3-d analog of the Catalan numbers: (3n)!/(n!(n+1)!(n+2)!) = A161581(n) = A006480(n) / ((n+1)^2*(n+2)), where A006480(n) = (3n)!/(n!)^3 De Bruijn's S(3,n). (End) For a relation to the inviscid Burgers's, or Hopf, equation, see A001764. - Tom Copeland, Feb 15 2014 One class of generalized Catalan numbers can be defined by g.f. A(x) = (1-sqrt(1-q*4*x*(1-(q-1)*x)))/(2*q*x) with nonzero parameter q. Recurrence: (n+3)*a(n+2) -2*q*(2*n+3)*a(n+1) +4*q*(q-1)*n*a(n) = 0 with a(0)=1, a(1)=1.
Asymptotic approximation for q >= 1: a(n) ~ (2*q+2*sqrt(q))^n*sqrt(2*q*(1+sqrt(q))) /sqrt(4*q^2*Pi*n^3).
For q <= -1, the g.f. defines signed sequences with asymptotic approximation: a(n) ~ Re(sqrt(2*q*(1+sqrt(q)))*(2*q+2*sqrt(q))^n) / sqrt(q^2*Pi*n^3), where Re denotes the real part. Due to Stokes' phenomena, accuracy of the asymptotic approximation deteriorates at/near certain values of n.
(End)
Number of sequences [s(0), s(1), ..., s(n)] with s(n)=0, Sum_{j=0..n} s(j) = n, and Sum_{j=0..k} s(j)-1 >= 0 for k < n-1 (and necessarily Sum_{j=0..n-1} s(j)-1 = 0). These are the branching sequences of the (ordered) trees with n non-root nodes, see example. - Joerg Arndt, Jun 30 2014 Number of stack-sortable permutations of [n], these are the 231-avoiding permutations; see the Bousquet-Mélou reference. - Joerg Arndt, Jul 01 2014 a(n) is the number of increasing strict binary trees with 2n-1 nodes that avoid 132. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014 In a one-dimensional medium with elastic scattering (zig-zag walk), first recurrence after 2n+1 scattering events has the probability C(n)/2^(2n+1). - Joachim Wuttke, Sep 11 2014 The o.g.f. C(x) = (1 - sqrt(1-4x))/2, for the Catalan numbers, with comp. inverse Cinv(x) = x*(1-x) and the functions P(x) = x / (1 + t*x) and its inverse Pinv(x,t) = -P(-x,t) = x / (1 - t*x) form a group under composition that generates or interpolates among many classic arrays, such as the Motzkin (Riordan, A005043), Fibonacci ( A000045), and Fine ( A000957) numbers and polynomials ( A030528), and enumerating arrays for Motzkin, Dyck, and Łukasiewicz lattice paths and different types of trees and non-crossing partitions ( A091867, connected to sums of the refined Narayana numbers A134264). - Tom Copeland, Nov 04 2014 Conjecture: All the rational numbers Sum_{i=j..k} 1/a(i) with 0 < min{2,k} <= j <= k have pairwise distinct fractional parts. - Zhi-Wei Sun, Sep 24 2015 The Catalan number series A000108(n+3), offset n=0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset n=0 (empirical observation). - Tony Foster III, Sep 05 2016 Hankel transforms of the Catalan numbers with the first 2, 4, and 5 terms omitted give A001477, A006858, and A091962, respectively, without the first 2 terms in all cases. More generally, the Hankel transform of the Catalan numbers with the first k terms omitted is H_k(n) = Product_{j=1..k-1} Product_{i=1..j} (2*n+j+i)/(j+i) [see Cigler (2011), Eq. (1.14) and references therein]; together they form the array A078920/ A123352/ A368025. - Andrey Zabolotskiy, Oct 13 2016 Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. See S. J. Miller, ed., 2015, p. 5. - N. J. A. Sloane, Feb 09 2017 Coefficients of the generating series associated to the Magmatic and Dendriform operadic algebras. Cf. p. 422 and 435 of the Loday et al. paper. - Tom Copeland, Jul 08 2018 Let M_n be the n X n matrix with M_n(i,j) = binomial(i+j-1,2j-2); then det(M_n) = a(n). - Tony Foster III, Aug 30 2018 Also the number of Catalan trees, or planted plane trees (Bona, 2015, p. 299, Theorem 4.6.3). - N. J. A. Sloane, Dec 25 2018 Number of coalescent histories for a caterpillar species tree and a matching caterpillar gene tree with n+1 leaves (Rosenberg 2007, Corollary 3.5). - Noah A Rosenberg, Jan 28 2019 Finding solutions of eps*x^2+x-1 = 0 for eps small, that is, writing x = Sum_{n>=0} x_{n}*eps^n and expanding, one finds x = 1 - eps + 2*eps^2 - 5*eps^3 + 14*eps^3 - 42*eps^4 + ... with x_{n} = (-1)^n*C(n). Further, letting x = 1/y and expanding y about 0 to