A105051 Define a(1)=0, a(2)=0, a(3)=15, a(4)=111 then a(n) = 254*a(n-2) - a(n-4) + 126 also sequence such that 7*a(n)*(a(n) + 1) + 1 = a square.

0, 0, 15, 111, 3936, 28320, 999855, 7193295, 253959360, 1827068736, 64504677711, 464068265775, 16383934179360, 117871512438240, 4161454776879855, 29938900091047311, 1056993129393303936, 7604362751613578880

Offset: 1

Author

Pierre CAMI, Apr 04 2005

Keywords

• nonn

Comments

This sequence is such that 7*a(n)*(a(n) + 1) + 1 = j^2 = a square integer positive root of a(n)^2 + a(n) - (j^2-1)/7 = 0 is such that 2*a(n) + 1 = sqrt((4*j^2+3)/7), ((4*j^2+3)/7) need to be a square so j^2 = (7*k^2-3)/4, put k=2*a(n)+1 you found that (7*k^2-3)/4 = 7*a(n)^2 + 7*a(n) + 1

Links

• G. C. Greubel, Table of n, a(n) for n = 1..825
• Index entries for linear recurrences with constant coefficients, signature (1,254,-254,-1,1).

Crossrefs

Cf. A105040.

Programs

Magma

R:=PowerSeriesRing(Integers(), 30); [0,0] cat Coefficients(R!( 3*x^2*(5+32*x+5*x^2)/((1-x)*(1-254*x^2+x^4)) )); // G. C. Greubel, Mar 13 2023

Mathematica

LinearRecurrence[{1,254,-254,-1,1}, {0,0,15,111,3936}, 30] (* G. C. Greubel, Mar 13 2023 *)

SageMath

@CachedFunction
def a(n): # a = A105051
    if (n<6): return (0,0,0,15,111,3936)[n]
    else: return a(n-1) +254*a(n-2) -254*a(n-3) -a(n-4) +a(n-5)
[a(n) for n in range(41)] # G. C. Greubel, Mar 13 2023

Formula

From R. J. Mathar, Aug 28 2008: (Start)

a(n) = A105040(n-2).

G.f.: 3*x^2*(5+32*x+5*x^2)/((1-x)*(1+16*x+x^2)*(1-16*x+x^2)). (End)

Extensions

Extended by R. J. Mathar, Aug 28 2008