Displaying 1-10 of 47 results found.
Fibonomial coefficients: column 5 of A010048.
(Formerly M4568 N1945)
+20
8
1, 8, 104, 1092, 12376, 136136, 1514513, 16776144, 186135312, 2063912136, 22890661872, 253854868176, 2815321003313, 31222272414424, 346260798314872, 3840089017377228, 42587248616222024, 472299787252290712, 5237885063192296801, 58089034826620525728
REFERENCES
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n) = A010048(5+n, 5) (or fibonomial(5+n, 5)). G.f.: 1/(1-8*x-40*x^2+60*x^3+40*x^4-8*x^5-x^6) = 1/((1-x-x^2)*(1+4*x-x^2)*(1-11*x-x^2)) (see Comments to A055870). a(n) = 11*a(n-1) + a(n-2) + ((-1)^n)*fibonomial(n+3, 3), n >= 2; a(0)=1, a(1)=8; fibonomial(n+3, 3)= A001655(n). a(n) = Fibonacci(n+3)*(Fibonacci(n+3)^4-1)/30. - Gary Detlefs, Apr 24 2012 a(n) = a(-6-n) * (-1)^n for all n in Z. - Michael Somos, Sep 19 2014 0 = a(n)*(-a(n+1) - 3*a(n+2)) + a(n+1)*(-8*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014
EXAMPLE
G.f. = 1 + 8*x + 104*x^2 + 1092*x^3 + 12376*x^4 + 136136*x^5 + 1514513*x^6 + ...
MAPLE
with(combinat) : a:=n-> 1/30*fibonacci(n)*fibonacci(n+1)*fibonacci(n+2)*fibonacci(n+3)*fibonacci(n+4): seq(a(n), n=1..19); # Zerinvary Lajos, Oct 07 2007
MATHEMATICA
LinearRecurrence[{8, 40, -60, -40, 8, 1}, {1, 8, 104, 1092, 12376, 136136}, 20] (* Harvey P. Dale, Nov 30 2019 *)
PROG
(PARI) b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
Row sums of Fibonomial triangle A010048.
+20
7
1, 2, 3, 6, 14, 42, 158, 756, 4594, 35532, 349428, 4370436, 69532964, 1407280392, 36228710348, 1186337370456, 49415178236344, 2618246576596392, 176462813970065208, 15128228719573952976, 1649746715671916095304
FORMULA
a(n) = Sum_{m=0..n} A010048(n, m), where A010048(n, m) = fibonomial(n, m). a(n) ~ c * ((1+sqrt(5))/2)^(n^2/4), where
c = EllipticTheta[3,0,1/GoldenRatio] / QPochhammer[-1/GoldenRatio^2] = 2.082828701647012450835512317685120373906427048806222527375... if n is even,
c = EllipticTheta[2,0,1/GoldenRatio] / QPochhammer[-1/GoldenRatio^2] = 2.082828691334156222136965926255238646603356514964103252122... if n is odd.
Or c = Sum_{j} ((1+sqrt(5))/2)^(-(j+(1-(-1)^n)/4)^2) / A062073, where A062073 = 1.2267420107203532444176302... is the Fibonacci factorial constant. (End)
MATHEMATICA
Table[Sum[Product[Fibonacci[j], {j, 1, n}] / Product[Fibonacci[j], {j, 1, k}] / Product[Fibonacci[j], {j, 1, n-k}], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 30 2015 *) (* Or, since version 10 *) Table[Sum[Fibonorial[n]/Fibonorial[k]/Fibonorial[n-k], {k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 30 2015 *) Round@Table[Sum[GoldenRatio^(k(n-k)) QBinomial[n, k, -1/GoldenRatio^2], {k, 0, n}], {n, 0, 20}] (* Round is equivalent to FullSimplify here, but is much faster - Vladimir Reshetnikov, Sep 25 2016 *)
PROG
(Maxima) ffib(n):=prod(fib(k), k, 1, n);
fibonomial(n, k):=ffib(n)/(ffib(k)*ffib(n-k));
makelist(sum(fibonomial(n, k), k, 0, n), n, 0, 30); /* Emanuele Munarini, Apr 02 2012 */
Diagonal sums of Fibonomial triangle A010048.
