Displaying 1-10 of 38 results found.
0, 6, 210, 7140, 242556, 8239770, 279909630, 9508687656, 323015470680
a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.
(Formerly M4217 N1760)
+10
193
0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559, 417501372047787720
COMMENTS
For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006 (a(n), b(n)) where b(n) = A082291(n) are the integer solutions of the equation 2*binomial(b,a) = binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003 This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n) = A001333(n)* A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s = a(n+1) = { A001652(n) + A046090(n) + A001653(n)}/2 and area rs = A029549(n) = 6* A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014] n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - Benoit Cloitre, May 10 2003 For n > 0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003 a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series). - Creighton Dement, Dec 16 2004 Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005 Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (= A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 2005 Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n > 0, define C(n) to be the smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005 Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance, t(20)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005 Tested for 2 < p < 27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006 If 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+3). - Kenneth J Ramsey, Jun 08 2006 If p is an odd prime, then if p == 1 or 7 (mod 8), then a((p-1)/2) == 0 (mod p) and a((p+1)/2) == 1 (mod p); if p == 3 or 5 (mod 8), then a((p-1)/2) == 1 (mod p) and a((p+1)/2) == 0 (mod p). Kenneth J Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007 a(n)*(a(n+b) - a(b-2)) = (a(n+1)+1)*(a(n+b-1) - a(b-1)). This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007 For n < 0, let a(n) = -a(-n). Then (a(n+j) + a(k+j)) * (a(n+b+k+j) - a(b-j-2)) = (a(n+j+1) + a(k+j+1)) * (a(n+b+k+j-1) - a(b-j-1)). - Charlie Marion, Mar 04 2011 Sequence gives y values of the Diophantine equation: 0+1+2+...+x = y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with a<c then a+b = c-d and ((d+b)^2, d^2-b^2) is a solution too. If (a,b), (c,d) and (e,f) are three consecutive solutions of the Diophantine equation 0+1+2+...+x = y^2 with a < c < e then (8*d^2, d*(f-b)) is a solution too. - Mohamed Bouhamida, Aug 29 2009 If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with p < r then r = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009 In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then
for n > 0, b(n) = a(n)*k-a(n-1); e.g.,
for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1, 64 = 35*2 - 6, 373 = 204*2 - 35; for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1; 99 = 35*3 - 6; 577 = 204*3 - 35; for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35; for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35. See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012 a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012 a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012 The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows: For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015 Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015 The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and 2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016 a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016 Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017 In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021 Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021
REFERENCES
Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012 A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, pp. 193, 197.
D. M. Burton, The History of Mathematics, McGraw Hill, (1991), p. 213.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 10.
P. Franklin, E. F. Beckenbach, H. S. M Coxeter, N. H. McCoy, K. Menger, and J. L. Synge, Rings And Ideals, No 8, The Carus Mathematical Monographs, The Mathematical Association of America, (1967), pp. 144-146.
A. Patra, G. K. Panda, and T. Khemaratchatakumthorn. "Exact divisibility by powers of the balancing and Lucas-balancing numbers." Fibonacci Quart., 59:1 (2021), 57-64; see B(n).
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).
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022
LINKS
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E. Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See v_n in Lemma 7.9 p. 21. Muniru A. Asiru, All square chiliagonal numbers, International Journal of Mathematical Education in Science and Technology, Volume 47, 2016 - Issue 7. Jeremiah Bartz, Bruce Dearden, Joel Iiams, and Julia Peterson, Powers of Two as Sums of Two Balancing Numbers, Combinatorics, Graph Theory and Computing (SEICCGTC 2021) Springer Proc. Math. Stat., Vol 448, pp. 383-392. See p. 384. Elwyn Berlekamp and Joe P. Buhler, Puzzle Column, Emissary, MSRI Newsletter, Fall 2005. Problem 1, (6 MB). Brian Hayes, Calculemus!, American Scientist, 96 (Sep-Oct 2008), 362-366.
FORMULA
G.f.: x / (1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation. a(n) = S(n-1, 6) = U(n-1, 3) with U(n, x) Chebyshev's polynomials of the second kind. S(-1, x) := 0. Cf. triangle A049310 for S(n, x). a(n) = sqrt(( A001541(n)^2-1)/8) (cf. Richardson comment). a(n) = 3*a(n-1) + sqrt(8*a(n-1)^2+1). - R. J. Mathar, Oct 09 2000 a(n) ~ (1/8)*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(2*n) = a(n)* A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003 a(k)*a(2*n+k) = a(n+k)^2 - a(n)^2; e.g., 204*7997214 = 40391^2 - 35^2. - Charlie Marion, Jan 15 2004 For j < n+1, a(k+j)*a(2*n+k-j) - Sum_{i = 0..j-1} a(2*n-(2*i+1)) = a(n+k)^2 - a(n)^2. - Charlie Marion, Jan 18 2004 a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))*sqrt(2)/8;
a(n) = Sum_{i=0..n} Sum_{j=0..n} A000129(i+j)*n!/(i!*j!*(n-i-j)!)/2. (End) E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Apr 21 2004 a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - Paul Barry, Oct 01 2004 a(n) = 7*(a(n-1) - a(n-2)) + a(n-3), a(1) = 0, a(2) = 1, a(3) = 6, n > 3. Also a(n) = ( (1 + sqrt(2) )^(2*n) - (1 - sqrt(2) )^(2*n) ) / (4*sqrt(2)). - Antonio Alberto Olivares, Oct 23 2003 Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),3), see second formula. - Marcos Carreira, Dec 27 2006 The perfect median m(n) can be expressed in terms of the Pell numbers P() = A000129() by m(n) = P(n + 2) * (P(n + 2) + (P(n + 1)) for n >= 0. - Winston A. Richards (ugu(AT)psu.edu), Jun 11 2007 a(n) = Sum_{k=0..n-1} 4^k*binomial(n+k,2*k+1). - Paul Barry, Apr 20 2009 Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(2).
