A247215 Integers k such that 3k+1 and 6k+1 are both squares.
0, 8, 280, 9520, 323408, 10986360, 373212840, 12678250208, 430687294240, 14630689753960, 497012764340408, 16883803297819920, 573552299361536880, 19483894374994434008, 661878856450449219400, 22484397224940279025600, 763807626791519037651008
Offset: 1
Examples
When n=1, a(1)=0, 3(0)+1=1, 6(0)+1=1. When n=2, a(2)=8, 3(8)+1=25, 6(8)+1=49. When n=3, a(3)=280, 3(280)+1=841=29^2, 6(280)+1=1681=41^2. When n=4, a(4)=9520, 3(9520)+1=28560=169^2, 6(9520)+1=57121=239^2.
Links
- Colin Barker, Table of n, a(n) for n = 1..650
- Index entries for linear recurrences with constant coefficients, signature (35,-35,1).
Programs
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Mathematica
LinearRecurrence[{35,-35,1},{0,8,280},20] (* Harvey P. Dale, Mar 25 2025 *) -
PARI
concat(0, Vec(-8*x^2/((x-1)*(x^2-34*x+1)) + O(x^100))) \\ Colin Barker, Nov 26 2014
Formula
a(n) = (1/72)*(3*(3*(17-12*sqrt(2))^n+2*sqrt(2)*(17-12*sqrt(2))^n+3*(17+12*sqrt(2))^n-2*sqrt(2)*(17+12*sqrt(2))^n)-18).
From Colin Barker, Nov 26 2014: (Start)
a(n) = 8*A029546(n).
a(n) = 35*a(n-1)-35*a(n-2)+a(n-3).
G.f.: -8*x^2 / ((x-1)*(x^2-34*x+1)).
(End)
Lim_{n -> infinity} a(n+1)/a(n) = 33.970562748... = (1+sqrt(2))^4 (the dominant root of x^2-34*x+1). - Joerg Arndt, Dec 01 2014
Extensions
More terms from Colin Barker, Nov 26 2014