+20
4
1, 1, 2, 2, 4, 6, 13, 27, 70, 191, 609, 2130, 8526, 38156, 194000, 1109673, 7176149, 52238676, 429004471, 3970438003, 41454181730, 488046132076, 6482590679282, 97134793638750, 1641654359781521, 31285014253070731, 672372121341768918, 16299021330860540657
COMMENTS
Cf. A000045 (Fibonacci) as diagonal sums of A007318 (Pascal's Triangle). For Fibonacci numbers, the ratio A000045(i+1)/ A000045(i) approaches the golden ratio (1+sqrt(5))/2 as i increases. For this sequence, it appears that (a(i+5)/a(i+4))/(a(i+1)/a(i)) approaches the golden ratio. - Dale Gerdemann, Apr 23 2015 This sequence can be interpreted as counting colored, square-domino tilings of a 1xn board, where the number of colors available for a domino with k squares to the left is Fib(k+1) and the number of colors available for a square with k dominoes to the left is Fib(k-1). "Fib(n)" here is A000045 (Fibonacci), extended so that Fib(-1) = 1, Fib(0) = 0,... . As an example, let d be a domino, s be a square an consider the uncolored tilings of length 5: sssss, sssd, ssds, sdss, dsss, sdd, dsd, dds. Then, after each 's' or 'd', write the number of colors available: s1s1s1s1s1, s1s1s1d3, s1s1d2s0, s1d1s0s0, d1s0s0s0, s1d1d1, d1s0d1, d1d1s1. So the number of colorings for these tilings is: 1,3,0,0,0,1,0,1 and the total number of colored tilings is 6 (= a(5)). - Dale Gerdemann, Apr 30 2015
FORMULA
a(n) = sum(fibonomial(k,n-k),k=ceiling(n/2)..n).
a(n) ~ c * ((1+sqrt(5))/2)^(n^2/8), where
c = 1.472885929099569314607134281503815932269629515265... if mod(n,4)=0,
c = 1.472782295338429619549807628338486893461428897618... if mod(n,4)=1 or 3,
c = 1.472678661577289942545896597162143392952724631588... if mod(n,4)=2.
Or c = Sum_{j} ((1+sqrt(5))/2)^(-2*(j+(1-cos(Pi*n/2))/4)^2) / A062073, where A062073 = 1.2267420107203532444176302... is the Fibonacci factorial constant. (End)
MATHEMATICA
Table[Sum[Product[Fibonacci[k-j+1]/Fibonacci[j], {j, 1, n-k}], {k, Ceiling[n/2], n}], {n, 0, 30}] (* Vaclav Kotesovec, Apr 29 2015 *) (* Or, since version 10 *) Table[Sum[Fibonorial[k]/Fibonorial[2k-n]/Fibonorial[n-k], {k, Ceiling[n/2], n}], {n, 0, 30}] (* Vaclav Kotesovec, Apr 30 2015 *) (* List of the multiplicative constants from an asymptotic formula: *) {N[EllipticTheta[3, 0, GoldenRatio^(-2)]/QPochhammer[-(GoldenRatio^2)^(-1)], 80], N[Sum[GoldenRatio^(-2*(j + 1/4)^2), {j, -Infinity, Infinity}]/QPochhammer[-(GoldenRatio^2)^(-1)], 80], N[EllipticTheta[2, 0, GoldenRatio^(-2)]/QPochhammer[-(GoldenRatio^2)^(-1)], 80]} (* Vaclav Kotesovec, Apr 30 2015 *)
PROG
(Maxima) ffib(n):=prod(fib(k), k, 1, n);
fibonomial(n, k):=ffib(n)/(ffib(k)*ffib(n-k));
makelist(sum(fibonomial(k, n-k), k, ceiling(n/2), n), n, 0, 30);
Triangle T read by rows: inverse of fibonomial triangle ( A010048).