Product_{n >= 2} (1 - 1/a(n)) = (1/3)*(1 + sqrt(2)). (End)
G.f.: G(0)*x/(2-6*x), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013 G.f.: H(0)*x/2, where H(k) = 1 + 1/( 1 - x*(6-x)/(x*(6-x) + 1/H(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 18 2014 a(n) = (a(n-1)^2 - a(n-3)^2)/a(n-2) + a(n-4) for n > 3. - Patrick J. McNab, Jul 24 2015 Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - Ilya Gutkovskiy, Jun 27 2016 a(n) = (a(n-3) + a(n+3))/198.
a(n) = Sum_{i=1..n} A001653(i), n>=1. a(n) = sinh( 2 * n * arccsch(1) ) / ( 2 * sqrt(2) ). - Federico Provvedi, Feb 01 2021 (End)
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - Peter Bala, Jul 17 2023
EXAMPLE
G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...
6 is in the sequence since 6^2 = 36 is a triangular number: 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. - Michael B. Porter, Jul 02 2016
MAPLE
a[0]:=1: a[1]:=6: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..26); # Emeric Deutschwith (combinat):seq(fibonacci(2*n, 2)/2, n=0..20); # Zerinvary Lajos, Apr 20 2008
MATHEMATICA
Transpose[NestList[Flatten[{Rest[#], ListCorrelate[{-1, 6}, #]}]&, {0, 1}, 30]][[1]] (* Harvey P. Dale, Mar 23 2011 *) CoefficientList[Series[x/(1-6x+x^2), {x, 0, 30}], x] (* Harvey P. Dale, Mar 23 2011 *) TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* Federico Provvedi, Feb 01 2021 *)
PROG
(PARI) {a(n) = imag((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */ (PARI) {a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* Michael Somos, Apr 07 2003 */ (PARI) {a(n) = polchebyshev( n-1, 2, 3)}; /* Michael Somos, Sep 02 2012 */ (Sage) [lucas_number1(n, 6, 1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008 (Sage) [chebyshev_U(n-1, 3) for n in (0..20)] # G. C. Greubel, Dec 23 2019 (Haskell)
a001109 n = a001109_list !! n :: Integer
a001109_list = 0 : 1 : zipWith (-)
(map (* 6) $ tail a001109_list) a001109_list
(Magma) [n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015 (GAP) a:=[0, 1];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Dec 18 2018
CROSSREFS
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), this sequence (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3.
(Formerly M3074 N1247)
+10
154
0, 3, 20, 119, 696, 4059, 23660, 137903, 803760, 4684659, 27304196, 159140519, 927538920, 5406093003, 31509019100, 183648021599, 1070379110496, 6238626641379, 36361380737780, 211929657785303, 1235216565974040, 7199369738058939, 41961001862379596, 244566641436218639
COMMENTS
Consider all Pythagorean triples (X, X+1, Z) ordered by increasing Z; sequence gives X values.
Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of two consecutive integers (cf. A097571). Members of Diophantine pairs. Solution to a(a+1) = 2b(b+1) in natural numbers including 0; a = a(n), b = b(n) = A053141(n); The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de). The index of all triangular numbers T(a(n)) for which 4T(n)+1 is a perfect square.