find large roots, that is, y = Sum_{n>=1} y_{n}*eps^n, one finds y = 0 - eps + eps^2 - 2*eps^3 + 5*eps^3 - ... with y_{n} = (-1)^n*C(n-1). - Derek Orr, Mar 15 2019 For n > 0, a random selection of n + 1 objects (the minimum number ensuring one pair by the pigeonhole principle) from n distinct pairs of indistinguishable objects contains only one pair with probability 2^(n-1)/a(n) = b(n-1)/ A098597(n), where b is the 0-offset sequence with the terms of A120777 repeated (1,1,4,4,8,8,64,64,128,128,...). E.g., randomly selecting 6 socks from 5 pairs that are black, blue, brown, green, and white, results in only one pair of the same color with probability 2^(5-1)/a(5) = 16/42 = 8/21 = b(4)/ A098597(5). - Rick L. Shepherd, Sep 02 2019 See Haran & Tabachnikov link for a video discussing Conway-Coxeter friezes. The Conway-Coxeter friezes with n nontrivial rows are generated by the counts of triangles at each vertex in the triangulations of regular n-gons, of which there are a(n). - Charles R Greathouse IV, Sep 28 2019 For connections to knot theory and scattering amplitudes from Feynman diagrams, see Broadhurst and Kreimer, and Todorov. Eqn. 6.12 on p. 130 of Bessis et al. becomes, after scaling, -12g * r_0(-y/(12g)) = (1-sqrt(1-4y))/2, the o.g.f. (expressed as a Taylor series in Eqn. 7.22 in 12gx) given for the Catalan numbers in Copeland's (Sep 30 2011) formula below. (See also Mizera p. 34, Balduf pp. 79-80, Keitel and Bartosch.) - Tom Copeland, Nov 17 2019 Number of permutations in S_n whose principal order ideals in the weak order are modular lattices. - Bridget Tenner, Jan 16 2020 Number of permutations in S_n whose principal order ideals in the weak order are distributive lattices. - Bridget Tenner, Jan 16 2020 Legendre gives the following formula for computing the square root modulo 2^m:
sqrt(1 + 8*a) mod 2^m = (1 + 4*a*Sum_{i=0..m-4} C(i)*(-2*a)^i) mod 2^m
as cited by L. D. Dickson, History of the Theory of Numbers, Vol. 1, 207-208. - Peter Schorn, Feb 11 2020 a(n) is the number of length n permutations sorted to the identity by a consecutive-132-avoiding stack followed by a classical-21-avoiding stack. - Kai Zheng, Aug 28 2020 Number of non-crossing partitions of a 2*n-set with n blocks of size 2. Also number of non-crossing partitions of a 2*n-set with n+1 blocks of size at most 3, and without cyclical adjacencies. The two partitions can be mapped by rotated Kreweras bijection. - Yuchun Ji, Jan 18 2021 Named by Riordan (1968, and earlier in Mathematical Reviews, 1948 and 1964) after the French and Belgian mathematician Eugène Charles Catalan (1814-1894) (see Pak, 2014). - Amiram Eldar, Apr 15 2021 For n >= 1, a(n-1) is the number of interpretations of x^n is an algebra where power-associativity is not assumed. For example, for n = 4 there are a(3) = 5 interpretations: x(x(xx)), x((xx)x), (xx)(xx), (x(xx))x, ((xx)x)x. See the link "Non-associate powers and a functional equation" from I. M. H. Etherington and the page "Nonassociative Product" from Eric Weisstein's World of Mathematics for detailed information. See also A001190 for the case where multiplication is commutative. - Jianing Song, Apr 29 2022 Number of states in the transition diagram associated with the Laplacian system over the complete graph K_N, corresponding to ordered initial conditions x_1 < x_2 < ... < x_N. - Andrea Arlette España, Nov 06 2022 a(n) is the number of 132-avoiding stabilized-interval-free permutations of size n+1. - Juan B. Gil, Jun 22 2023 Number of rooted polyominoes composed of n triangular cells of the hyperbolic regular tiling with Schläfli symbol {3,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {3,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024 a(n) is the number of extremely lucky Stirling permutations of order n; i.e., the number of Stirling permutations of order n that have exactly n lucky cars. (see Colmenarejo et al. reference) - Bridget Tenner, Apr 16 2024
REFERENCES
The large number of references and links demonstrates the ubiquity of the Catalan numbers.
