cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A251688 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} T(n,k)^2 * x^k] / A(x)^n * x^n/n ), where T(n,k) is the coefficient of x^k in (1+2*x)^n*(1+3*x)^n.

Original entry on oeis.org

1, 1, 25, 61, 336, 1200, 3600, 13500, 32400, 118800, 259200, 939600, 1944000, 6998400, 13996800, 50155200, 97977600, 349920000, 671846400, 2393452800, 4534963200, 16124313600, 30233088000, 107327462400, 199538380800, 707454259200, 1306069401600, 4625662464000, 8489451110400
Offset: 0

Views

Author

Paul D. Hanna, Feb 25 2015

Keywords

Comments

More generally, if G(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} T(n,k)^2 * x^k] / G(x)^n * x^n/n ), where T(n,k) is the coefficient of x^k in (1 + a*x)^n*(1 + b*x)^n, then G(x) = (1+x)*(1 + a^2*x^2)*(1 + b^2*x^2)*(1 + a^2*b^2*x^3) / (1 - a*b*x^2)^2; here a=2, b=3.
More generally, if G(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} T(n,k)^2 * x^k] / G(x)^n * x^n/n ), where T(n,k) is the coefficient of x^k in (p + q*x)^n*(r + s*x)^n, then G(x) = (1 + p^2*r^2*x)*(1 + p^2*s^2*x^2)*(1 + q^2*r^2*x^2)*(1 + q^2*s^2*x^3) / (1 - p*q*r*s*x^2)^2.

Examples

			G.f.: A(x) = 1 + x + 25*x^2 + 61*x^3 + 336*x^4 + 1200*x^5 + 3600*x^6 +...
where
log(A(x)) = (1 + 5^2*x + 6^2*x^2)/A(x) * x +
(1 + 10^2*x + 37^2*x^2 + 60^2*x^3 + 36^2*x^4)/A(x)^2 * x^2/2 +
(1 + 15^2*x + 93^2*x^2 + 305^2*x^3 + 558^2*x^4 + 540^2*x^5 + 216^2*x^6)/A(x)^3 * x^3/3 +
(1 + 20^2*x + 174^2*x^2 + 860^2*x^3 + 2641^2*x^4 + 5160^2*x^5 + 6264^2*x^6 + 4320^2*x^7 + 1296^2*x^8)/A(x)^4 * x^4/4 +
(1 + 25^2*x + 280^2*x^2 + 1850^2*x^3 + 7985^2*x^4 + 23525^2*x^5 + 47910^2*x^6 + 66600^2*x^7 + 60480^2*x^8 + 32400^2*x^9 + 7776^2*x^10)/A(x)^5 * x^5/5 +...
which involves the squares of coefficients in (1 + 5*x + 6*x^2)^n.
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff( (1+x)*(1+4*x^2)*(1+9*x^2)*(1+36*x^3) / ((1-6*x^2)^2 +x*O(x^n)), n)}
    for(n=0, 40, print1(a(n), ", "))
    
  • PARI
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=0, n, polcoeff(((1+2*x)*(1+3*x) +x*O(x^k))^m, k)^2 *x^k) *x^m/(A+x*O(x^n))^m/m)+x*O(x^n))); polcoeff(A, n)}
    for(n=0, 40, print1(a(n), ", "))

Formula

G.f.: (1+x)*(1+4*x^2)*(1+9*x^2)*(1+36*x^3) / (1-6*x^2)^2.

A248876 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3*n} T(n,k)^2 * x^k] / A(x)^n * x^n/n ) where T(n,k) equals the coefficient of x^k in (1+x+x^2+x^3)^n.

Original entry on oeis.org

1, 1, 1, 2, 4, 8, 13, 24, 45, 85, 161, 305, 582, 1116, 2149, 4152, 8049, 15653, 30528, 59695, 117012, 229880, 452565, 892703, 1764099, 3492029, 6923494, 13747483, 27335873, 54427621, 108505081, 216568556, 432740907, 865610375, 1733227339, 3473805680, 6968708734, 13991916510, 28116598325
Offset: 0

Views

Author

Paul D. Hanna, Mar 04 2015

Keywords

Comments

Compare the definition of this sequence to G(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} T(n,k)^2 * x^k] / G(x)^n * x^n/n ) where T(n,k) = [x^k] (1+x+x^2)^n, which is satisfied by the rational function: G(x) = (1+x^3)*(1+x^6)/((1-x)*(1-x^4)).