All Pythagorean triples {X(n), Y(n)=X(n)+1, Z(n)} with X<Y<Z, may be recursively generated through the mapping W(n) -> M*W(n), where W(n)=transpose of vector [X(n) Y(n) Z(n)] and M a 3 X 3 matrix given by [2 1 2 / 1 2 2 / 2 2 3]. - Lekraj Beedassy, Aug 14 2006 In general, if b(n) = A053141(n), then a(n)*b(n+k) = a(n+k)*b(n)+b(k); e.g., 3*84 = 119*2+14; 3*2870 = 4059*2+492; 20*2870 = 5741*14+84. - Charlie Marion, Nov 19 2007 If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p<r then s-r=p+q+1. - Mohamed Bouhamida, Aug 29 2009 If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = y^2 with p<r then r=3p+2q+1 and s=4p+3q+2. - Mohamed Bouhamida, Sep 02 2009 The numbers 3* A001652 = (0, 9, 60, 357, 2088, 12177, 70980, ...) are all the nonnegative values of X such that X^2 + (X+3)^2 = Z^2 (Z is in A075841). - Bruno Berselli, Aug 26 2010 Let T(n) = n*(n+1)/2 (the n-th triangular number). For n > 0,
Set a(n)=p; a(n)+1=q; the generated triple x=p^2+pq; y=q^2+pq; k=p^2+q^2 satisfies x^2+y^2=k(x+y). - Carmine Suriano, Dec 17 2013 The arms of the triangle are found with (b(n),c(n)) for 2*b(n)*c(n) and c(n)^2 - b(n)^2. Let b(1) = 1 and c(1) = 2, then b(n) = c(n-1) and c(n) = 2*c(n-1) + b(n-1). Alternatively, b(n) = c(n-1) and c(n) equals the nearest integer to b(n)*(1+sqrt(2)). - J. M. Bergot, Oct 09 2014 Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015 Numbers m such that Product_{k=1..m} (4*k^4+1) is a square (see A274307). - Chai Wah Wu, Jun 21 2016 Numbers m such that m^2+(m+1)^2 is a square. - César Aguilera, Aug 14 2017 For integers a and d, let P(a,d,1) = a, P(a,d,2) = a+d, and, for n>2, P(a,d,n) = 2*P(a,d,n-1) + P(a,d,n-2). Further, let p(n) = Sum_{i=1..2n} P(a,d,i). Then p(n)^2 + (p(n)+d)^2 + a^2 = P(a,d,2n+1)^2 + d^2. When a = 1 and d = 1, p(n) = a(n) and P(a,d,n) = A000129(n), the n-th Pell number. - Charlie Marion, Dec 08 2018 The terms of this sequence satisfy the Diophantine equation k^2 + (k+1)^2 = m^2, which is equivalent to (2k+1)^2 - 2*m^2 = -1. Now, with x=2k+1 and y=m, we get the Pell-Fermat equation x^2 - 2*y^2 = -1. The solutions (x,y) of this equation are respectively in A002315 and A001653. The relation k = (x-1)/2 explains Lekraj Beedassy's Nov 25 2003 formula. Thus, the corresponding numbers m = y, which express the length of the hypotenuse of these right triangles (k,k+1,m) are in A001653. - Bernard Schott, Mar 10 2019 Members of Diophantine pairs. Related to solutions of p^2 = 2q^2 + 2 in natural numbers; p = p(n) = 2*sqrt(4T(a(n))+1), q = q(n) = sqrt(8*T(a(n))+1). Note that this implies that 4*T(a(n))+1 is a perfect square (numbers of the form 8*T(n)+1 are perfect squares for all n); these T(a(n))'s are the only solutions to the given Diophantine equation. - Steven Blasberg, Mar 04 2021
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
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
G.f.: x *(3 - x) / ((1 - 6*x + x^2) * (1 - x)). - Simon Plouffe in his 1992 dissertation a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). a_{n} = -1/2 + ((1-2^{1/2})/4)*(3 - 2^{3/2})^n + ((1+2^{1/2})/4)*(3 + 2^{3/2})^n. - Antonio Alberto Olivares, Oct 13 2003 a(n) = a(n-2) + 4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). - Pierre CAMI, Mar 30 2005 a(n) = (sinh((2*n+1)*log(1+sqrt(2)))-1)/2 = (sqrt(1+8* A029549)-1)/2. - Bill Gosper, Feb 07 2010 Let b(n) = A046090(n) and c(n) = A001653(n). Then for k>j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n<0, a(n) = -b(-n-1). Also a(n)*a(n+2*k+1) + b(n)*b(n+2*k+1) + c(n)*c(n+2*k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2*k) + b(n)*b(n+2*k) + c(n)*c(n+2*k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003 a(n+1) = 3*a(n) + sqrt(8*a(n)^2 + 8*a(n) +4) + 1, a(1)=0. - Richard Choulet, Sep 18 2007 As noted (Sep 20 2006), a(n) = 5*(a(n-1) + a(n-2)) - a(n-3) + 4. In general, for n > 2*k, a(n) = A001653(k)*(a(n-k) + a(n-k-1) + 1) - a(n-2*k-1) - 1. Also a(n) = 7*(a(n-1) - a(n-2)) + a(n-3). In general, for n > 2*k, A002378(k)*(a(n-k)-a(n-k-1)) + a(n-2*k-1). - Charlie Marion, Dec 26 2007 a(n) = ( (1 + sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 2)/4 = ( A001333(2n+1) - 1)/2. a(2*n+k-1) = Pell(2*n-1)*Pell(2*n+2*k) + Pell(2*n-2)*Pell(2*n+2*k+1) + A001108(k+1); a(2*n+k) = Pell(2*n)*Pell(2*n+2*k+1) + Pell(2*n-1)*Pell(2*n+2*k+2) - A055997(k+2). (End) E.g.f.: (1/4)*(-2*exp(x) - (sqrt(2) - 1)*exp((3-2*sqrt(2))*x) + (1 + sqrt(2))*exp((3+2*sqrt(2))*x)). - Ilya Gutkovskiy, Apr 11 2016 a(n) = ((a(n-1)+1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020 In general, for each k >= 0, a(n) = ((a(n-k)+a(k-1)+1)*(a(n-k)-a(k)))/a(n-2*k) for n > 2*k. - Charlie Marion, Dec 27 2020
EXAMPLE
The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ...