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LINKS
P. C. Allaart and K. Kawamura, The Takagi function: a survey, Real Analysis Exchange, 37 (2011/12), 1-54; arXiv:1110.1691 [math.CA]. See Section 3.2. Drew Armstrong, Generalized Noncrossing Partitions and Combinatorics of Coxeter Groups, Mem. Amer. Math. Soc. 202 (2009), no. 949, x+159. MR 2561274 16; See Table 2.8. Also arXiv: math/0611106, 2006-2007. C. Banderier, M. Bousquet-Mélou, A. Denise, P. Flajolet, D. Gardy and D. Gouyou-Beauchamps, INRIA report 3661, preprint for FPSAC 99, Generating Functions for Generating Trees, Discrete Mathematics 246(1-3), March 2002, pp. 29-55. L. W. Beineke and R. E. Pippert, Enumerating labeled k-dimensional trees and ball dissections, pp. 12-26 of Proceedings of Second Chapel Hill Conference on Combinatorial Mathematics and its Applications, University of North Carolina, Chapel Hill, 1970. Reprinted in Math. Annalen 191 (1971), 87-98. M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Alg. Applications, 226-228 (1995), 57-72; erratum 320 (2000), 210. [Link to Lin. Alg. Applic. version together with omitted figures]. Aubrey Blecher, Charlotte Brennan and Arnold Knopfmacher, Water capacity of Dyck paths, Advances in Applied Mathematics (2019) Vol. 112, 101945. M. Bousquet-Mélou and Gilles Schaeffer, Walks on the slit plane, Probability Theory and Related Fields, Vol. 124, no. 3 (2002), 305-344. David Callan and Emeric Deutsch, The Run Transform, Discrete Math. 312 (2012), no. 19, 2927-2937, arXiv:1112.3639 [math.CO], 2011. Peter J. Cameron, Some treelike objects, The Quarterly Journal of Mathematics, Vol. 38, No. 2 (1987), 155-183. See pp. 155, 162. A. Cayley, On the partitions of a polygon, Proc. London Math. Soc., 22 (1891), 237-262 = Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 93ff. Julie Christophe, Jean-Paul Doignon and Samuel Fiorini, Counting Biorders, J. Integer Seqs., Vol. 6, 2003. Kai Lai Chung and W. Feller, On Fluctuations in Coin-Tossing, Proceedings of the National Academy of Sciences of the United States of America, Vol. 35, No. 10 (1949), 605-608. Laura Colmenarejo, Aleyah Dawkins, Jennifer Elder, Pamela E. Harris, Kimberly J. Harry, Selvi Kara, Dorian Smith, and Bridget Eileen Tenner, On the lucky and displacement statistics of Stirling permutations, arXiv:2403.03280 [math.CO], 2024. Dennis E. Davenport, Lara K. Pudwell, Louis W. Shapiro and Leon C. Woodson, The Boundary of Ordered Trees, Journal of Integer Sequences, Vol. 18 (2015), Article 15.5.8. I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. [Annotated scanned copy]. Part II [not scanned] is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue. S. Gilliand, C. Johnson, S. Rush, D. Wood, The sock matching problem, Involve, a Journal of Mathematics, Vol. 7 (2014), No. 5, 691-697. F. Harary & R. W. Robinson, The number of achiral trees, Jnl. Reine Angewandte Mathematik 278 (1975), 322-335. (Annotated scanned copy) G. Kreweras, Sur les éventails de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #15 (1970), 3-41. [Annotated scanned copy] Colin L. Mallows and Lou Shapiro, Balls on the Lawn, J. Integer Sequences, Vol. 2, 1999, #5. D. Merlini, R. Sprugnoli and M. C. Verri, Waiting patterns for a printerDiscrete Applied Mathematics, 144 (2004), 359-373; FUN with algorithm'01, Isola d'Elba, 2001. R. J. Nowakowski, G. Renault, E. Lamoureux, S. Mellon and T. Miller, The Game of timber!, 2013. C. D. Olds (Proposer) and H. W. Becker (Discussion), Problem 4277, Amer. Math. Monthly 56 (1949), 697-699. [Annotated scanned copy] Vincent Pilaud, Pebble trees, arXiv:2205.06686 [math.CO], 2022. L. W. Shapiro, A Catalan triangle, Discrete Math. 14 (1976), no. 1, 83-90. [Annotated scanned copy] P. J. Stockmeyer, The charm bracelet problem and its applications, pp. 339-349 of Graphs and Combinatorics (Washington, Jun 1973), Ed. by R. A. Bari and F. Harary. Lect. Notes Math., Vol. 406. Springer-Verlag, 1974. [Scanned annotated and corrected copy] Eric Weisstein's World of Mathematics, Dyck Path.