+20
1
1, -1, 1, 0, -1, 1, 1, 0, -2, 1, -1, 3, 0, -3, 1, -6, -5, 15, 0, -5, 1, 35, -48, -40, 60, 0, -8, 1, 181, 455, -624, -260, 260, 0, -13, 1, -6056, 3801, 9555, -6552, -1820, 1092, 0, -21, 1, -3741, -205904, 129234, 162435, -74256, -12376, 4641, 0, -34, 1
COMMENTS
Conjecture: all rows except the first add to zero.
FORMULA
Conjecture: T(n+k, n) = A010048(n+k-1, k)*T(k, 1), n>1. a(n,k) = A010048(n,k) * (Sum[s=1..n-k;(-1)^s * Sum[k1+k2+..+ks=n-k,ki>=1; C(n-k; k1,k2,...,ks)] ]) where C(n; k1,k2,...,ks) is a multi-F-nomial coefficient. - Maciej Dziemianczuk, Dec 21 2008
EXAMPLE
1,
-1,1,
0,-1,1,
1,0,-2,1,
-1,3,0,-3,1,
-6,-5,15,0,-5,1,
35,-48,-40,60,0,-8,1,
181,455,-624,-260,260,0,-13,1,
-6056,3801,9555,-6552,-1820,1092,0,-21,1,
Triangle T(n, k) = A010048(n, k)* A010048(n, k-1)/Fibonacci(n), read by rows.
+20
1
1, 1, 1, 1, 2, 1, 1, 6, 6, 1, 1, 15, 45, 15, 1, 1, 40, 300, 300, 40, 1, 1, 104, 2080, 5200, 2080, 104, 1, 1, 273, 14196, 94640, 94640, 14196, 273, 1, 1, 714, 97461, 1689324, 4504864, 1689324, 97461, 714, 1, 1, 1870, 667590, 30375345, 210602392, 210602392, 30375345, 667590, 1870, 1
COMMENTS
These numbers are called the FiboNarayana numbers by Garrett and Killpatrick. - Michel Marcus, Oct 23 2019
EXAMPLE
Triangle begins as:
1;
1, 1;
1, 2, 1;
1, 6, 6, 1;
1, 15, 45, 15, 1;
1, 40, 300, 300, 40, 1;
1, 104, 2080, 5200, 2080, 104, 1;
1, 273, 14196, 94640, 94640, 14196, 273, 1;
1, 714, 97461, 1689324, 4504864, 1689324, 97461, 714, 1;
1, 1870, 667590, 30375345, 210602392, 210602392, 30375345, 667590, 1870, 1;
MATHEMATICA
A010048[n_, k_]:= Product[Fibonacci[n-j+1]/Fibonacci[j], {j, k}];
Table[T[n, k], {n, 12}, {k, n}]//Flatten (* G. C. Greubel, May 08 2021 *)
PROG
(PARI) fibonomial(n, k) = prod(j=0, k-1, fibonacci(n-j))/prod(j=1, k, fibonacci(j)); \\ A010048T(n, k) = fibonomial(n, k)*fibonomial(n, k-1)/fibonacci(n);
matrix(10, 10, n, k, T(n, k)) \\ to see the triangle \\ Michel Marcus, Oct 23 2019 (Sage)
@CachedFunction
def A010048(n, q): return product( fibonacci(n-j+1)/fibonacci(j) for j in (1..k) ) flatten([[T(n, k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, May 08 2021
Row square-sums of Fibonomial triangle A010048.
+20
1
1, 2, 3, 10, 56, 502, 6930, 157172, 5847270, 350430420, 33789991248, 5280020814732, 1338210835193414, 548265785425359340, 363248986031094300018, 389399454403643525265020, 675824289510077938157099920
FORMULA
a(n) = sum(fibonomial(n,k)^2,k=0..n).