MAPLE
option remember;
if n <= 1 then
op(n+1, [0, 3]) ;
else
6*procname(n-1)-procname(n-2)+2 ;
end if;
MATHEMATICA
LinearRecurrence[{7, -7, 1}, {0, 3, 20}, 30] (* Harvey P. Dale, Aug 19 2011 *) With[{c=3+2*Sqrt[2]}, NestList[Floor[c*#]+3&, 3, 30]] (* Harvey P. Dale, Oct 22 2012 *) CoefficientList[Series[x (3 - x)/((1 - 6 x + x^2) (1 - x)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 21 2014 *) Table[(LucasL[2*n + 1, 2] - 2)/4, {n, 0, 30}] (* G. C. Greubel, Jul 15 2018 *)
PROG
(PARI) {a(n) = subst( poltchebi(n+1) - poltchebi(n) - 2, x, 3) / 4}; /* Michael Somos, Aug 11 2006 */ (PARI) concat(0, Vec(x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^50))) \\ Altug Alkan, Nov 08 2015 (PARI) {a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
for(n=0, 30, print1(round((Q(2*n+1) - 2)/4), ", ")) \\ G. C. Greubel, Jul 15 2018 (Magma) Z<x>:=PolynomialRing(Integers()); N<r2>:=NumberField(x^2-2); S:=[ (-2+(r2+1)*(3+2*r2)^n-(r2-1)*(3-2*r2)^n)/4: n in [1..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Feb 17 2009 (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3-x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018 (Haskell)
a001652 n = a001652_list !! n
a001652_list = 0 : 3 : map (+ 2)
(zipWith (-) (map (* 6) (tail a001652_list)) a001652_list)
(GAP) a:=[0, 3];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]+2; od; a; # Muniru A Asiru, Dec 08 2018 (Sage) (x*(3-x)/((1-6*x+x^2)*(1-x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 08 2019
CROSSREFS
Cf. A000129, A001333, A001542, A002024, A002378, A029549, A048739, A053142, A055997, A075841, A084159, A084703, A097571, A123737.
a(0)=0, a(1)=2 then a(n) = a(n-2) + 2*sqrt(8*a(n-1)^2 + 8*a(n-1) + 1).
+10
47
0, 2, 14, 84, 492, 2870, 16730, 97512, 568344, 3312554, 19306982, 112529340, 655869060, 3822685022, 22280241074, 129858761424, 756872327472, 4411375203410, 25711378892990, 149856898154532, 873430010034204, 5090723162050694, 29670908962269962, 172934730611569080
COMMENTS
Solution to b(b+1) = 2a(a+1) in natural numbers including 0; a = a(n), b = b(n) = A001652(n). The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
Also the indices of triangular numbers that are half other triangular numbers [a of T(a) such that 2T(a)=T(b)]. The T(a)'s are in A075528, the T(b)'s are in A029549 and the b's are in A001652. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002 Sequences A053141 (this entry), A016278, A077259, A077288 and A077398 are part of an infinite series of sequences. Each depends upon the polynomial p(n) = 4k*n^2 + 4k*n + 1, when 4k is not a perfect square. Equivalently, they each depend on the equation k*t(x)=t(z) where t(n) is the triangular number formula n(n+1)/2. The dependencies are these: they are the sequences of positive integers n such that p(n) is a perfect square and there exists a positive integer m such that k*t(n)=t(m). A053141 is for k=2, A016278 is for k=3, A077259 is for k=5. - Robert Phillips (bobanne(AT)bellsouth.net), Oct 11 2007, Nov 27 2007 The values x(n)= A001652(n), y(n)= A046090(n) and z(n)= A001653(n) form a nearly isosceles Pythagorean triple since y(n)=x(n)+1 and x(n)^2 + y(n)^2 = z(n)^2; e.g., for n=2, 20^2 + 21^2 = 29^2. In a similar fashion, if we define b(n)= A011900(n) and c(n)= A001652(n), a(n), b(n) and c(n) form a nearly isosceles anti-Pythagorean triple since b(n)=a(n)+1 and a(n)^2 + b(n)^2 = c(n)^2 + c(n) + 1; i.e., the value a(n)^2 + b(n)^2 lies almost exactly between two perfect squares; e.g., 2^2 + 3^2 = 13 = 4^2 - 3 = 3^2 + 4; 14^2 + 15^2 = 421 = 21^2 - 20 = 20^2 + 21. - Charlie Marion, Jun 12 2009 Behera & Panda call these the balancers and A001109 are the balancing numbers. - Michel Marcus, Nov 07 2017
FORMULA
a(n) = 6*a(n-1) - a(n-2) + 2, a(0) = 0, a(1) = 2.
G.f.: 2*x/((1-x)*(1-6*x+x^2)).