Examples

			G.f.: A(x) = 1 + x + x^2 + 2*x^3 + 4*x^4 + 8*x^5 + 13*x^6 + 24*x^7 +...
where
log(A(x)) = (1 + x + x^2 + x^3)/A(x) * x +
(1 + 2^2*x + 3^2*x^2 + 4^2*x^3 + 3^2*x^4 + 2^2*x^5 + x^6)/A(x)^2 * x^2/2 +
(1 + 3^2*x + 6^2*x^2 + 10^2*x^3 + 12^2*x^4 + 12^2*x^5 + 10^2*x^6 + 6^2*x^7 + 3^2*x^8 + x^9)/A(x)^3 * x^3/3 +
(1 + 4^2*x + 10^2*x^2 + 20^2*x^3 + 31^2*x^4 + 40^2*x^5 + 44^2*x^6 + 40^2*x^7 + 31^2*x^8 + 20^2*x^9 + 10^2*x^10 + 4^2*x^11 + x^12)/A(x)^4 * x^4/4 +
which involves the squares of the coefficients in (1 + x + x^2 + x^3)^n.
		

Crossrefs

Programs

  • PARI
    /* By Definition: */
    {T(n,k)=polcoeff((1 + x + x^2 + x^3 + x*O(x^k))^n, k)}
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=0, min(3*m,n-m), T(m,k)^2 * x^k) / (A +x*O(x^n))^m * x^m/m)+x*O(x^n))); polcoeff(A, n)}
    for(n=0, 40, print1(a(n), ", "))

A255839 G.f. satisfies: A(x) = exp( Sum_{n>=1} [Sum_{k=0..3*n} binomial(3*n,k)^2 * x^k] / A(x)^n * x^n/n ).

Original entry on oeis.org

1, 1, 9, 18, 64, 172, 477, 1368, 3681, 10485, 28701, 80829, 225090, 632160, 1778553, 5010948, 14181849, 40161357, 114151716, 324873027, 926918784, 2649218580, 7585705665, 21758756931, 62508649059, 179859399129, 518234494662, 1495275239115, 4319808231645, 12495043092609, 36183457564425
Offset: 0

Views

Author

Paul D. Hanna, Mar 07 2015

Keywords

Comments

Compare the definition of this sequence to G(x) = exp( Sum_{n>=1} [Sum_{k=0..2*n} binomial(2*n,k)^2 * x^k] / G(x)^n * x^n/n ), which is satisfied by the rational function: G(x) = (1+x^2)^2*(1+x^3)/((1-x)*(1-x^2)).

Examples

			G.f.: A(x) = 1 + x + 9*x^2 + 18*x^3 + 64*x^4 + 172*x^5 + 477*x^6 +...
where
log(A(x)) = (1 + 3^2*x + 3^2*x^2 + x^3)/A(x) * x +
(1 + 6^2*x + 15^2*x^2 + 20^2*x^3 + 15^2*x^4 + 6^2*x^5 + x^6)/A(x)^2 * x^/2 +
(1 + 9^2*x + 36^2*x^2 + 84^2*x^3 + 126^2*x^4 + 126^2*x^5 + 84^2*x^6 + 36^2*x^7 + 9^2*x^8 + x^9)/A(x)^3 * x^3/3 +
(1 + 12^2*x + 66^2*x^2 + 220^2*x^3 + 495^2*x^4 + 792^2*x^5 + 924^2*x^6 + 792^2*x^7 + 495^2*x^8 + 220^2*x^9 + 66^2*x^10 + 12^2*x^11 + x^12)/A(x)^4 * x^4/4 +...
which involves the squares of the coefficients in (1 + x)^(3*n).
		

Crossrefs

Programs

  • PARI
    /* By Definition: */
    {a(n)=local(A=1+x); for(i=1, n, A=exp(sum(m=1, n, sum(k=0, min(3*m,n-m), binomial(3*m,k)^2 * x^k) / (A +x*O(x^n))^m * x^m/m)+x*O(x^n))); polcoeff(A, n)}
    for(n=0, 40, print1(a(n), ", "))
Showing 1-3 of 3 results.