FORMULA
a(n) = binomial(2*n, n)/(n+1) = (2*n)!/(n!*(n+1)!) = A000984(n)/(n+1). Recurrence: a(n) = 2*(2*n-1)*a(n-1)/(n+1) with a(0) = 1.
Recurrence: a(n) = Sum_{k=0..n-1} a(k)a(n-1-k).
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x), and satisfies A(x) = 1 + x*A(x)^2.
a(n) = Product_{k=2..n} (1 + n/k).
a(n+1) = Sum_{i} binomial(n, 2*i)*2^(n-2*i)*a(i). - Touchard
It is known that a(n) is odd if and only if n=2^k-1, k=0, 1, 2, 3, ... - Emeric Deutsch, Aug 04 2002, corrected by M. F. Hasler, Nov 08 2015 Using the Stirling approximation in A000142 we get the asymptotic expansion a(n) ~ 4^n / (sqrt(Pi * n) * (n + 1)). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 13 2001 Integral representation: a(n) = (1/(2*Pi))*Integral_{x=0..4} x^n*sqrt((4-x)/x). - Karol A. Penson, Apr 12 2001 E.g.f.: exp(2*x)*(I_0(2*x)-I_1(2*x)), where I_n is Bessel function. - Karol A. Penson, Oct 07 2001 a(n) = polygorial(n, 6)/polygorial(n, 3). - Daniel Dockery (peritus(AT)gmail.com), Jun 24 2003
G.f. A(x) satisfies ((A(x) + A(-x)) / 2)^2 = A(4*x^2). - Michael Somos, Jun 27, 2003 G.f. A(x) satisfies Sum_{k>=1} k(A(x)-1)^k = Sum_{n>=1} 4^{n-1}*x^n. - Shapiro, Woan, Getu
a(n+1) = (1/(n+1))*Sum_{k=0..n} a(n-k)*binomial(2k+1, k+1). - Philippe Deléham, Jan 24 2004 a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005 a(n) = Sum_{k=0..floor(n/2)} ((n-2*k+1)*binomial(n, n-k)/(n-k+1))^2, which is equivalent to: a(n) = Sum_{k=0..n} A053121(n, k)^2, for n >= 0. - Paul D. Hanna, Apr 23 2005 E.g.f. Sum_{n>=0} a(n) * x^(2*n) / (2*n)! = BesselI(1, 2*x) / x. - Michael Somos, Jun 22 2005 Given g.f. A(x), then B(x) = x * A(x^3) satisfies 0 = f(x, B(X)) where f(u, v) = u - v + (u*v)^2 or B(x) = x + (x * B(x))^2 which implies B(-B(x)) = -x and also (1 + B^3) / B^2 = (1 - x^3) / x^2. - Michael Somos, Jun 27 2005 a(n) = (1/(s-n))*Sum_{k=0..n} (-1)^k (k+s-n)*binomial(s-n,k) * binomial(s+n-k,s) with s a nonnegative free integer [H. W. Gould].