Let c(n) = A001109(n). Then a(n+1) = a(n)+2*c(n+1), a(0)=0. This gives a generating function (same as existing g.f.) leading to a closed form: a(n) = (1/8)*(-4+(2+sqrt(2))*(3+2*sqrt(2))^n + (2-sqrt(2))*(3-2*sqrt(2))^n). - Bruce Corrigan (scentman(AT)myfamily.com), Oct 30 2002 a(n) = 2*Sum_{k = 0..n} A001109(k). - Mario Catalani (mario.catalani(AT)unito.it), Mar 22 2003 a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3).
a(n) = -(1/2) - (1-sqrt(2))/(4*sqrt(2))*(3-2*sqrt(2))^n + (1+sqrt(2))/(4*sqrt(2))*(3+2*sqrt(2))^n. (End)
Let G(n,m) = (2*m+1)*a(n)+ m and H(n,m) = (2*m+1)*b(n)+m where b(n) is from the sequence A001652 and let T(a) = a*(a+1)/2. Then T(G(n,m)) + T(m) = 2*T(H(n,m)). - Kenneth J Ramsey, Aug 16 2007 Let S(n) equal the average of two adjacent terms of G(n,m) as defined immediately above and B(n) be one half the difference of the same adjacent terms. Then for T(i) = triangular number i*(i+1)/2, T(S(n)) - T(m) = B(n)^2 (setting m = 0 gives the square triangular numbers). - Kenneth J Ramsey, Aug 16 2007 a(n) = (ChebyshevU(n, 3) - ChebyshevU(n-1, 3) - 1)/2 = (Pell(2*n+1) - 1)/2. - G. C. Greubel, Apr 27 2020 E.g.f.: (exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 2*exp(x))/4. - Stefano Spezia, Mar 16 2024
MAPLE
option remember;
if n <= 1 then
op(n+1, [0, 2]) ;
else
6*procname(n-1)-procname(n-2)+2 ;
end if;
MATHEMATICA
a[n_] := Floor[1/8*(2+Sqrt[2])*(3+2*Sqrt[2])^n]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Nov 28 2013 *)
PROG
(Haskell)
a053141 n = a053141_list !! n
a053141_list = 0 : 2 : map (+ 2)
(zipWith (-) (map (* 6) (tail a053141_list)) a053141_list)
(PARI) {x=1+sqrt(2); y=1-sqrt(2); P(n) = (x^n - y^n)/(x-y)};
a(n) = round((P(2*n+1) - 1)/2);
(Magma) R<x>:=PowerSeriesRing(Integers(), 30); Coefficients(R!(2*x/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018 (Sage) [(lucas_number1(2*n+1, 2, -1)-1)/2 for n in range(30)] # G. C. Greubel, Apr 27 2020
Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.
+10
38
1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041, 7199369738058940, 41961001862379597, 244566641436218640
COMMENTS
Solution to a*(a-1) = 2b*(b-1) in natural numbers: a = a(n), b = b(n) = A011900(n). n such that n^2 = (1/2)*(n+floor(sqrt(2)*n*floor(sqrt(2)*n))). - Benoit Cloitre, Apr 15 2003 Place a(n) balls in an urn, of which b(n) = A011900(n) are red; draw 2 balls without replacement; 2*Probability(2 red balls) = Probability(2 balls); this is equivalent to the Pell equation A(n)^2-2*B(n)^2 = -1 with a(n) = (A(n)+1)/2; b(n) = (B(n)+1)/2; and the fundamental solution (7;5) and the solution (3;2) for the unit form. - Paul Weisenhorn, Aug 03 2010 Find base x in which repdigit yy has a square that is repdigit zzzz, corresponding to Diophantine equation zzzz_x = (yy_x)^2; then, solution z = a(n) with x = A002315(n) and y = A001653(n+1) for n >= 1 (see Maurice Protat reference). - Bernard Schott, Dec 21 2022
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
Maurice Protat, Des Olympiades à l'Agrégation, De zzzz_x = (yy_x)^2 à Pell-Fermat, Problème 23, pp. 52-54, Ellipses, Paris, 1997.
FORMULA
a(n) = (-1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n). [corrected by Michel Marcus, Dec 23 2022] a(n) = 6*a(n-1) - a(n-2) - 2, n >= 2, a(0) = 1, a(1) = 4.
a(n) = (A(n+1) - 3*A(n) + 2)/4 with A(n) = A001653(n). G.f.: (1-3*x)/((1-6*x+x^2)*(1-x)).
a(n) = partial sums of A001541(n). (End) Let a(n) = A001652(n), b(n) = this sequence and c(n) = A001653(n). Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. (End) a(n+1) = round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^n)/2);
b(n+1) = round((2+(10+7*sqrt(2))*(3+2*sqrt(2))^n)/4) = A011900(n+1). (End)
a(n)*(a(n)-1)/2 = b(n)*b(n+1) and 2*a(n) - 1 = b(n) + b(n+1), where b(n) = A001109. - Kenneth J Ramsey, Apr 24 2011 a(0)=1, a(1)=4, a(2)=21, a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Harvey P. Dale, Apr 13 2012 E.g.f.: (exp(x) + exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2. - Stefano Spezia, Mar 16 2024
EXAMPLE
For n=4: a(4)=697; b(4)=493; 2*binomial(493,2)=485112=binomial(697,2). - Paul Weisenhorn, Aug 03 2010
MAPLE
Digits:=100: seq(round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^(n-1))/2)/2, n=0..20); # Paul Weisenhorn, Aug 03 2010
MATHEMATICA
Join[{1}, #+1&/@With[{c=3+2Sqrt[2]}, NestList[Floor[c #]+3&, 3, 20]]] (* Harvey P. Dale, Aug 19 2011 *) LinearRecurrence[{7, -7, 1}, {1, 4, 21}, 25] (* Harvey P. Dale, Apr 13 2012 *) a[n_] := (2-ChebyshevT[n, 3]+ChebyshevT[n+1, 3])/4; Array[a, 21, 0] (* Jean-François Alcover, Jul 10 2016, adapted from PARI *)
PROG
(PARI) a(n)=(2-subst(poltchebi(abs(n))-poltchebi(abs(n+1)), x, 3))/4
(PARI) x='x+O('x^30); Vec((1-3*x)/((1-6*x+x^2)*(1-x))) \\ G. C. Greubel, Jul 15 2018 (Haskell)
a046090 n = a046090_list !! n
a046090_list = 1 : 4 : map (subtract 2)
(zipWith (-) (map (* 6) (tail a046090_list)) a046090_list)
(Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-3*x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018
Triangular numbers that are half other triangular numbers.
