A273053 Numbers n such that 15*n^2 + 16 is a square.
0, 4, 32, 252, 1984, 15620, 122976, 968188, 7622528, 60012036, 472473760, 3719778044, 29285750592, 230566226692, 1815244062944, 14291386276860, 112515846151936, 885835382938628, 6974167217357088, 54907502355918076, 432285851629987520, 3403379310683982084
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (8,-1).
Programs
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Magma
[n: n in [0..2*10^7] |IsSquare(15*n^2+16)];
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Maple
a:=proc(n) option remember; if n=1 then 0 elif n=2 then 4 else 8*a(n-1) - a(n-2); fi; end: seq(a(n), n=1..30); # Wesley Ivan Hurt, May 14 2016
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Mathematica
LinearRecurrence[{8, -1}, {0, 4}, 30] -
PARI
concat(0, Vec(4*x^2/(1-8*x+x^2) + O(x^50))) \\ Colin Barker, May 14 2016
Formula
O.g.f.: 4*x^2/(1 - 8*x + x^2).
E.g.f.: 4*(1 + (4*sqrt(15)*sinh(sqrt(15)*x) - 15*cosh(sqrt(15)*x))*exp(4*x)/15). - Ilya Gutkovskiy, May 14 2016
a(n) = 8*a(n-1) - a(n-2) for n>2.
a(n) = -(2*((4-sqrt(15))^n*(4+sqrt(15))+(-4+sqrt(15))*(4+sqrt(15))^n))/sqrt(15). - Colin Barker, May 14 2016
a(n+2) - a(n+1) = 4*070997(n) for n>0. - Wesley Ivan Hurt, May 14 2016