A007750 Nonnegative integers n such that n^2*(n+1)*(2*n+1)^2*(7*n+1)/36 is a square.
0, 1, 7, 24, 120, 391, 1921, 6240, 30624, 99457, 488071, 1585080, 7778520, 25261831, 123968257, 402604224, 1975713600, 6416405761, 31487449351, 102259887960, 501823476024, 1629741801607, 7997688167041, 25973608937760
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- K. R. S. Sastry, Problem 533 The College Mathematics Journal, 25, issue 4, 1994, p. 334.
- K. R. S. Sastry, Square Products of Sums of Squares The College Mathematics Journal, 26, issue 4, 1995, p. 333.
- Index entries for linear recurrences with constant coefficients, signature (1,16,-16,-1,1).
Programs
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GAP
a:=[0,1,7,24,120];; for n in [6..30] do a[n]:=a[n-1]+16*a[n-2]-16*a[n-3] -a[n-4]+a[n-5]; od; a; # G. C. Greubel, Feb 10 2020
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Magma
R
:=PowerSeriesRing(Integers(), 30); Coefficients(R!( x*(1+6*x+x^2)/((1-x)*(1-16*x^2+x^4)) )); // G. C. Greubel, Feb 10 2020 -
Maple
m:=30; S:=series(x*(1+6*x+x^2)/((1-x)*(1-16*x^2+x^4)), x, m+1): seq(coeff(S, x, j), j=0..m); # G. C. Greubel, Feb 10 2020
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Mathematica
CoefficientList[Series[x*(1+6*x+x^2)/((1-x)*(1-16*x^2+x^4)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Jan 15 2017 *) Table[If[EvenQ[n], (4*ChebyshevU[n/2,8] -11*ChebyshevU[(n-2)/2,8] -4)/7, (11*ChebyshevU[(n-1)/2,8] -4*ChebyshevU[(n-3)/2,8] -4)/7], {n,0,30}] (* G. C. Greubel, Feb 10 2020 *) -
PARI
{a(n) = if( n<0, a(-1-n), if( n<2, n>0, 16 * a(n-2) - a(n-4) + 8))} /* Michael Somos, Jul 27 2002 */ -
PARI
{a(n) = local(w); if( n<0, 0, w = 8 + 3*quadgen(28); n = ((n+1)\2) * (-1)^(n%2); imag(w^n) + 4 * (real(w^n) - 1) / 7)} /* Michael Somos, Jul 27 2002 */ -
Sage
def A007750_list(prec): P.
= PowerSeriesRing(ZZ, prec) return P( x*(1+6*x+x^2)/((1-x)*(1-16*x^2+x^4)) ).list() A007750_list(30) # G. C. Greubel, Feb 10 2020
Formula
From Michael Somos, Jul 27 2002: (Start)
G.f.: x * (1 + 6*x + x^2) / ((1 - x) * (1 - 16*x^2 + x^4)).
a(n) = 16 * a(n-2) - a(n-4) + 8. (End)
From G. C. Greubel, Feb 10 2020: (Start)
a(2*n) = (4*ChebyshevU(n,8) - 11*ChebyshevU(n-1,8) - 4)/7 = A007751(n).
a(2*n+1) = (11*ChebyshevU(n,8) - 4*ChebyshevU(n-1,8) - 4)/7 = A007752(n+1). (End)
Extensions
Edited by Michael Somos, Jul 27 2002
Comments