# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a046195 Showing 1-1 of 1 %I A046195 #43 Jun 13 2015 00:49:49 %S A046195 1,6,49,961,8214,70225,1385329,11844150,101263969,1997643025, %T A046195 17079255654,146022572641,2880599856289,24628274808486, %U A046195 210564448483921,4153822995125281,35513955194580726,303633788691241009,5989809878370798481,51211098762310597974 %N A046195 Indices of heptagonal numbers (A000566) which are also squares (A000290). %C A046195 (10 * a(n) - 3)^2 - 40 * (A046196(n))^2 = 9. - _Ant King_, Nov 12 2011 %C A046195 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 A046195 Also indices of heptagonal numbers (A000566) which are also centered octagonal numbers (A016754). - _Colin Barker_, Jan 19 2015 %C A046195 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 A046195 Colin Barker, Table of n, a(n) for n = 1..950 %H A046195 Eric Weisstein's World of Mathematics, Heptagonal Square Number. %H A046195 Index entries for linear recurrences with constant coefficients, signature (1,0,1442,-1442,0,-1,1). %F A046195 From _Paul Weisenhorn_, May 01 2009: (Start) %F A046195 Pell equations: r^2-10*s^2=1 with solution (19,6) %F A046195 (10*n-3)^2-10*(2*m)^2=9; basic solutions: (7,-2); (7,+2)((57,18); %F A046195 with x=10*n-3; y=2*m; A=(19+6*sqrt(10))^2; B=(19-6*sqrt(10))^2 one get %F A046195 x(3*k)+sqrt(10)*y(3*k)=(7-2*sqrt(10))*A^k; %F A046195 x(3*k+1)+sqrt(10)*y(3*k+1)=(7+2*sqrt(10))*A^k; %F A046195 x(3*k+2)+sqrt(10)*y(3*k+2)=(57+18*sqrt(10))*A^k; %F A046195 with the eigenvalues A=721+228*sqrt(10); B=721-228*sqrt(10) %F A046195 one get the recurrences with 1442=4*19*19-2 %F A046195 x(k+6)=1442*x(k+3)-x(k); y(k+6)=1442*y(k+3)-y(k); %F A046195 m(k+6)=1442*m(k+3)-m(k); n(k+6)=1442*n(k+3)-n(k)-432; %F A046195 and the explicit formulas %F A046195 x(3*k+1)=(7*(A^k+B^k)+2*sqrt(10)*(A^k-B^k))/2; %F A046195 x(3*k+2)=(57*(A^k+B^k)+18*sqrt(10)*(A^k-B^k))/2; %F A046195 x(3*k)=(7*(A^k+B^k)-2*sqrt(10)*(A^k-B^k))/2; %F A046195 y(3*k+1)=(7*(A^k-B^k)/sqrt(10)+2*(A^k+B^k)/2; %F A046195 y(3*k+2)=(57*(A^k-B^k)/sqrt(10)+18*(A^k+B^k))/2; %F A046195 y(3*k)=(7*(A^k-B^k)/sqrt(10)-2*(A^k+B^k))/2; %F A046195 n(k)=(x(k)+3)/10; m(k)=y(k)/2; %F A046195 (End) %F A046195 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 A046195 a(n) = 1442*a(n-3) - a(n-6) - 432. - _Ant King_, Nov 12 2011 %p A046195 for n from 1 to 10000 do m:=sqrt((5*n^2-3*n)/2): %p A046195 if (trunc(m)=m) then print(n,m): end if: end do: # _Paul Weisenhorn_, May 01 2009 %t A046195 LinearRecurrence[{1 ,0, 1442, -1442, 0, -1, 1}, {1, 6, 49, 961, 8214, 70225, 1385329}, 17] (* _Ant King_, Nov 12 2011 *) %Y A046195 Cf. A000566, A036254, A046196, A251927, A253920. %K A046195 nonn,easy %O A046195 1,2 %A A046195 _Eric W. Weisstein_ %E A046195 More terms from _Colin Barker_, Dec 11 2014 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE