A253920 -id:A253920 - cp's OEIS frontend

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

Eric W. Weisstein

(10 * a(n) - 3)^2 - 40 * (A046196(n))^2 = 9. - Ant King, Nov 12 2011
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
Also indices of heptagonal numbers (A000566) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 19 2015
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

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).

Cf. A000566, A036254, A046196, A251927, A253920.

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

LinearRecurrence[{1 ,0, 1442, -1442, 0, -1, 1}, {1, 6, 49, 961, 8214, 70225, 1385329}, 17] (* Ant King, Nov 12 2011 *)

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

More terms from Colin Barker, Dec 11 2014

A036354 Heptagonal square numbers.

Original entry on oeis.org

1, 81, 5929, 2307361, 168662169, 12328771225, 4797839017609, 350709705290025, 25635978392186449, 9976444135331412025, 729252434211108535809, 53306479301521270428241, 20744638830126197732344369, 1516379800105728357531817761, 110843467413344235941816109721

Offset: 1

Jean-Francois Chariot (jeanfrancois.chariot(AT)afoc.alcatel.fr)

From Ant King, Nov 11 2011: (Start)
This sequence is also the union of the three sequences defined by:
a(3n-2) = ((10 - sqrt(10)) * (3 + sqrt(10))^(4*n-3) - (10 + sqrt(10)) * (-3 + sqrt(10))^(4*n-3))^2 / 1600.
a(3n-1) =  9/160 * ((3 + sqrt(10))^(4*n-2) - (-3 + sqrt(10))^(4*n-2))^2.
a(3n) =  ((20 - 7*sqrt(10)) * (3 + sqrt(10))^(4*n) + (20 + 7*sqrt(10)) * (-3 + sqrt(10))^(4*n))^2 / 1600.
Equivalent short forms for these subsequences are:
a(3n-2) = floor((10 - sqrt(10))^2 * (3 + sqrt(10))^(8*n - 6) / 1600).
a(3n-1) = floor( 9/160 * (3 + sqrt(10))^(8*n - 4)).
a(3n) = floor((20 - 7*sqrt(10))^ 2 * (3 + sqrt(10))^(8*n) / 1600).
(End)
Also heptagonal numbers (A000566) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 19 2015

Colin Barker, Table of n, a(n) for n = 1..475
Eric Weisstein's World of Mathematics, Heptagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (1,0,2079362,-2079362,0,-1,1).

Cf. A046195, A046196, A253920.

A036354 := proc(n)
if n <= 7 then
    op(n,[1, 81, 5929, 2307361, 168662169, 12328771225, 4797839017609]);
else
    procname(n-1) +2079362 *(procname(n-3)-procname(n-4)) -procname(n-6) +procname(n-7) ;
end if;
end proc:
seq(A036354(n),n=1..12) ;

LinearRecurrence[{ 1, 0, 2079362, -2079362, 0, -1, 1 }, {1, 81, 5929, 2307361, 168662169, 12328771225, 4797839017609 }, 13] (* Ant King, Nov 11 2011 *)

Vec(-x*(x^6+80*x^5+5848*x^4+222070*x^3+5848*x^2+80*x+1)/((x-1)*(x^6-2079362*x^3+1)) + O(x^100)) \\ Colin Barker, Jan 19 2015

O.g.f.: -x*(1 + 80*x + 5848*x^2 + 222070*x^3 + 5848*x^4 + 80*x^5 + x^6) / ( (x-1)*(x^6 - 2079362*x^3 + 1) ).
From Richard Choulet, May 08 2009: (Start)
With the first values, for n>=0, a(n+9) = 2079363*(a(n+6) - a(n+3)) + a(n).
On every bisection modulo 2: a(n+1) = 1039681*a(n) + 116964 + 164388*sqrt(40*a(n)^2 + 9*a(n)).
On every bisection modulo 2: a(n+2) = 2079362*a(n+1) - a(n) + 233928. (End)
From Ant King, Nov 11 2011: (Start)
a(n) = a(n-1) + 2079362*a(n-3) - 2079362*a(n-4) - a(n-6) + a(n-7).
a(n) = 2079362*a(n-3) - a(n-6) + 233928.
(End)
From Jonathan Pappas, Jan 16 2022: (Start)
Define the three sequences
  b(n) = 1442*b(n-1) - b(n-2) for n >= 2, with b(0) = -77, b(1) =  1;
  c(n) = 1442*c(n-1) - c(n-2) for n >= 2, with c(0) =  -9, c(1) =  9; and
  d(n) = 1442*d(n-1) - d(n-2) for n >= 2, with d(0) =  -1, d(1) = 77.
Then, for n >= 1,
  a(3n - 2) = b(n)^2,
  a(3n - 1) = c(n)^2, and
  a(3n)     = d(n)^2.
(End)

More terms from Eric W. Weisstein

One more term from Richard Choulet, May 08 2009

cp's OEIS Frontend

A046195 Indices of heptagonal numbers (A000566) which are also squares (A000290).

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