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 1-6 of 6 results.

A245049 Number A(n,k) of hybrid k-ary trees with n internal nodes; square array A(n,k), n>=0, k>=1, read by antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 2, 7, 5, 1, 2, 11, 31, 8, 1, 2, 15, 81, 154, 13, 1, 2, 19, 155, 684, 820, 21, 1, 2, 23, 253, 1854, 6257, 4575, 34, 1, 2, 27, 375, 3920, 24124, 60325, 26398, 55, 1, 2, 31, 521, 7138, 66221, 331575, 603641, 156233, 89, 1, 2, 35, 691, 11764, 148348, 1183077, 4736345, 6210059, 943174, 144
Offset: 0

Views

Author

Alois P. Heinz, Jul 10 2014

Keywords

Examples

			Square array A(n,k) begins:
   1,    1,     1,      1,       1,       1,       1, ...
   2,    2,     2,      2,       2,       2,       2, ...
   3,    7,    11,     15,      19,      23,      27, ...
   5,   31,    81,    155,     253,     375,     521, ...
   8,  154,   684,   1854,    3920,    7138,   11764, ...
  13,  820,  6257,  24124,   66221,  148348,  290305, ...
  21, 4575, 60325, 331575, 1183077, 3262975, 7585749, ...
		

Crossrefs

Rows n=0-2 give: A000012, A007395, A004767(k-1).
Main diagonal gives A245054.

Programs

  • Maple
    A:= (n, k)-> add(binomial((k-1)*n+i, i)*
        binomial((k-1)*n+i+1, n-i), i=0..n)/((k-1)*n+1):
    seq(seq(A(n, 1+d-n), n=0..d), d=0..12);
  • Mathematica
    A[n_, k_] := Sum[Binomial[(k-1)*n+i, i]*Binomial[(k-1)*n+i+1, n-i], {i, 0, n}]/((k-1)*n+1); Table[A[n, 1+d-n], {d, 0, 12}, {n, 0, d}] // Flatten (* Jean-François Alcover, Feb 18 2017, translated from Maple *)

Formula

A(n,k) = 1/((k-1)*n+1) * Sum_{i=0..n} C((k-1)*n+i,i)*C((k-1)*n+i+1,n-i).
A(n,k) = [x^n] ((1+x)/(1-x-x^2))^((k-1)*n+1) / ((k-1)*n+1).
G.f. for column k satisfies: A_k(x) = (1+x*A_k(x)^(k-1)) * (1+x*A_k(x)^k).

A364339 G.f. satisfies A(x) = (1 + x) * (1 + x*A(x)^6).

Original entry on oeis.org

1, 2, 13, 150, 1978, 28603, 438273, 6992052, 114915180, 1932233883, 33081722359, 574755965137, 10107627041697, 179576579730534, 3218352405778284, 58114340679967608, 1056284029850962674, 19310039426151335622, 354818596435147647654, 6549556302551204621664, 121394125733645986376838
Offset: 0

Views

Author

Seiichi Manyama, Jul 19 2023

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 21; A[] = 0; Do[A[x] = (1+x)(1+x*A[x]^6) + O[x]^terms // Normal, terms]; CoefficientList[A[x], x] (* Stefano Spezia, Mar 24 2025 *)
  • PARI
    a(n) = sum(k=0, n, binomial(6*k+1, k)*binomial(6*k+1, n-k)/(6*k+1));

Formula

a(n) = Sum_{k=0..n} binomial(6*k+1,k) * binomial(6*k+1,n-k) / (6*k+1).

A239108 Number of hybrid 5-ary trees with n internal nodes.

Original entry on oeis.org

1, 2, 19, 253, 3920, 66221, 1183077, 21981764, 420449439, 8223704755, 163727846678, 3307039145618, 67600147666909, 1395822347989531, 29070233296701815, 609950649080323320, 12881240945694949696, 273590092192962485985, 5840400740191969187922
Offset: 0

Views

Author

N. J. A. Sloane, Mar 26 2014

Keywords

Crossrefs

Programs

  • Mathematica
    (1/x InverseSeries[x(1 - x - x^2)^4/(1 + x)^4 + O[x]^20])^(1/4) // CoefficientList[#, x]& (* Jean-François Alcover, Oct 02 2019 *)
  • PARI
    a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1 + x*A^4)*(1 + x*A^5)); polcoeff(A, n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
    
  • PARI
    a(n)=polcoeff( ((1/x)*serreverse( x*(1-x-x^2)^4/(1+x +x*O(x^n))^4))^(1/4), n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
    
  • PARI
    a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*A^j)*x^m*A^(3*m)/m))); polcoeff(A, n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
    
  • PARI
    a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2/A^j)*x^m*A^(4*m)/m))); polcoeff(A, n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
    
  • PARI
    a(n)=polcoeff(((1+x)/(1-x-x^2 +x*O(x^n)))^(4*n+1)/(4*n+1), n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

