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.

A094601 G.f. satisfies: A(x) = F(x*A(x)), where F(x) is the g.f. of A094600.

Original entry on oeis.org

1, 1, 3, 12, 50, 234, 1125, 5620, 28753, 150106, 796240, 4279232, 23251672, 127518750, 704957715, 3924307492, 21978740682, 123758612644, 700204091361, 3978636187708, 22694470914700, 129904466979030, 745949776425002
Offset: 0

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Author

Paul D. Hanna, May 13 2004

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 12*x^3 + 50*x^4 + 234*x^5 + 1125*x^6 + 5620*x^7 +...
where
A(x) = Sum_{n>=1} A094600(2*n)*x^n/(n+1), and
log(A(x)) = Sum_{n>=1} A094600(2*n-1)*x^n/n,
log(A(x)) = x + 5*x^2/2 + 28*x^3/3 + 145*x^4/4 + 831*x^5/5 + 4664*x^6/6 +...
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x); for(i=1,n, A=subst(A+x*A', x, x^2*A^2)+x*A*subst(A', x, x^2*A^2)/subst(A, x, x^2*A^2) +x*O(x^n)); polcoeff(A, n)}
    for(n=0, 30, print1(a(n), ", "))

Formula

a(n) = A094600(2*n)/(n+1) for n>=0.
G.f.: A(x) = exp( Sum_{n>=1} A094600(2*n-1)*x^n/n ).
G.f. satisfies: A(x) = A(y) + x*A(x)*A'(y)/A(y) + x^2*A(x)^2*A'(y) where y = x^2*A(x)^2.

Extensions

Entry revised by Paul D. Hanna, Apr 17 2013

A094557 a(2*n) equals coefficient of x^n in A(x)^(n+1) and a(2*n+1) equals coefficient of x^n in A(x)^(n+2), for n>=0.

Original entry on oeis.org

1, 1, 2, 3, 9, 14, 40, 65, 210, 339, 1080, 1764, 5775, 9448, 30992, 50931, 168849, 277920, 925240, 1525887, 5106288, 8431260, 28309440, 46796334, 157627548, 260788843, 880639004, 1458096900, 4934715105, 8175734400, 27721876064
Offset: 0

Views

Author

Paul D. Hanna, May 11 2004

Keywords

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 3*x^3 + 9*x^4 + 14*x^5 + 40*x^6 + 65*x^7 +...
Terms are produced by main and secondary diagonals in the table of successive self-convolutions of this sequence:
  [(1), 1, 2, 3, 9, 14, 40, 65, 210, 339, 1080, ...];
  [(1),(2), 5, 10, 28, 58, 153, 320, 875, 1850, ...];
  [1, (3),(9), 22, 63, 153, 410, 978, 2607, 6222, ...];
  [1, 4, (14),(40), 121, 328, 918, 2392, 6504, 16708, ...];
  [1, 5, 20, (65),(210), 621, 1830, 5110, 14395, 39085, ...];
  [1, 6, 27, 98, (339),(1080), 3356, 9942, 29163, 83008, ...];
  [1, 7, 35, 140, 518, (1764),(5775), 18040, 55160, 163863, ...];
  [1, 8, 44, 192, 758, 2744, (9448),(30992), 98729, 305240, ...];
  [1, 9, 54, 255, 1071, 4104, 14832, (50931),(168849), 542164, ...];
  [1, 10, 65, 330, 1470, 5942, 22495, 80660, (277920),(925240), ...]; ...
from which A094558 may be formed from the main diagonal:
  [1/1, 2/2, 9/3, 40/4, 210/5, 1080/6, 5775/7, 30992/8, 168849/9, 925240/10,...].
Let G(x) be the g.f. of A094558:
G(x) = 1 + x + 3*x^2 + 10*x^3 + 42*x^4 + 180*x^5 + 825*x^6 + 3874*x^7 +...
then the coefficients of G(x)^2 generates the secondary diagonal:
[1*2/2, 3*2/3, 14*2/4, 65*2/5, 339*2/6, 1764*2/7, 9448*2/8, 50931*2/9,...]
and may be derived from the odd-indexed terms of this sequence.
		

Crossrefs

Programs

  • PARI
    {a(n)=local(A=1+x,G);for(i=1,n,G=serreverse(x/A+x*O(x^n));A=subst(deriv(G),x,x^2)+subst(deriv(G^2/2),x,x^2)/x);polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))

Formula

a(2*n) = (n+1)*A094558(n).
G.f. satisfies: A(x) = G(x/A(x)) where G(x) = A(x*G(x)) is the g.f. of A094558.
G.f. satisfies: A(x) = F'(x^2)*(1 + F(x^2)/x) where F(x) = Series_Reversion(x/A(x)) and F(x)/x is the g.f. of A094558.

Extensions

Entry revised by Paul D. Hanna, Apr 17 2013

A139796 Last term of A139687(n) with a fourth leading 1 = 1, 1, 1, 1, 2, 2, 1, 3, 5, 5 rows.

Original entry on oeis.org

1, 1, 2, 5, 9, 28, 48, 165, 275, 1001, 1638, 6188
Offset: 0

Views

Author

Paul Curtz, Jun 14 2008

Keywords

Comments

Based on sequences identical to their p-th differences.

Crossrefs

Cf. A094600.

Formula

Odd rows are 1; 1, 2, 2; 1, 3, 6, 9, 9; 1, 4, 10, 20, 34, 48, 48; 1, 5, 15, 35, 70, 125, 200, 275, 275; 1, 6, 21, 56, 126, 252, 461, 780, 1209, 1638, 1638; with respective sums second bisection a(2n+1)= 1, 5, 28, 165, 1001, 6188 = A025174(n+1)?
Showing 1-3 of 3 results.