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%I #43 Jun 13 2015 00:49:49
%S 1,6,49,961,8214,70225,1385329,11844150,101263969,1997643025,
%T 17079255654,146022572641,2880599856289,24628274808486,
%U 210564448483921,4153822995125281,35513955194580726,303633788691241009,5989809878370798481,51211098762310597974
%N Indices of heptagonal numbers (A000566) which are also squares (A000290).
%C (10 * a(n) - 3)^2 - 40 * (A046196(n))^2 = 9. - _Ant King_, Nov 12 2011
%C Also numbers n such that the n-th heptagonal number is equal to the sum of two consecutive triangular numbers. - _Colin Barker_, Dec 11 2014
%C Also indices of heptagonal numbers (A000566) which are also centered octagonal numbers (A016754). - _Colin Barker_, Jan 19 2015
%C Also nonnegative integers y in the solutions to 2*x^2-5*y^2+4*x+3*y+2+2 = 0, the corresponding values of x being A251927. - _Colin Barker_, Dec 11 2014
%H Colin Barker, <a href="/A046195/b046195.txt">Table of n, a(n) for n = 1..950</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HeptagonalSquareNumber.html">Heptagonal Square Number.</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,1442,-1442,0,-1,1).
%F From _Paul Weisenhorn_, May 01 2009: (Start)
%F Pell equations: r^2-10*s^2=1 with solution (19,6)
%F (10*n-3)^2-10*(2*m)^2=9; basic solutions: (7,-2); (7,+2)((57,18);
%F with x=10*n-3; y=2*m; A=(19+6*sqrt(10))^2; B=(19-6*sqrt(10))^2 one get
%F x(3*k)+sqrt(10)*y(3*k)=(7-2*sqrt(10))*A^k;
%F x(3*k+1)+sqrt(10)*y(3*k+1)=(7+2*sqrt(10))*A^k;
%F x(3*k+2)+sqrt(10)*y(3*k+2)=(57+18*sqrt(10))*A^k;
%F with the eigenvalues A=721+228*sqrt(10); B=721-228*sqrt(10)
%F one get the recurrences with 1442=4*19*19-2
%F x(k+6)=1442*x(k+3)-x(k); y(k+6)=1442*y(k+3)-y(k);
%F m(k+6)=1442*m(k+3)-m(k); n(k+6)=1442*n(k+3)-n(k)-432;
%F and the explicit formulas
%F x(3*k+1)=(7*(A^k+B^k)+2*sqrt(10)*(A^k-B^k))/2;
%F x(3*k+2)=(57*(A^k+B^k)+18*sqrt(10)*(A^k-B^k))/2;
%F x(3*k)=(7*(A^k+B^k)-2*sqrt(10)*(A^k-B^k))/2;
%F y(3*k+1)=(7*(A^k-B^k)/sqrt(10)+2*(A^k+B^k)/2;
%F y(3*k+2)=(57*(A^k-B^k)/sqrt(10)+18*(A^k+B^k))/2;
%F y(3*k)=(7*(A^k-B^k)/sqrt(10)-2*(A^k+B^k))/2;
%F n(k)=(x(k)+3)/10; m(k)=y(k)/2;
%F (End)
%F 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
%F a(n) = 1442*a(n-3) - a(n-6) - 432. - _Ant King_, Nov 12 2011
%p for n from 1 to 10000 do m:=sqrt((5*n^2-3*n)/2):
%p if (trunc(m)=m) then print(n,m): end if: end do: # _Paul Weisenhorn_, May 01 2009
%t LinearRecurrence[{1 ,0, 1442, -1442, 0, -1, 1}, {1, 6, 49, 961, 8214, 70225, 1385329}, 17] (* _Ant King_, Nov 12 2011 *)
%Y Cf. A000566, A036254, A046196, A251927, A253920.
%K nonn,easy
%O 1,2
%A _Eric W. Weisstein_
%E More terms from _Colin Barker_, Dec 11 2014
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