A220186 Numbers n >= 0 such that n^2 + n*(n+1)/2 is a square.
0, 8, 800, 78408, 7683200, 752875208, 73774087200, 7229107670408, 708378777612800, 69413891098384008, 6801852948864020000, 666512175097575576008, 65311391306613542428800, 6399849835873029582446408, 627119972524250285537319200
Offset: 1
Links
- Index entries for linear recurrences with constant coefficients, signature (99,-99,1).
Crossrefs
Programs
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C
#include
#include int main() { unsigned long long a, i, t; for (i=0; i < (1L<<32); ++i) { a = (i*i) + ((i+1)*i/2); t = sqrt(a); if (a == t*t) printf("%llu\n", i); } return 0; } -
Mathematica
a[n_]:=Floor[(1/12)*(49 + 20*Sqrt[6])^n]; Table[a[n],{n,0,10}] (* Giovanni Resta, Apr 12 2013 *) LinearRecurrence[{99,-99,1},{0,8,800},20] (* Harvey P. Dale, Nov 01 2022 *)
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PARI
lista(nn) = for(n=0, nn, if(issquare(n^2 + n*(n+1)/2), print1(n, ", "))); \\ Altug Alkan, Mar 05 2016
Formula
a(n) = A098308(2*n-2).
a(1) = 0, a(2) = 8, a(3) = 800 and a(n) = 99*a(n-1)-99*a(n-2)+a(n-3) for n>3. - Giovanni Resta, Apr 12 2013
G.f.: -8*x^2*(x+1) / ((x-1)*(x^2-98*x+1)). - Colin Barker, May 31 2013
a(n) = (49+20*sqrt(6))^(-n)*(49+20*sqrt(6)-2*(49+20*sqrt(6))^n+(49-20*sqrt(6))*(49+20*sqrt(6))^(2*n))/12. - Colin Barker, Mar 05 2016
a(n) = 8*A108741(n). - R. J. Mathar, Feb 19 2017
Comments