Formula

From Paul D. Hanna, Mar 30 2014: (Start)
G.f. A(x) satisfies:
(1) A(x) = (1 + x*A(x)^4) * (1 + x*A(x)^5).
(2) A(x) = ( (1/x)*Series_Reversion( x*(1-x-x^2)^4/(1+x)^4 ) )^(1/4).
(3) A(x) = exp( Sum_{n>=1} x^n*A(x)^(3*n)/n * Sum_{k=0..n} C(n,k)^2 * A(x)^k ).
(4) A(x) = exp( Sum_{n>=1} x^n*A(x)^(4*n)/n * Sum_{k=0..n} C(n,k)^2 / A(x)^k ).
(5) A(x) = Sum_{n>=0} Fibonacci(n+2) * x^n * A(x)^(4*n).
(6) A(x) = G(x*A(x)^3) where G(x) = A(x/G(x)^3) is the g.f. of A007863 (number of hybrid binary trees with n internal nodes).
The formal inverse of g.f. A(x) is (sqrt(1-2*x+5*x^2) - (1+x))/(2*x^5).
a(n) = [x^n] ( (1+x)/(1-x-x^2) )^(4*n+1) / (4*n+1).
(End)

A239107 Number of hybrid 4-ary trees with n internal nodes.

Original entry on oeis.org

1, 2, 15, 155, 1854, 24124, 331575, 4736345, 69616637, 1046054129, 15995716263, 248111418112, 3894303176880, 61737213540306, 987116931080661, 15899835212249761, 257758369219909534, 4202381519278498915, 68859442092723799788, 1133401910867109123200
Offset: 0

Views

Author

N. J. A. Sloane, Mar 26 2014

Keywords

Crossrefs

Programs

  • Mathematica
    (1/x InverseSeries[x(1 - x - x^2)^3/(1 + x)^3 + O[x]^21])^(1/3) // CoefficientList[#, x]& (* Jean-François Alcover, Oct 02 2019 *)
  • PARI
    a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=(1 + x*A^3)*(1 + x*A^4)); polcoeff(A, n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
    
  • PARI
    a(n)=polcoeff( ((1/x)*serreverse( x*(1-x-x^2)^3/(1+x +x*O(x^n))^3))^(1/3), n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
    
  • PARI
    a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*A^j)*x^m*A^(2*m)/m))); polcoeff(A, n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014
    
  • PARI
    a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2/A^j)*x^m*A^(3*m)/m))); polcoeff(A, n) \\ Paul D. Hanna, Mar 30 2014
    for(n=0, 20, print1(a(n), ", "))
    
  • PARI
    a(n)=polcoeff(((1+x)/(1-x-x^2 +x*O(x^n)))^(3*n+1)/(3*n+1), n)
    for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Mar 30 2014

Formula

From Paul D. Hanna, Mar 30 2014: (Start)
G.f. A(x) satisfies:
(1) A(x) = (1 + x*A(x)^3) * (1 + x*A(x)^4).
(2) A(x) = ( (1/x)*Series_Reversion( x*(1-x-x^2)^3/(1+x)^3 ) )^(1/3).
(3) A(x) = exp( Sum_{n>=1} x^n*A(x)^(2*n)/n * Sum_{k=0..n} C(n,k)^2 * A(x)^k ).
(4) A(x) = exp( Sum_{n>=1} x^n*A(x)^(3*n)/n * Sum_{k=0..n} C(n,k)^2 / A(x)^k ).
(5) A(x) = Sum_{n>=0} Fibonacci(n+2) * x^n * A(x)^(3*n).
(6) A(x) = G(x*A(x)^2) where G(x) = A(x/G(x)^2) is the g.f. of A007863 (number of hybrid binary trees with n internal nodes).
The formal inverse of g.f. A(x) is (sqrt(1-2*x+5*x^2) - (1+x))/(2*x^4).
a(n) = [x^n] ( (1+x)/(1-x-x^2) )^(3*n+1) / (3*n+1).
(End)
a(n) = 1/(3*n+1) * Sum_{i=0..n} C(3*n+i,i)*C(3*n+i+1,n-i). - Alois P. Heinz, Jul 10 2014

A364333 G.f. satisfies A(x) = (1 + x*A(x)^2) * (1 + x*A(x)^6).

Original entry on oeis.org

1, 2, 17, 216, 3224, 52640, 910452, 16392140, 303996224, 5767278431, 111401778266, 2183535060362, 43319505976084, 868220464851417, 17552981176788200, 357544690982030744, 7330803752675100908, 151172599088871911072, 3133367418601958989295, 65242183918761533467216
Offset: 0

Views

Author

Seiichi Manyama, Jul 18 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(2*n+4*k+1, k)*binomial(2*n+4*k+1, n-k)/(2*n+4*k+1));

Formula

a(n) = Sum_{k=0..n} binomial(2*n+4*k+1,k) * binomial(2*n+4*k+1,n-k) / (2*n+4*k+1).

A364340 G.f. satisfies A(x) = (1 + x*A(x)) * (1 + x*A(x)^6).

Original entry on oeis.org

1, 2, 15, 179, 2502, 38262, 619991, 10459410, 181771289, 3231782239, 58505593456, 1074766446526, 19984671314164, 375414901633692, 7113886504446443, 135820770971898805, 2610186429457347486, 50452256583633573513, 980187901557594671335, 19130197594133100828170, 374894511736219913097375
Offset: 0

Views

Author

Seiichi Manyama, Jul 19 2023

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n+5*k+1, k)*binomial(n+5*k+1, n-k)/(n+5*k+1));

Formula

a(n) = Sum_{k=0..n} binomial(n+5*k+1,k) * binomial(n+5*k+1,n-k) / (n+5*k+1).
Showing 1-6 of 6 results.