+10
24
0, 3, 105, 3570, 121278, 4119885, 139954815, 4754343828, 161507735340, 5486508657735, 186379786627653, 6331426236682470, 215082112260576330, 7306460390622912753, 248204571168918457275, 8431648959352604634600, 286427860046819639119128
COMMENTS
This is the sequence of 1/2 the areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n)=x(n)+1, z(n)) with x(0)=0, y(0)=1, z(0)=1, a(0)=0 and x(1)=3, y(1)=4, z(1)=5, a(1)=3. - George F. Johnson, Aug 24 2012
FORMULA
G.f.: 3*x/((1-x)*(1-34*x+x^2)).
a(n) = ((3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1) - 6)/64.
128*a(n)^2 + 24*a(n) + 1 is a perfect square.
a(n+1) = 17*a(n) + 3/2 + 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1) = 17*a(n) + 3/2 - 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1)*a(n+1) = a(n)*(a(n)-3); a(n+1) = 34*a(n) - a(n-1) + 3.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2); a(n) = A096979(2*n)/2. Lim_{n->infinity} a(n)/a(n-1) = 17 + 12*sqrt(2).
Lim_{n->infinity} a(n)/a(n-2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Lim_{n->infinity} a(n)/a(n-r) = (17 + 12*sqrt(2))^r.
Lim_{n->infinity} a(n-r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r.
(End)
a(n) = 34*a(n-1) - a(n-2) + 3, n >= 2. - R. J. Mathar, Nov 07 2015
MATHEMATICA
CoefficientList[ Series[ 3x/(1 - 35 x + 35 x^2 - x^3), {x, 0, 15}], x] (* Robert G. Wilson v, Jun 24 2011 *)
PROG
(PARI) concat(0, Vec(3*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 18 2015
CROSSREFS
Cf. A000129, A000217, A001652, A002315, A029546, A029549, A046090, A046729, A053141, A084159, A096979.
a(n) = 6*a(n-1) - a(n-2) - 2 with a(0) = 1, a(1) = 3.
+10
21
1, 3, 15, 85, 493, 2871, 16731, 97513, 568345, 3312555, 19306983, 112529341, 655869061, 3822685023, 22280241075, 129858761425, 756872327473, 4411375203411, 25711378892991, 149856898154533, 873430010034205, 5090723162050695, 29670908962269963, 172934730611569081
COMMENTS
Members of Diophantine pairs.
Solution to b*(b-1) = 2*a*(a-1) in natural numbers; a = a(n), b = b(n) = A046090(n). Also the indices of centered octagonal numbers which are also centered square numbers. - Colin Barker, Jan 01 2015 Also positive integers y in the solutions to 4*x^2 - 8*y^2 - 4*x + 8*y = 0. - Colin Barker, Jan 01 2015 Also the number of perfect matchings on a triangular lattice of width 3 and length n. - Sergey Perepechko, Jul 11 2019
REFERENCES
Mario Velucchi "The Pell's equation ... an amusing application" in Mathematics and Informatics Quarterly, to appear 1997.
FORMULA
a(n) = (((1+sqrt(2))^(2*n-1) - (1-sqrt(2))^(2*n-1))/sqrt(8) + 1)/2.
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3); a(1) = 1, a(2) = 3, a(3) = 15. Also a(n) = 1/2 + ( (1-sqrt(2))/(-4*sqrt(2)) )*(3-2*sqrt(2))^n + ( (1+sqrt(2))/(4*sqrt(2)) )*(3+2*sqrt(2))^n. - Antonio Alberto Olivares, Dec 23 2003 Sqrt(2) = Sum_{n>=0} 1/a(n); a(n) = a(n-1) + floor(1/(sqrt(2) - Sum_{k=0..n-1} 1/a(k))) (n>0) with a(0)=1. - Paul D. Hanna, Jan 25 2004 a(n+1) = 3*a(n) - 1 + sqrt(8*a(n)^2 - 8*a(n) + 1), a(1)=1. - Richard Choulet, Sep 18 2007 a(n+1) = a(n) * (a(n) + 2) / a(n-1) for n>=1 with a(0)=1 and a(1)=3. - Paul D. Hanna, Apr 08 2012 G.f.: (1 - 4*x + x^2)/((1-x)*(1 - 6*x + x^2)). - R. J. Mathar, Oct 26 2009 E.g.f.: (2*exp(x) + exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/4. - Stefano Spezia, Mar 16 2024
EXAMPLE
G.f. = 1 + 3*x + 15x^2 + 85*x^3 + 493*x^4 + 2871*x^5 + 16731*x^6 + ... - Michael Somos, Feb 23 2019
MAPLE
f:= gfun:-rectoproc({a(n)=6*a(n-1)-a(n-2)-2, a(0)=1, a(1)=3}, a(n), remember):
MATHEMATICA
a[0] = 1; a[1] = 3; a[n_]:= a[n]= 6 a[n-1] -a[n-2] -2; Table[a@ n, {n, 0, 40}] (* Michael De Vlieger, Dec 05 2015 *) LinearRecurrence[{7, -7, 1}, {1, 3, 15}, 40] (* Harvey P. Dale, Feb 16 2017 *) a[ n_] := (4 + ChebyshevT[n, 3] + ChebyshevT[n + 1, 3])/8; (* Michael Somos, Feb 23 2019 *)
PROG
(PARI) Vec((1-4*x+x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Altug Alkan, Dec 06 2015 (Magma) I:=[1, 3]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2) - 2: n in [1..40]]; // Vincenzo Librandi, Dec 05 2015 (SageMath) [(1+lucas_number1(2*n+1, 2, -1))//2 for n in range(41)] # G. C. Greubel, Oct 17 2024
CROSSREFS
Cf. A000194, A001109, A001541, A001653, A002024, A006062, A011906, A029549, A046090, A053141, A075528, A156035.
AUTHOR
Mario Velucchi (mathchess(AT)velucchi.it)
Expansion of 4*x/((1+x)*(1-6*x+x^2)).
+10
15
0, 4, 20, 120, 696, 4060, 23660, 137904, 803760, 4684660, 27304196, 159140520, 927538920, 5406093004, 31509019100, 183648021600, 1070379110496, 6238626641380, 36361380737780, 211929657785304, 1235216565974040, 7199369738058940, 41961001862379596, 244566641436218640
COMMENTS
Related to Pythagorean triples: alternate terms of A001652 and A046090. Even-valued legs of nearly isosceles right triangles: legs differ by 1. 0 is smaller leg of degenerate triangle with legs 0 and 1 and hypotenuse 1. - Charlie Marion, Nov 11 2003 The complete (nearly isosceles) primitive Pythagorean triple is given by {a(n), a(n)+(-1)^n, A001653(n)}. - Lekraj Beedassy, Feb 19 2004 Note also that A046092 is the even leg of this other class of nearly isosceles Pythagorean triangles { A005408(n), A046092(n), A001844(n)}, i.e., {2n+1, 2n(n+1), 2n(n+1)+1} where longer sides (viz. even leg and hypotenuse) are consecutive. - Lekraj Beedassy, Apr 22 2004
REFERENCES
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
W. Sierpiński, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, p. 17. MR2002669.
FORMULA
a(n) = ((1+sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) + 2*(-1)^(n+1))/4.
a(n) = ceiling((sqrt(2)+1)^(2*n+1) - (sqrt(2)-1)^(2*n+1) - 2*(-1)^n)/4. - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 12 2004
a(n) is the k-th entry among the complete near-isosceles primitive Pythagorean triple A114336(n), where k = (3*(2n-1) - (-1)^n)/2, i.e., a(n) = A114336( A047235(n)), for positive n. - Lekraj Beedassy, Jun 04 2006 2*a(n)*(a(n) + (-1)^n) + 1 = ( A000129(2*n+1))^2; n > 0, 2*a(n)*(a(n) + (-1)^n) + 1 = ((a(n+1) - a(n-1))/4)^2, a perfect square.
a(n+1) = (3*a(n) + 2*(-1)^n) + 2*sqrt(2*a(n)*(a(n) + (-1)^n)+ 1).
a(n-1) = (3*a(n) + 2*(-1)^n) - 2*sqrt(2*a(n)*(a(n) + (-1)^n)+ 1).
a(n+1) = 6*a(n) - a(n-1) + 4*(-1)^n.
a(n+1) = 5*a(n) + 5*a(n-1) - a(n-2).
a(n+1) *a(n-1) = a(n)*(a(n) + 4*(-1)^n).
a(n) = (sqrt(1 + 8* A029549(n)) - (-1)^n)/2. Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2).
Limit_{n->oo} a(n)/a(n-2) = 17 + 12*sqrt(2).
Limit_{n->oo} a(n)/a(n-r) = (3 + 2*sqrt(2))^r.
Limit_{n->oo} a(n-r)/a(n) = (3 - 2*sqrt(2))^r. (End)
a(n) = ( A001333(2*n+1) - 2*(-1)^n)/4. E.g.f.: exp(-x)*(exp(4*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 1)/2. - Stefano Spezia, Aug 03 2024
EXAMPLE
[1,0,1]*[1,2,2; 2,1,2; 2,2,3]^0 gives (degenerate) primitive Pythagorean triple [1, 0, 1], so a(0) = 0. [1,0,1]*[1,2,2; 2,1,2; 2,2,3]^7 gives primitive Pythagorean triple [137903, 137904, 195025] so a(7) = 137904.
G.f. = 4*x + 20*x^2 + 120*x^3 + 696*x^4 + 4060*x^5 + 23660*x^6 + ...
PROG
(PARI) a(n)=n%2+(real((1+quadgen(8))^(2*n+1))-1)/2
(PARI) a(n)=if(n<0, -a(-1-n), polcoeff(4*x/(1+x)/(1-6*x+x^2)+x*O(x^n), n))
(Magma) [4*Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)-2*(-1)^n) / 16): n in [0..30]]; // Vincenzo Librandi, Jul 29 2019 (SageMath) [(lucas_number2(2*n+1, 2, -1) -2*(-1)^n)/4 for n in range(41)] # G. C. Greubel, Feb 11 2023
Numbers k such that (k+1)*(2*k+1) is a perfect square.
+10
14
0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024
COMMENTS
Equivalently, both k+1 and 2*k+1 are perfect squares.
The square roots of (k+1)*(2*k+1) are in A046176. The sequence of areas k(n)*q(n)/2, of the ordered Pythagorean triples (k(n), q(n) = k(n) + 2, c(n)) with k(1)=0, q(1)=2, c(1)=0, a(1)=0, and k(2)=6, q(2)=8, c(2)=10, a(2)=24 (conjectured).
Conjecture: let f(n) be a sequence of form x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + v, z(n)) with x(1)=0, y(1)=v, z(1)=0, f(1)=0, where v is an even number. Then there exists such subset p(i) that p(1) = 0, p(2) = 24*(v/2)^2, for any i > 2, p(i) = 34*p(i-1) - p(i-2) + 24*(v/2)^2, and any p(i) is a term of the above sequence f(n) (see also the first formula by Benoit Cloitre in the Formula section). (End)
FORMULA
a(1) = 0, a(2) = 24; for n > 2, a(n) = 34*a(n-1) - a(n-2) + 24.
a(n) = floor(A*B^n), where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2).
G.f.: 24*x^2/( (1-x)*(1-34*x+x^2) ).
a(n) = (3/4)*(ChebyshevT(n, 17) - 16*Chebyshev(n-1, 17) - 1). - G. C. Greubel, Jan 13 2020 Sum_{n>=2} 1/a(n) = (3 - 2*sqrt(2))/4. (End)
MAPLE
seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020
MATHEMATICA
RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n, 20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x, 0, 20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *) Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n, 20}] (* G. C. Greubel, Jan 13 2020 *)
PROG
(PARI) concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016 (Magma) I:=[0, 24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019 (Sage)
P.<x> = PowerSeriesRing(ZZ, prec)
return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
(GAP) a:=[0, 24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020
CROSSREFS
Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.
Numbers m such that 3*m+1 and 8*m+1 are both squares.
+10
13
0, 1, 21, 120, 2080, 11781, 203841, 1154440, 19974360, 113123361, 1957283461, 11084934960, 191793804840, 1086210502741, 18793835590881, 106437544333680, 1841604094101520, 10429793134197921, 180458407386358101, 1022013289607062600
COMMENTS
Naturally, all terms are triangular numbers.
Numbers m such that k*m+1 and 8*m+1 are both squares:
k=3: this sequence;
k=5: 0, 3, 231, 4560, 333336, 6575751, ...
Numbers m such that 3*m+1 and k*m+1 are both squares:
k=8: this sequence.
FORMULA
G.f.: x^2*(1 + 20*x + x^2)/((1 - x)*(1 - 10*x + x^2)*(1 + 10*x + x^2)).
a(n) = a(n-1) + 98*a(n-2) - 98*a(n-3) - a(n-4) + a(n-5).
G.f. of the quadrisections:
a(4k+1): 40*x*(52 + 3*x)/((1 - x)*(1 - 9602*x + x^2));
a(4k+2): (1 + 2178*x + 21*x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+3): (21 + 2178*x + x^2)/((1 - x)*(1 - 9602*x + x^2));
a(4k+4): 40*(3 + 52*x)/((1 - x)*(1 - 9602*x + x^2)).
MATHEMATICA
LinearRecurrence[{1, 98, -98, -1, 1}, {0, 1, 21, 120, 2080}, 20] (* or *) CoefficientList[Series[x (1 + 20 x + x^2)/((1 - x) (1 - 10 x + x^2) (1 + 10 x + x^2)), {x, 0, 20}], x]
PROG
(PARI) a=vector(20); a[1]=0; a[2]=1; a[3]=21; a[4]=120; a[5]=2080; for(i=6, #a, a[i]=a[i-1]+98*a[i-2]-98*a[i-3]-a[i-4]+a[i-5]); a
(Maxima) a[1]:0$ a[2]:1$ a[3]:21$ a[4]:120$ a[5]:2080$ a[n]:=a[n-1]+98*a[n-2]-98*a[n-3]-a[n-4]+a[n-5]$ makelist(a[n], n, 1, 20);
(Magma) I:=[0, 1, 21, 120, 2080]; [n le 5 select I[n] else Self(n-1)+98*Self(n-2)-98*Self(n-3)-Self(n-4)+Self(n-5): n in [1..20]];
EXTENSIONS
Changed offset from 0 to 1 and adapted formulas by Bruno Berselli, Mar 03 2016
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