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.

A363308 Expansion of g.f. C(x*C(x)^3), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).

Original entry on oeis.org

1, 1, 5, 26, 141, 790, 4542, 26668, 159333, 966038, 5930678, 36801660, 230491410, 1455283172, 9253674120, 59209786992, 380961295445, 2463303690790, 15998687418030, 104325569140156, 682768883525830, 4483232450501492, 29527005540912660, 195006621974036808
Offset: 0

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Author

Paul D. Hanna, May 28 2023

Keywords

Comments

Compare the g.f. A(x) = C(x*C(x)^3) to the identity C(-x*C(x)^3) = 1/C(x), where C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
Conjecture: a(n) is odd iff n is a power of 2 or n = 0.

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 26*x^3 + 141*x^4 + 790*x^5 + 4542*x^6 + 26668*x^7 + 159333*x^8 + 966038*x^9 + 5930678*x^10 + ...
such that A(x) = C(x*C(x)^3), where
C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + ... + A000108(n)*x^n + ...
x*C(x)^3 = x + 3*x^2 + 9*x^3 + 28*x^4 + 90*x^5 + ... + A000245(n)*x^n + ...
Note that x*C(x)^3 = (C(x) - 1)*(1-x)/x - 1.
Also, the g.f. of related sequence A033296 begins
B(x) = 1 + x + 6*x^2 + 42*x^3 + 326*x^4 + 2706*x^5 + 23526*x^6 + ...
where A(x) = B(x/A(x)), B(x) = A(x*B(x)) = C(x*B(x)*C(x*B(x))^3).
		

Crossrefs

Cf. A127632, A153294, A033296, A000108 (C(x)), A000245 (x*C(x)^3).

Programs

  • PARI
    {a(n) = if(n==0,1, sum(k=1,n, 3*k* binomial(2*k+1,k) * binomial(2*n+k,n-k) / ((2*k+1)*(2*n+k)) ) )}
    for(n=0, 30, print1(a(n), ", "))
    
  • PARI
    /* G.f. A(x) = C(x*C(x)^3), where C(x) = 1 + x*C(x)^2 */
    {a(n) = my(C = (1 - sqrt(1 - 4*x +x^2*O(x^n)))/(2*x)); polcoeff( subst(C, x, x*C^3), n)}
    for(n=0, 30, print1(a(n), ", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n may be defined as follows; here, C(x) is the g.f. of the Catalan numbers (A000108).
(1) A(x) = C(x*C(x)^3), where C(x) = (1 - sqrt(1-4*x))/(2*x).
(2) A(x) = B(x/A(x)) where B(x) = A(x*B(x)) = C( x*B(x) * C(x*B(x))^3 ) is the g.f. of A033296.
(3) a(n) = Sum_{k=1..n} 3*k* binomial(2*k+1,k) * binomial(2*n+k,n-k) / ((2*k+1)*(2*n+k)) for n > 0, with a(0) = 1.
D-finite with recurrence 4*n*(n-1)*(21687905*n +56141583)*(n+1)*a(n) +2*n*(n-1) *(43375810*n^2 -6022811713*n +10976463649)*a(n-1) -(n-1) *(15429963345*n^3 -200018809315*n^2 +658353214412*n -632905646028)*a(n-2) +(102558230760*n^4 -1409936457473*n^3 +6909548744112*n^2 -14414518702669*n +10812683474490)*a(n-3) +(-212869593020*n^4 +3377685007909*n^3 -20069314453381*n^2 +52902205420466*n -52146873039204)*a(n-4) +4*(2*n-11) *(10773532140*n^3 -171459615587*n^2 +902576783797*n -1572525214995)*a(n-5) +4*(n-6) *(208500820*n -955419151)*(2*n-11) *(2*n-13)*a(n-6)=0. - R. J. Mathar, Nov 22 2024

A110682 A convolution triangle of numbers based on A027307.

Original entry on oeis.org

1, 2, 1, 10, 4, 1, 66, 24, 6, 1, 498, 172, 42, 8, 1, 4066, 1360, 326, 64, 10, 1, 34970, 11444, 2706, 536, 90, 12, 1, 312066, 100520, 23526, 4672, 810, 120, 14, 1, 2862562, 911068, 211546, 42024, 7410, 1156, 154, 16, 1
Offset: 0

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Author

Philippe Deléham, Sep 15 2005

Keywords

Comments

Triangle T(n,k) for A(x)^k = Sum_{n>=k} T(n,k)*x^n, where o.g.f. A(x) satisfies A(x) = (1+x*A(x)^2)/(1-x*A(x)^2). - Vladimir Kruchinin, Mar 16 2011

Crossrefs

Columns: A027307, A032349, A033296.

Programs

  • Mathematica
    T[n_, k_] := (k/(2*n - k))*Sum[Binomial[2*n - k, n - k - j]*Binomial[2*n - k + j - 1, 2*n - k - 1], {j, 0, n - k}]; Table[T[n, k], {n, 0, 25}, {k, 1, n}] // Flatten (* G. C. Greubel, Sep 05 2017 *)
  • PARI
    for(n=0,25, for(k=1,n, print1((k/(2*n-k))*sum(i=0,n-k, binomial(2*n-k,n-k-i)*binomial(2*n-k+i-1,2*n-k-1)), ", "))) \\ G. C. Greubel, Sep 05 2017

Formula

T(0, 0) = 1; T(n, k) = 0 if k<0 or if k>n; T(n, k) = Sum_{j, j>=0} T(n-1, k-1+j)*A006318(j).
Sum_{k, k>=0} T(n, k) = A108442(n+1).
T(n,k) = k/(2*n-k)*Sum_{i=0,n-k} binomial(2*n-k,n-k-i)*binomial(2*n-k+i-1,2*n-k-1), n >= k > 0. - Vladimir Kruchinin, Mar 16 2011

A378840 G.f. A(x) satisfies A(x) = ( 1 + x * A(x)^(4/3) * (1 + A(x)^(1/3)) )^3.

Original entry on oeis.org

1, 6, 66, 902, 13794, 225990, 3878946, 68854278, 1253647938, 23283474310, 439394508162, 8401507608966, 162413310158626, 3169029168475206, 62330703549363810, 1234503404283308038, 24599422679682518658, 492824963618477891334, 9920626149798702401730
Offset: 0

Views

Author

Seiichi Manyama, Dec 09 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n, r=3, t=4, u=1) = r*sum(k=0, n, binomial(n, k)*binomial(t*n+u*k+r, n)/(t*n+u*k+r));

Formula

a(n) = 3 * Sum_{k=0..n} binomial(n,k) * binomial(4*n+k+3,n)/(4*n+k+3).
Conjecture: g.f.: B(x)^3, where B(x) is the g.f. of A260332.
Showing 1-3 of 3 results.