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, the present sequence is Phi([1]) (also Phi([1,1])). - Gary W. Adamson, Oct 27 2008 a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}...Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n) where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i < l_(i+1) and l_(i+1) <> 0 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009 Let A(x) be the g.f., then B(x)=x*A(x) satisfies the differential equation B'(x)-2*B'(x)*B(x)-1=0. - Vladimir Kruchinin, Jan 18 2011 G.f.: 1/(1-x/(1-x/(1-x/(...)))) (continued fraction). - Joerg Arndt, Mar 18 2011 With F(x) = (1-2*x-sqrt(1-4*x))/(2*x) an o.g.f. in x for the Catalan series, G(x) = x/(1+x)^2 is the compositional inverse of F (nulling the n=0 term). - Tom Copeland, Sep 04 2011 With H(x) = 1/(dG(x)/dx) = (1+x)^3 / (1-x), the n-th Catalan number is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)), and H(x) is the o.g.f. for A115291. - Tom Copeland, Sep 04 2011 With F(x) = (1-sqrt(1-4*x))/2 an o.g.f. in x for the Catalan series, G(x)= x*(1-x) is the compositional inverse and this relates the Catalan numbers to the row sums of A125181. With H(x) = 1/(dG(x)/dx) = 1/(1-2x), the n-th Catalan number (offset 1) is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)). (End)
G.f.: (1-sqrt(1-4*x))/(2*x) = G(0) where G(k) = 1 + (4*k+1)*x/(k+1-2*x*(k+1)*(4*k+3)/(2*x*(4*k+3)+(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011 E.g.f.: exp(2*x)*(BesselI(0,2*x) - BesselI(1,2*x)) = G(0) where G(k) = 1 + (4*k+1)*x/((k+1)*(2*k+1)-x*(k+1)*(2*k+1)*(4*k+3)/(x*(4*k+3)+(k+1)*(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011 E.g.f.: Hypergeometric([1/2],[2],4*x) which coincides with the e.g.f. given just above, and also by Karol A. Penson further above. - Wolfdieter Lang, Jan 13 2012 G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = k*(4*x+1) + 2*x + 2 - x*(2*k+3)*(2*k+4)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 20 2012 G.f.: hypergeom([1/2,1],[2],4*x). - Joerg Arndt, Apr 06 2013 Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,1,-1/2-n,-1)/(n+1). - Karol A. Penson, Jul 28 2013 a(n-1) = Sum_{t1+2*t2+...+n*tn=n} (-1)^(1+t1+t2+...+tn)*multinomial(t1+t2 +...+tn,t1,t2,...,tn)*a(1)^t1*a(2)^t2*...*a(n)^tn. - Mircea Merca, Feb 27 2014 a(n) = -2^(2*n+1) * binomial(n-1/2, -3/2). - Peter Luschny, May 06 2014 a(n) = (2*n)!*[x^(2*n)]hypergeom([],[2],x^2). - Peter Luschny, Jan 31 2015 a(n) = 4^(n-1)*hypergeom([3/2, 1-n], [3], 1). - Peter Luschny, Feb 03 2015 a(n) = Sum_{t=1..n+1} n^(t-1)*abs(Stirling1(n+1, t)) / Sum_{t=1..n+1} abs(Stirling1(n+1, t)), for n > 0, see (10) in Cereceda link. - Michel Marcus, Oct 06 2015 a(n) ~ 4^(n-2)*(128 + 160/N^2 + 84/N^4 + 715/N^6 - 10180/N^8)/(N^(3/2)*Pi^(1/2)) where N = 4*n+3. - Peter Luschny, Oct 14 2015 a(n) = Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*binomial(n+1-k,k)*a(n-k) if n > 0; and a(0) = 1. - David Pasino, Jun 29 2016 Sum_{n>=0} (-1)^n/a(n) = 14/25 - 24*arccsch(2)/(25*sqrt(5)) = 14/25 - 24* A002390/(25*sqrt(5)) = 0.353403708337278061333... - Ilya Gutkovskiy, Jun 30 2016 C(n) = (1/n) * Sum_{i+j+k=n-1} C(i)*C(j)*C(k)*(k+1), n >= 1. - Yuchun Ji, Feb 21 2016 C(n) = 1 + Sum_{i+j+k<n-1} C(i)*C(j)*C(k). - Yuchun Ji, Sep 01 2016 G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x) = 2F1(1/2,1;2;4*x). G.f. A(x) satisfies A = 1 + x*A^2. - R. J. Mathar, Nov 17 2018 (1+sqrt(1+4*x))/2 = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4; and sqrt(x+sqrt(x+sqrt(x+...))) = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4 and x <> 0. (End)
