A077259 First member of the Diophantine pair (m,k) that satisfies 5*(m^2 + m) = k^2 + k; a(n) = m.

0, 2, 6, 44, 116, 798, 2090, 14328, 37512, 257114, 673134, 4613732, 12078908, 82790070, 216747218, 1485607536, 3889371024, 26658145586, 69791931222, 478361013020, 1252365390980, 8583840088782, 22472785106426, 154030760585064, 403257766524696, 2763969850442378

Offset

0,2

References

Mohammad K. Azarian, Diophantine Pair, Problem B-881, Fibonacci Quarterly, Vol. 37, No. 3, August 1999, pp. 277-278. Solution appeared in Vol. 38, No. 2, May 2000, pp. 183-184.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.

Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.

Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.

Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.

Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.

Hermann Stamm-Wilbrandt, 6 interlaced bisections

Index entries for linear recurrences with constant coefficients, signature (1,18,-18,-1,1).

Formula

Let b(n) be A007805(n). Then with a(0)=0, a(1)=2, a(2*n+2) = 2*a(2*n+1) - a(2*n) + 2*b(n), a(2*n+3) = 2*a(2*n+2) - a(2*n+1) + 2*b(n+1).

a(n) = (A000045(A007310(n+1))-1)/2. - Vladeta Jovovic, Nov 02 2002 [corrected by R. J. Mathar, Sep 16 2009]

a(0)=0, a(1)=2, a(n+2) = 4 + 9*a(n) + 2*sqrt(1 +20*a(n) +20*a(n)^2). - Herbert Kociemba, May 12 2008

a(0)=0, a(1)=2, a(2)=6, a(3)=44, a(n) = 18*a(n-2) - a(n-4) + 8. - Robert Phillips, Sep 01 2008

G.f.: 2*x*(1+x)^2/((1-x)*(1+4*x-x^2)*(1-4*x-x^2)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009

a(n) = a(-1-n) for all n in Z. - Michael Somos, Jul 15 2018

a(2*n) = A049651(2*n); a(2*n+1) = A110679(2*n+1). See "6 interlaced bisections" link. - Hermann Stamm-Wilbrandt, Apr 18 2019

a(n) = a(n-1) + 18*a(n-2) - 18*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020

From Vladimir Pletser, Feb 07 2021: (Start)

a(n) = ((5+sqrt(5))*(2+sqrt(5))^n + (5-sqrt(5))*(2-sqrt(5))^n)/20 - 1/2 for even n;

a(n) = ((5+3*sqrt(5))*(2+sqrt(5))^n + (5-3*sqrt(5))*(2-sqrt(5))^n)/20 - 1/2 for odd n. (End)

Example

a(3) = (2*6) - 2 + (2*17) = 12 - 2 + 34 = 44.
G.f. = 2*x + 6*x^2 + 44*x^3 + 116*x^4 + 798*x^5 + 2090*x^6 + 14328*x^7 + ... - Michael Somos, Jul 15 2018

Maple

f := gfun:-rectoproc({a(-2) = 2, a(-1) = 0, a(0) = 0, a(1) = 2, a(n) = 18*a(n - 2) - a(n - 4) + 8}, a(n), remember): map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 24 2020

Mathematica

LinearRecurrence[{1, 18, -18, -1, 1}, {0, 2, 6, 44, 116}, 30] (* G. C. Greubel, Jul 15 2018 *)
a[ n_] := With[{m = Max[n, -1 - n]}, SeriesCoefficient[ 2 x (x + 1)^2 / ((1 - x) (x^2 - 4 x - 1) (x^2 + 4 x - 1)), {x, 0, m}]]; (* Michael Somos, Jul 15 2018 *)

Prog

(PARI) my(x='x+O('x^30)); concat([0], Vec(2*x*(x+1)^2/((1-x)*(x^2-4*x-1)*(x^2+4*x-1)))) \\ G. C. Greubel, Jul 15 2018
(Magma) R<x>:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients(R!(2*x*(x+1)^2/((1-x)*(x^2-4*x-1)*(x^2+4*x-1)))); // G. C. Greubel, Jul 15 2018

Crossrefs

Cf. A007805, A077260, A077261, A077262.

Cf. A053141.

Sequence in context: A296828 A135815 A055564 * A319112 A342986 A136589

Adjacent sequences: A077256 A077257 A077258 * A077260 A077261 A077262

Keyword

easy,nonn

Author

Bruce Corrigan (scentman(AT)myfamily.com), Nov 01 2002

Extensions

More terms from Colin Barker, Mar 23 2014

Status

approved