MATHEMATICA
FiboFactorial[n_] := Product[Fibonacci[k], {k, 1, n}]
Fibonomial[n_, k_] :=
If[k > n, 0, FiboFactorial[n]/(FiboFactorial[k] FiboFactorial[n - k])
]
Table[Sum[Fibonomial[n, k]^2, {k, 0, n}], {n, 0, 100}]
PROG
(Maxima) ffib(n):=prod(fib(k), k, 1, n);
fibonomial(n, k):=ffib(n)/(ffib(k)*ffib(n-k));
makelist(sum(fibonomial(n, k)^2, k, 0, n), n, 0, 30);
Golden rectangle numbers: F(n)*F(n+1), where F(n) = A000045(n) (Fibonacci numbers).
(Formerly M1606 N0628)
+10
122
0, 1, 2, 6, 15, 40, 104, 273, 714, 1870, 4895, 12816, 33552, 87841, 229970, 602070, 1576239, 4126648, 10803704, 28284465, 74049690, 193864606, 507544127, 1328767776, 3478759200, 9107509825, 23843770274, 62423800998, 163427632719
COMMENTS
a(n)/ A007598(n) ~= golden ratio, especially for larger n. - Robert Happelberg (roberthappelberg(AT)yahoo.com), Jul 25 2005 Let phi be the golden ratio (cf. A001622). Then 1/phi = phi - 1 = Sum_{n>=1} (-1)^(n-1)/a(n), an alternating infinite series consisting solely of unit fractions. - Franz Vrabec, Sep 14 2005 A more exact name would be: Golden convergents to rectangle numbers. These rectangles are not actually golden (ratio of sides is not phi) but are golden convergents (sides are numerator and denominator of convergents in the continued fraction expansion of phi, whence ratio of sides converges to phi). - Daniel Forgues, Nov 29 2009 Numbers m such that m(5m+2)+1 or m(5m-2)+1 is a square. - Bruno Berselli, Oct 22 2012 In pairs, these numbers are important in finding binomial coefficients that appear in at least six places in Pascal's triangle. For instance, the pair (m,n) = (40, 104) finds the numbers binomial(n-1,m) = binomial(n,m-1). Two additional numbers are found on the other side of the triangle. The final two numbers appear in row binomial(n-1,m). See A003015. - T. D. Noe, Mar 13 2013 For n>1, a(n) is one-half the area of the trapezoid created by the four points (F(n),L(n)), (L(n),F(n)), (F(n+1), L(n+1)), (L(n+1), F(n+1)) where F(n) = A000045(n) and L(n) = A000032(n). - J. M. Bergot, May 14 2014 [Note on how to calculate: take the two points (a,b) and (c,d) with a<b, c<d and a<d then subtract a from each: a-a=0, b-a=B, c-a=C, and d-a=D. The area is (D-(C-B)^2)/2.]
Can be obtained (up to signs) by setting x = F(n)/F(n+1) in g.f. for Fibonacci numbers - see Pongsriiam. - N. J. A. Sloane, Mar 23 2017
REFERENCES
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., 58:2 (2020), 140-142.
A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 9.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n) = A010048(n+1, 2) = Fibonomial(n+1, 2). For n > 0, 1 - 1/a(n+1) = Sum_{k=1..n} 1/(F(k)*F(k+2)) where F(k) is the k-th Fibonacci number. - Benoit Cloitre, Aug 31 2002. G.f.: x/(1-2*x-2*x^2+x^3) = x/((1+x)*(1-3*x+x^2)). ( Simon Plouffe in his 1992 dissertation; see Comments to A055870), a(n) = 3*a(n-1) - a(n-2) - (-1)^n = -a(-1-n).