a(3n+1)*a(5n+4)*a(15n+10) = a(3n+2)*a(5n+2)*a(15n+11). The first case of Catalan product equation of a triple partition of 23n+15. - Yuchun Ji, Sep 27 2020 a(n) = 4^n * (-1)^(n+1) * 3F2[{n + 1,n + 1/2,n}, {3/2,1}, -1], n >= 1. - Sergii Voloshyn, Oct 22 2020 a(n) = 2^(1 + 2 n) * (-1)^(n)/(1 + n) * 3F2[{n, 1/2 + n, 1 + n}, {1/2, 1}, -1], n >= 1. - Sergii Voloshyn, Nov 08 2020 a(n) = (1/Pi)*4^(n+1)*Integral_{x=0..Pi/2} cos(x)^(2*n)*sin(x)^2 dx. - Greg Dresden, May 30 2021 G.f. A(x) satisfies A(x) = 1/sqrt(1 - 4*x) * A( -x/(1 - 4*x) ) and (A(x) + A(-x))/2 = 1/sqrt(1 - 4*x) * A( -2*x/(1 - 4*x) ); these are the cases k = 0 and k = -1 of the general formula 1/sqrt(1 - 4*x) * A( (k-1)*x/(1 - 4*x) ) = Sum_{n >= 0} ((k^(n+1) - 1)/(k - 1))*Catalan(n)*x^n.
2 - sqrt(1 - 4*x)/A( k*x/(1 - 4*x) ) = 1 + Sum_{n >= 1} (1 + (k + 1)^n) * Catalan(n-1)*x^n. (End)
0 = a(n)*(16*a(n+1) - 10*a(n+2)) + a(n+1)*(2*a(n+1) + a(n+2)) for all n>=0. - Michael Somos, Dec 12 2022 G.f.: (offset 1) 1/G(x), with G(x) = 1 - 2*x - x^2/G(x) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 01 2023 a(n) = K^(2n+1, n, 1) for all n >= 0, where K^(n, s, x) is the Krawtchouk polynomial defined to be Sum_{k=0..s} (-1)^k * binomial(n-x, s-k) * binomial(x, k). - Vladislav Shubin, Aug 17 2023 The g.f. A(x) satisfies the following functional equations:
A(x) = 1 + x/(1 - 4*x) * A(-x/(1 - 4*x))^2,
A(x^2) = 1/(1 - 2*x) * A(- x/(1 - 2*x))^2 and, for arbitrary k,
1/(1 - k*x) * A(x/(1 - k*x))^2 = 1/(1 - (k+4)*x) * A(-x/(1 - (k+4)*x))^2. (End)
EXAMPLE
From Joerg Arndt and Greg Stevenson, Jul 11 2011: (Start) The following products of 3 transpositions lead to a 4-cycle in S_4:
(1,2)*(1,3)*(1,4);
(1,2)*(1,4)*(3,4);
(1,3)*(1,4)*(2,3);
(1,4)*(2,3)*(2,4);
(1,4)*(2,4)*(3,4). (End)
G.f. = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + ...
For n=3, a(3)=5 since there are exactly 5 binary sequences of length 7 in which the number of ones first exceed the number of zeros at entry 7, namely, 0001111, 0010111, 0011011, 0100111, and 0101011. - Dennis P. Walsh, Apr 11 2012 The a(4) = 14 branching sequences of the (ordered) trees with 4 non-root nodes are (dots denote zeros):
01: [ 1 1 1 1 . ]
02: [ 1 1 2 . . ]
03: [ 1 2 . 1 . ]
04: [ 1 2 1 . . ]
05: [ 1 3 . . . ]
06: [ 2 . 1 1 . ]
07: [ 2 . 2 . . ]
08: [ 2 1 . 1 . ]
09: [ 2 1 1 . . ]
10: [ 2 2 . . . ]
11: [ 3 . . 1 . ]
12: [ 3 . 1 . . ]
13: [ 3 1 . . . ]
14: [ 4 . . . . ]
(End)
MAPLE
A000108 := n->binomial(2*n, n)/(n+1);
G000108 := (1 - sqrt(1 - 4*x)) / (2*x);
spec := [ A, {A=Prod(Z, Sequence(A))}, unlabeled ]: [ seq(combstruct[count](spec, size=n+1), n=0..42) ];
