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 1-4 of 4 results.

A365856 Expansion of (1/x) * Series_Reversion( x*(1+x)^2*(1-x)^5 ).

Original entry on oeis.org

1, 3, 17, 115, 863, 6903, 57687, 497683, 4398980, 39630305, 362562226, 3359252039, 31457036708, 297247495745, 2830725974514, 27140465365203, 261768686779800, 2538061348959000, 24724191398850125, 241862002342417585, 2374978445599884762
Offset: 0

Views

Author

Seiichi Manyama, Sep 20 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^k*binomial(2*n+k+1, k)*binomial(6*n-k+4, n-k))/(n+1);
    
  • SageMath
    def A365856(n):
        h = binomial(6*n + 4, n) * hypergeometric([-n, 2*n + 2], [-6 * n - 4], -1) / (n + 1)
        return simplify(h)
    print([A365856(n) for n in range(21)])  # Peter Luschny, Sep 20 2023

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(2*n+k+1,k) * binomial(6*n-k+4,n-k).
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(2*n+k+1,k) * binomial(4*n-2*k+2,n-2*k). - Seiichi Manyama, Jan 18 2024
a(n) = (1/(n+1)) * [x^n] 1/( (1+x)^2 * (1-x)^5 )^(n+1). - Seiichi Manyama, Feb 16 2024

A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_(2*n).

Original entry on oeis.org

2, 14, 92, 654, 4752, 35204, 264112, 2000526, 15264866, 117161264, 903533380, 6995547780, 54343476072, 423360920528, 3306313730592, 25876855432846, 202909132368942, 1593755466338030, 12537009118650016, 98753463725849904, 778825917274945408, 6149069826564738780
Offset: 1

Views

Author

Peter Bala, Feb 22 2022

Keywords

Comments

This is a companion sequence to A348410.
Suppose n identical objects are distributed in 4*n labeled baskets, 2*n colored white and 2*n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.

Examples

			n = 2: 14 distributions of 2 identical objects in 4 white and 4 black baskets
             White             Black
   1)   (0) (0) (0) (0)   [2] [0] [0] [0]
   2)   (0) (0) (0) (0)   [0] [2] [0] [0]
   3)   (0) (0) (0) (0)   [0] [0] [2] [0]
   4)   (0) (0) (0) (0)   [0] [0] [0] [2]
   5)   (2) (0) (0) (0)   [0] [0] [0] [0]
   6)   (0) (2) (0) (0)   [0] [0] [0] [0]
   7)   (0) (0) (2) (0)   [0] [0] [0] [0]
   8)   (0) (0) (0) (2)   [0] [0] [0] [0]
   9)   (1) (1) (0) (0)   [0] [0] [0] [0]
  10)   (1) (0) (1) (0)   [0] [0] [0] [0]
  11)   (1) (0) (0) (1)   [0] [0] [0] [0]
  12)   (0) (1) (1) (0)   [0] [0] [0] [0]
  13)   (0) (1) (0) (1)   [0] [0] [0] [0]
  14)   (0) (0) (1) (1)   [0] [0] [0] [0]
		

References

  • R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

Programs

  • Maple
    seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(2*n+k-1,k), k = 0..floor(n/2) ), n = 0..20);
  • PARI
    a(n) = sum(k = 0, n\2, binomial(3*n-2*k-1, n-2*k)*binomial(2*n+k-1, k)); \\ Michel Marcus, Feb 27 2022

Formula

a(n) = [x^n] ( 1/((1 - x)*(1 - x^2)) )^(2*n).
a(n) = Sum_{k = 0..floor(n/2)} C(3*n-2*k-1,n-2*k)*C(2*n+k-1,k).
a(n) = Sum_{k = 0..n} (-1)^k*C(5*n-k-1,n-k)*C(2*n+k-1,k).
1024*n*(n-1)*(2*n-1)*(2*n-3)*(4*n-1)*(4*n-3)*P(n-1)*a(n) = 8*(n-1)*(2*n-3)*Q(n)*a(n-1) + 7*(7*n-8)*(7*n-9)*(7*n-10)*(7*n-11)*(7*n-12)*(7*n-13)*P(n)*a(n-2), with a(1) = 2, a(2) = 14, P(n) = 1744*n^4-3815*n^3+ 2920*n^2-912*n+96 and Q(n) = 46599680*n^8-381534880*n^7+1306363456*n^6- 2428492279*n^5+2661904813*n^4 -1747232452*n^3+664205312*n^2- 132046848*n+10321920.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
The o.g.f. A(x) = 2*x + 14*x^2 + 92*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)*(1 - x^2))^2 ) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
Let F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2)^2 ) = A365855(x). Then A(x) = x*(d/dx log(F(x))). [corrected by Jason Yuen, Mar 22 2025]

A365878 Expansion of (1/x) * Series_Reversion( x*(1+x)^3*(1-x)^4 ).

Original entry on oeis.org

1, 1, 5, 17, 83, 381, 1939, 9905, 52544, 282315, 1545130, 8552557, 47880020, 270401515, 1539288570, 8821594865, 50860072024, 294774097800, 1716506373521, 10037592274363, 58920231785426, 347051995986538, 2050627029532225, 12151336260368205
Offset: 0

Views

Author

Seiichi Manyama, Sep 21 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^k*binomial(3*n+k+2, k)*binomial(5*n-k+3, n-k))/(n+1);
    
  • SageMath
    def A365878(n):
        h = binomial(5*n + 3, n) * hypergeometric([-n, 3*(n + 1)], [-5 * n - 3], -1) / (n + 1)
        return simplify(h)
    print([A365878(n) for n in range(24)])  # Peter Luschny, Sep 21 2023

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(3*n+k+2,k) * binomial(5*n-k+3,n-k).
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(3*n+k+2,k) * binomial(2*n-2*k,n-2*k). - Seiichi Manyama, Jan 18 2024
a(n) = (1/(n+1)) * [x^n] 1/( (1+x)^3 * (1-x)^4 )^(n+1). - Seiichi Manyama, Feb 16 2024

A365854 Expansion of (1/x) * Series_Reversion( x*(1+x)^2*(1-x)^3 ).

Original entry on oeis.org

1, 1, 4, 13, 55, 232, 1052, 4869, 23206, 112519, 554560, 2767336, 13959941, 71060356, 364569352, 1883143669, 9785481498, 51118097686, 268294595396, 1414106565611, 7481787454031, 39721596068000, 211549545257760, 1129912319370600, 6050931114958080
Offset: 0

Views

Author

Seiichi Manyama, Sep 20 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, (-1)^k*binomial(2*n+k+1, k)*binomial(4*n-k+2, n-k))/(n+1);
    
  • SageMath
    def A365854(n):
        h = binomial(2*(2*n + 1), n) * hypergeometric([-n, 2*(n + 1)], [-2*(2*n + 1)], -1) / (n + 1)
        return simplify(h)
    print([A365854(n) for n in range(25)])  # Peter Luschny, Sep 20 2023

Formula

a(n) = (1/(n+1)) * Sum_{k=0..n} (-1)^k * binomial(2*n+k+1,k) * binomial(4*n-k+2,n-k).
a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(2*n+k+1,k) * binomial(2*n-2*k,n-2*k). - Seiichi Manyama, Jan 18 2024
a(n) = (1/(n+1)) * [x^n] 1/( (1+x)^2 * (1-x)^3 )^(n+1). - Seiichi Manyama, Feb 16 2024
Showing 1-4 of 4 results.