Let M = the 3 X 3 matrix [1 2 1 / 1 1 0 / 1 0 0]; then a(n) = the center term in M^n *[1 0 0]. E.g., a(5) = 40 since M^5 * [1 0 0] = [64 40 25]. - Gary W. Adamson, Oct 10 2004 a(n) = Sum{k=0..n} Fibonacci(k)^2. The proof is easy. Start from a square (1*1). On the right side, draw another square (1*1). On the above side draw a square ((1+1)*(1+1)). On the left side, draw a square ((1+2)*(1+2)) and so on. You get a rectangle (F(n)*F(1+n)) which contains all the squares of side F(1), F(2), ..., F(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 19 2007
With phi = (1+sqrt(5))/2, a(n) = round((phi^(2*n+1))/5) = floor((1/2) + (phi^(2*n+1))/5), n >= 0. - Daniel Forgues, Nov 29 2009 a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3), a(1)=1, a(2)=2, a(3)=6. - Sture Sjöstedt, Feb 06 2010 a(n) = 1/|F(n+1)/F(n) - F(n)/F(n-1)| where F(n) = Fibonacci numbers A000045. b(n) = F(n+1)/F(n) - F(n)/F(n-1): 1/1, -1/2, 1/6, -1/15, 1/40, -1/104, ...; c(n) = 1/b(n) = a(n)*(-1)^(n+1): 1, -2, 6, -15, 40, -104, ... (n=1,2,...). - Thomas Ordowski, Nov 04 2010 a(n) = (Fibonacci(n+2)^2 - Fibonacci(n-1)^2)/4. - Gary Detlefs, Dec 03 2010 Let d(n) = n mod 2, a(0)=0 and a(1)=1. For n > 1, a(n) = d(n) + 2*a(n-1) + Sum_{k=0..n-2} a(k). - L. Edson Jeffery, Mar 20 2011 a(n+1) = ((2+sqrt(5))*((3+sqrt(5))/2)^n+(2-sqrt(5))*((3-sqrt(5))/2)^n+(-1)^n)/5.
a(n) = ((1+sqrt(5))*((3+sqrt(5))/2)^n+(1-sqrt(5))*((3-sqrt(5))/2)^n-2*(-1)^n)/10. (End)
a(n) = (-1)^n * Sum_{k=0..n} (-1)^k*F(2*k), n >= 0. - Wolfdieter Lang, Aug 11 2012 0 = a(n)*(+a(n+1) - a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 19 2014 E.g.f.: ((3 + sqrt(5))*exp((5+sqrt(5))*x/2) - 2*exp((2*x)/(3+sqrt(5))+x) - 1 - sqrt(5))*exp(-x)/(5*(1 + sqrt(5))). - Ilya Gutkovskiy, Apr 15 2016 a(n) = A061646(n) - Fibonacci(n-1)^2.
EXAMPLE
G.f. = x + 2*x^2 + 6*x^3 + 15*x^4 + 40*x^5 + 104*x^6 + 273*x^7 + 714*x^8 + ...
MAPLE
with(combinat): A001654:=n->fibonacci(n)*fibonacci(n+1):
MATHEMATICA
Times@@@Partition[Fibonacci[Range[0, 30]], 2, 1] (* Harvey P. Dale, Aug 18 2011 *) Accumulate[Fibonacci[Range[0, 30]]^2] (* Paolo Xausa, May 31 2024 *)
PROG
(PARI) A001654(n)=fibonacci(n)*fibonacci(n+1); (PARI) b(n, k)=prod(j=1, k, fibonacci(n+j)/fibonacci(j));
(Haskell)
a001654 n = a001654_list !! n
a001654_list = zipWith (*) (tail a000045_list) a000045_list
(Python)
from sympy import fibonacci as F
def a(n): return F(n)*F(n + 1)
(Python)
from math import prod
from gmpy2 import fib2
(Magma) I:=[0, 1, 2]; [n le 3 select I[n] else 2*Self(n-1) + 2*Self(n-2) - Self(n-3): n in [1..30]]; // G. C. Greubel, Jan 17 2018
Signed Fibonomial triangle.