with(combstruct): bin := {B=Union(Z, Prod(B, B))}: seq(count([B, bin, unlabeled], size=n+1), n=0..25); # Zerinvary Lajos, Dec 05 2007 gser := series(G000108, x=0, 42): seq(coeff(gser, x, n), n=0..41); # Zerinvary Lajos, May 21 2008 seq((2*n)!*coeff(series(hypergeom([], [2], x^2), x, 2*n+2), x, 2*n), n=0..30); # Peter Luschny, Jan 31 2015 A000108List := proc(m) local A, P, n; A := [1, 1]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), A[-1]]);
A := [op(A), P[-1]] od; A end: A000108List(31); # Peter Luschny, Mar 24 2022
MATHEMATICA
Table[(2 n)!/n!/(n + 1)!, {n, 0, 20}]
Table[4^n Gamma[n + 1/2]/(Sqrt[Pi] Gamma[n + 2]), {n, 0, 20}] (* Eric W. Weisstein, Oct 31 2024 *) CoefficientList[InverseSeries[Series[x/Sum[x^n, {n, 0, 31}], {x, 0, 31}]]/x, x] (* Mats Granvik, Nov 24 2013 *) CoefficientList[Series[(1 - Sqrt[1 - 4 x])/(2 x), {x, 0, 20}], x] (* Stefano Spezia, Aug 31 2018 *)
PROG
(PARI) a(n)=binomial(2*n, n)/(n+1) \\ M. F. Hasler, Aug 25 2012 (PARI) a(n) = (2*n)! / n! / (n+1)!
(PARI) a(n) = my(A, m); if( n<0, 0, m=1; A = 1 + x + O(x^2); while(m<=n, m*=2; A = sqrt(subst(A, x, 4*x^2)); A += (A - 1) / (2*x*A)); polcoeff(A, n));
(PARI) {a(n) = if( n<1, n==0, polcoeff( serreverse( x / (1 + x)^2 + x * O(x^n)), n))}; /* Michael Somos */ (PARI) (recur(a, b)=if(b<=2, (a==2)+(a==b)+(a!=b)*(1+a/2), (1+a/b)*recur(a, b-1))); a(n)=recur(n, n); \\ R. J. Cano, Nov 22 2012 (PARI) x='x+O('x^40); Vec((1-sqrt(1-4*x))/(2*x)) \\ Altug Alkan, Oct 13 2015 (MuPAD) combinat::dyckWords::count(n) $ n = 0..38 // Zerinvary Lajos, Apr 14 2007 (Magma) C:= func< n | Binomial(2*n, n)/(n+1) >; [ C(n) : n in [0..60]];
(Haskell)
import Data.List (genericIndex)
a000108 n = genericIndex a000108_list n
a000108_list = 1 : catalan [1] where
catalan cs = c : catalan (c:cs) where
c = sum $ zipWith (*) cs $ reverse cs
a000108 = map last $ iterate (scanl1 (+) . (++ [0])) [1]
(Sage) [catalan_number(i) for i in range(27)] # Zerinvary Lajos, Jun 26 2008 (Sage) # Generalized algorithm of L. Seidel
D = [0]*(n+1); D[1] = 1
b = True; h = 1; R = []
for i in range(2*n-1) :
if b :
for k in range(h, 0, -1) : D[k] += D[k-1]
h += 1; R.append(D[1])
else :
for k in range(1, h, 1) : D[k] += D[k+1]
b = not b
return R
(Python)
from gmpy2 import divexact
for n in range(1, 10**3):
(Python)
# Works in Sage also.
for n in range(1000):
CROSSREFS
Cf. A000142, A000245, A000344, A000588, A000957, A000984, A001392, A001453, A001791, A002057, A002420, A003046, A003517, A003518, A003519, A006480, A008276, A008549, A014137, A014138, A014140, A022553 (inv. Eul. trans.), A024492, A032357, A032443, A039599, A048990, A059288, A068875, A069640, A086117, A088327 (Eul. trans.), A094216, A094638, A094639, A098597, A099731, A119822, A120304, A124926, A129763, A137697, A154559, A161581, A167892, A167893, A179277, A211611, A275431 (multisets). Enumerates objects encoded by A014486. Cf. A051168 (diagonal of the square array described). Cf. A120303 (largest prime factor of Catalan number). Cf. A002390 (decimal expansion of natural logarithm of golden ratio). Cf. A332602 (conjectured production matrix).
KEYWORD
core,nonn,easy,eigen,nice
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