A180926 Numbers k such that 6*k and 10*k are triangular numbers.

0, 1, 63, 3906, 242110, 15006915, 930186621, 57656563588, 3573776755836, 221516502298245, 13730449365735355, 851066344173293766, 52752382889378478138, 3269796672797292350791, 202674641330542747270905

Offset: 1

Author

Vladimir Pletser, Sep 25 2010

Keywords

• easy
• nonn

Comments

From Klaus Purath, Jul 25 2024: (Start)

Numbers k such that 48k + 1 is a square as well as the sum of two consecutive terms.

a(n) = t(n-1)*t(n)/(8*t(1)^2) where (t) is any recurrence t(k) = 8*t(k-1) - t(k-2) with t(0) = 0 and arbitrary t(1) != 0. (End)

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..500
• Index entries for linear recurrences with constant coefficients, signature (63,-63,1).

Crossrefs

Subsequence of A154293.

Cf. A069498, A154293.

Programs

Mathematica

a[1] = 0; a[n_] := a[n] = (62 a[n - 1] + 1 + Sqrt[(48 a[n - 1] + 1)*(80 a[n - 1] + 1)])/2; Array[a, 14] (* Robert G. Wilson v, Sep 27 2010 *)
Rest[CoefficientList[Series[-x^2/((x - 1) (x^2 - 62 x + 1)), {x, 0, 30}], x]] (* Vincenzo Librandi, Jun 26 2014 *)
LinearRecurrence[{63,-63,1},{0,1,63},20] (* Harvey P. Dale, Dec 25 2019 *)

PARI

isok(n) = ispolygonal(6*n, 3) && ispolygonal(10*n, 3); \\ Michel Marcus, Jun 25 2014

Formula

a(n) = (62*a(n-1) + 1 + ((48*a(n-1) + 1)*(80*a(n-1) + 1))^(1/2))/2 with a(1)=0.

G.f.: -x^2 / ((x-1)*(x^2-62*x+1)). - Colin Barker, Jun 25 2014

a(n) = (-8+(4+sqrt(15))*(31+8*sqrt(15))^(-n) - (-4+sqrt(15))*(31+8*sqrt(15))^n)/480. - Colin Barker, Mar 03 2016

Extensions

a(8) onwards from Robert G. Wilson v, Sep 27 2010