+10
27
1, 1, -1, 1, -1, -1, 1, -2, -2, 1, 1, -3, -6, 3, 1, 1, -5, -15, 15, 5, -1, 1, -8, -40, 60, 40, -8, -1, 1, -13, -104, 260, 260, -104, -13, 1, 1, -21, -273, 1092, 1820, -1092, -273, 21, 1, 1, -34, -714, 4641, 12376, -12376, -4641, 714, 34, -1, 1, -55, -1870, 19635, 85085, -136136, -85085, 19635, 1870, -55, -1
COMMENTS
Row n+1 (n >= 1) of the signed triangle lists the coefficients of the recursion relation for the n-th power of Fibonacci numbers A000045: Sum_{m=0..n+1} T(n+1,m)*(Fibonacci(k-m))^n = 0, k >= n+1; inputs: (Fibonacci(k))^n, k=0..n. The inverse of the row polynomial p(n,x) := Sum_{m=0..n} T(n,m)*x^m is the g.f. for the column m=n-1 of the Fibonomial triangle A010048. The row polynomials p(n,x) factorize according to p(n,x) = G(n-1)*p(n-2,-x), with inputs p(0,x)= 1, p(1,x)= 1-x and G(n):= 1 - A000032(n)*x + (-1)^n*x^2. (Derived from Riordan's result and Knuth's exercise). The row polynomials are the characteristic polynomials of product of the binomial matrix binomial(i,j) and the exchange matrix J_n (matrix with 1's on the antidiagonal, 0 elsewhere). - Paul Barry, Oct 05 2004
REFERENCES
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, pp. 84-5 and 492.
FORMULA
T(n, m) = (-1)^floor((m+1)/2)* A010048(n, m), where A010048(n, m) := fibonomial(n, m). G.f. for column m: (-1)^floor((m+1)/2)*x^m/p(m+1, x) with the row polynomial of the (signed) triangle: p(n, x) := Sum_{m=0..n} T(n, m)*x^m.
EXAMPLE
Row polynomial for n=4: p(4,x) = 1-3*x-6*x^2+3*x^3+x^4 = (1+x-x^2)*(1-4*x-x^2). 1/p(4,x) is G.f. for A010048(n+3,3), n >= 0: {1,3,15,60,...} = A001655(n). n=3: 1*(Fibonacci(k))^3 - 3*(Fibonacci(k-1))^3 - 6*(Fibonacci(k-2))^3 + 3*(Fibonacci(k-3))^3 + 1*(Fibonacci(k-4))^3 = 0, k >= 4; inputs: (Fibonacci(k))^3, k=0..3.
The triangle begins:
n\m 0 1 2 3 4 5 6 7 8 9
0 1
1 1 -1
2 1 -1 -1
3 1 -2 -2 1
4 1 -3 -6 3 1
5 1 -5 -15 15 5 -1
6 1 -8 -40 60 40 -8 -1
7 1 -13 -104 260 260 -104 -13 1
8 1 -21 -273 1092 1820 -1092 -273 21 1
9 1 -34 -714 4641 12376 -12376 -4641 714 34 -1
MAPLE
(-1)^floor((k+1)/2)* A010048(n, k) ;
MATHEMATICA
T[n_, m_]:= {1, -1, -1, 1}[[Mod[m, 4] + 1]] * Product[ Fibonacci[n-j+1]/Fibonacci[j], {j, m}];
PROG
(Magma)
Fibonomial:= func< n, k | k eq 0 select 1 else (&*[Fibonacci(n-j+1)/Fibonacci(j): j in [1..k]]) >;
[(-1)^Floor((k+1)/2)*Fibonomial(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 20 2024 (SageMath)
def fibonomial(n, k): return 1 if k==0 else product(fibonacci(n-j+1)/fibonacci(j) for j in range(1, k+1))
flatten([[(-1)^((k+1)//2)*fibonomial(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 20 2024
CROSSREFS
Cf. A000032, A000045, A001654, A001655, A001656, A001657, A001658, A010048, A051159, A056565, A056566, A056567.
Triangle T(n,k) = t(n-k, k); t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.
