A046195 Indices of heptagonal numbers (A000566) which are also squares (A000290).
1, 6, 49, 961, 8214, 70225, 1385329, 11844150, 101263969, 1997643025, 17079255654, 146022572641, 2880599856289, 24628274808486, 210564448483921, 4153822995125281, 35513955194580726, 303633788691241009, 5989809878370798481, 51211098762310597974
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..950
- Eric Weisstein's World of Mathematics, Heptagonal Square Number.
- Index entries for linear recurrences with constant coefficients, signature (1,0,1442,-1442,0,-1,1).
Programs
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Maple
for n from 1 to 10000 do m:=sqrt((5*n^2-3*n)/2): if (trunc(m)=m) then print(n,m): end if: end do: # Paul Weisenhorn, May 01 2009
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Mathematica
LinearRecurrence[{1 ,0, 1442, -1442, 0, -1, 1}, {1, 6, 49, 961, 8214, 70225, 1385329}, 17] (* Ant King, Nov 12 2011 *)
Formula
From Paul Weisenhorn, May 01 2009: (Start)
Pell equations: r^2-10*s^2=1 with solution (19,6)
(10*n-3)^2-10*(2*m)^2=9; basic solutions: (7,-2); (7,+2)((57,18);
with x=10*n-3; y=2*m; A=(19+6*sqrt(10))^2; B=(19-6*sqrt(10))^2 one get
x(3*k)+sqrt(10)*y(3*k)=(7-2*sqrt(10))*A^k;
x(3*k+1)+sqrt(10)*y(3*k+1)=(7+2*sqrt(10))*A^k;
x(3*k+2)+sqrt(10)*y(3*k+2)=(57+18*sqrt(10))*A^k;
with the eigenvalues A=721+228*sqrt(10); B=721-228*sqrt(10)
one get the recurrences with 1442=4*19*19-2
x(k+6)=1442*x(k+3)-x(k); y(k+6)=1442*y(k+3)-y(k);
m(k+6)=1442*m(k+3)-m(k); n(k+6)=1442*n(k+3)-n(k)-432;
and the explicit formulas
x(3*k+1)=(7*(A^k+B^k)+2*sqrt(10)*(A^k-B^k))/2;
x(3*k+2)=(57*(A^k+B^k)+18*sqrt(10)*(A^k-B^k))/2;
x(3*k)=(7*(A^k+B^k)-2*sqrt(10)*(A^k-B^k))/2;
y(3*k+1)=(7*(A^k-B^k)/sqrt(10)+2*(A^k+B^k)/2;
y(3*k+2)=(57*(A^k-B^k)/sqrt(10)+18*(A^k+B^k))/2;
y(3*k)=(7*(A^k-B^k)/sqrt(10)-2*(A^k+B^k))/2;
n(k)=(x(k)+3)/10; m(k)=y(k)/2;
(End)
a(n) = +a(n-1) +1442*a(n-3) -1442*a(n-4) -a(n-6) +a(n-7). G.f.: -x*(1+5*x+43*x^2-530*x^3+43*x^4+5*x^5+x^6) / ( (x-1)*(x^6-1442*x^3+1) ). - R. J. Mathar, Aug 01 2010
a(n) = 1442*a(n-3) - a(n-6) - 432. - Ant King, Nov 12 2011
Extensions
More terms from Colin Barker, Dec 11 2014
Comments