+10
24
1, 2, 2, 4, 12, 4, 8, 52, 52, 8, 16, 196, 416, 196, 16, 32, 684, 2644, 2644, 684, 32, 64, 2276, 14680, 26440, 14680, 2276, 64, 128, 7340, 74652, 220280, 220280, 74652, 7340, 128, 256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172, 256, 512, 72076, 1637860, 10978444, 27227908, 27227908, 10978444, 1637860, 72076, 512
COMMENTS
Related triangles may be found by varying the function f(x). If f(x) is a linear function, it can be parameterized as f(x) = a*x + b. With different values for a and b, the following triangles are obtained:
a\b 1.......2.......3.......4.......5.......6
The row sums of these, and similarly constructed number triangles, are shown in the following table:
a\b 1.......2.......3.......4.......5.......6.......7.......8.......9
The formula can be further generalized to: t(n,m) = f(m+s)*t(n-1,m) + f(n-s)*t(n,m-1), where f(x) = a*x + b. The following table specifies triangles with nonzero values for s (given after the slash).
a\ b 0 1 2 3
-1
0
If f(x) = A000045(x) (Fibonacci) and s = 1, the result is A010048 (Fibonomial). In the notation of Carlitz and Scoville, this is the triangle of generalized Eulerian numbers A(r, s | alpha, beta) with alpha = beta = 2. Also the array A(2,1,4) in the notation of Hwang et al. (see page 31). - Peter Bala, Dec 27 2019
FORMULA
T(n,k) = t(n-k, k); t(0,0) = 1, t(n,m) = 0 if n < 0 or m < 0 else t(n,m) = f(m)*t(n-1,m) + f(n)*t(n,m-1), where f(x) = x + 2.
T(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(j+3,j)*binomial(n+4,k-j)*(j+2)^n. - Peter Bala, Dec 27 2019 Modified rule of Pascal: T(0,0) = 1, T(n,k) = 0 if k < 0 or k > n else T(n,k) = f(n-k) * T(n-1,k-1) + f(k) * T(n-1,k), where f(x) = x + 2. - Georg Fischer, Nov 11 2021 T(n, n-k) = T(n, k).
EXAMPLE
Array, t(n, k), begins as:
1, 2, 4, 8, 16, 32, 64, ...;
2, 12, 52, 196, 684, 2276, 7340, ...;
4, 52, 416, 2644, 14680, 74652, 357328, ...;
8, 196, 2644, 26440, 220280, 1623964, 10978444, ...;
16, 684, 14680, 220280, 2643360, 27227908, 251195000, ...;
32, 2276, 74652, 1623964, 27227908, 381190712, 4677894984, ...;
64, 7340, 357328, 10978444, 251195000, 4677894984, 74846319744, ...;
Triangle, T(n, k), begins as:
1;
2, 2;
4, 12, 4;
8, 52, 52, 8;
16, 196, 416, 196, 16;
32, 684, 2644, 2644, 684, 32;
64, 2276, 14680, 26440, 14680, 2276, 64;
128, 7340, 74652, 220280, 220280, 74652, 7340, 128;
256, 23172, 357328, 1623964, 2643360, 1623964, 357328, 23172, 256;
MATHEMATICA
Table[Sum[(-1)^(k-j)*Binomial[j+3, j] Binomial[n+4, k-j] (j+2)^n, {j, 0, k}], {n, 0, 9}, {k, 0, n}]//Flatten (* Michael De Vlieger, Dec 27 2019 *)
PROG
(PARI) t(n, m) = if ((n<0) || (m<0), 0, if ((n==0) && (m==0), 1, (m+2)*t(n-1, m) + (n+2)*t(n, m-1)));
tabl(nn) = {for (n=0, nn, for (k=0, n, print1(t(n-k, k), ", "); ); print(); ); } \\ Michel Marcus, Apr 14 2015 (Magma)
A256890:= func< n, k | (&+[(-1)^(k-j)*Binomial(j+3, j)*Binomial(n+4, k-j)*(j+2)^n: j in [0..k]]) >;
(SageMath)
def A256890(n, k): return sum((-1)^(k-j)*Binomial(j+3, j)*Binomial(n+4, k-j)*(j+2)^n for j in range(k+1))
Coefficient triangle of certain polynomials.
+10
18
1, 1, -1, 1, -2, -1, 1, -4, -4, 1, 1, -7, -16, 7, 1, 1, -12, -53, 53, 12, -1, 1, -20, -166, 318, 166, -20, -1, 1, -33, -492, 1784, 1784, -492, -33, 1, 1, -54, -1413, 9288, 17840, -9288, -1413, 54, 1, 1, -88, -3960, 46233, 163504, -163504, -46233, 3960, 88, -1
COMMENTS
The row polynomials p(n,x) := sum(a(n,m)*x^m) occur as numerators of the g.f. for the (n+1)-th power of Fibonacci numbers A000045. The corresponding denominator polynomials are the row polynomials q(n+2,x) = Sum_{m=0..n+2} A055870(n+2, m)*x^m (signed Fibonomial triangle). The row polynomials p(n,x) and the companion denominator polynomials q(n,x) can be deduced from Riordan's recursion result.
The explicit formula is found from the recursion relation for powers of Fibonacci numbers (see Knuth's exercise with solution). - Roger L. Bagula, Apr 03 2010
REFERENCES
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1969, Vol. 1, p. 84, (exercise 1.2.8. Nr. 30) and p. 492 (solution).
FORMULA
a(n, m)=0 if n<m; a(n, 0)=1; a(n, m) = F(m+1)^(n+1) + sum(sfibonomial(n+2, j)*(F(m+1-j)^(n+1)), j=1..m) m=1..n, with F(n)= A000045(n) (Fibonacci) and sfibonomial(n, m) := A055870(n, m). p(x,n) = Sum_{k>=0} (((1 + sqrt(5))^k - (1 - sqrt(5))^k)/(2^k*sqrt(5)))^n*x^k;
t(n,m) = Numerator_coefficients(p(x,n)/x)/2^(1 + floor(n/2));
out(n,m) = t(n,m)/t(n,1). (End)
EXAMPLE
Row polynomial for n=4: p(4,x) = 1 - 7*x - 16*x^2 + 7*x^3 + x^4. x*p(4,x) is the numerator of the g.f. for A056572(n), n >= 0 (fifth power of Fibonacci numbers) {0,1,1,32,243,...}. The denominator polynomial is Sum_{m=0..6} A055870(6,m)*x^m (n=6 row polynomial of signed fibonomial triangle). 1;
1, -1;
1, -2, -1;
1, -4, -4, 1;
1, -7, -16, 7, 1;
1, -12, -53, 53, 12, -1;
1, -20, -166, 318, 166, -20, -1;
1, -33, -492, 1784, 1784, -492, -33, 1;
1, -54, -1413, 9288, 17840, -9288, -1413, 54, 1;
1, -88, -3960, 46233, 163504, -163504, -46233, 3960, 88, -1; (End)
MAPLE
if k = 0 then
1;
elif k >n then
0;
else
combinat[fibonacci](k+1)^(n+1)+add( A055870(n+2, j)*(combinat[fibonacci](k+1-j)^(n+1)), j=1..k) ; end if;
MATHEMATICA
p[x_, n_] = Sum[(((1 + Sqrt[5])^k - (1 - Sqrt[5])^k)/(2^k*Sqrt[5]))^n*x^k, {k, 0, Infinity}];
a = Table[CoefficientList[FullSimplify[Numerator[p[ x, n]]/x], x]/2^(1 + Floor[n/2]), {n, 1, 10}];
Table[a[[n]]/a[[n]][[1]], {n, 1, 10}];
PROG
(PARI) S(n, k) = (-1)^floor((k+1)/2)*(prod(j=0, k-1, fibonacci(n-j))/prod(j=1, k, fibonacci(j)));
T(n, k) = sum(j=0, k, fibonacci(k+1-j)^(n+1) * S(n+2, j));
tabl(m) = for (n=0, m, for (k=0, n, print1(T(n, k), ", ")); print);
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