A366012 -id:A366012 - cp's OEIS frontend

A107111 Number array whose rows are the series reversions of x(1-x)/(1+x)^k, read by antidiagonals.

1, 1, 1, 1, 2, 2, 1, 3, 6, 5, 1, 4, 13, 22, 14, 1, 5, 23, 67, 90, 42, 1, 6, 36, 156, 381, 394, 132, 1, 7, 52, 305, 1162, 2307, 1806, 429, 1, 8, 71, 530, 2833, 9192, 14589, 8558, 1430, 1, 9, 93, 847, 5919, 27916, 75819, 95235, 41586, 4862, 1, 10, 118, 1272, 11070, 70098, 286632, 644908, 636925, 206098, 16796

Offset: 0

Paul Barry, May 12 2005

First row is the Catalan numbers A000108, second row is the large Schroeder numbers A006318, third row is A062992, fourth row is A007297. As a number triangle, this is T(n,k)=if(k<=n,sum{j=0..k, binomial((n-k)(k+1),k-j)*binomial(k+j,j)}/(k+1),0) with row sums A107112 and diagonal sums A107113.

			Array begins
1,1,2,5,14,42,132,...
1,2,6,22,90,394,1806,...
1,3,13,67,381,2307,14589,...
1,4,23,156,1162,9192,75819,...

Cf. A366012.

A107111 := proc(n,k)
    add(binomial(n*(k+1),k-j)*binomial(k+j,j),j=0..k);
    %/(k+1) ;
end proc: # R. J. Mathar, Aug 02 2016

T[n_, k_] := Sum[Binomial[n (k + 1), k - j] Binomial[k + j, j], {j, 0, k}]/(k + 1);
Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 21 2020 *)

T(n, k)=sum{j=0..k, binomial(n(k+1), k-j)*binomial(k+j, j)}/(k+1)

A366038 a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(n(n+1),n-k) n^k.

Original entry on oeis.org

1, 2, 25, 658, 27193, 1548526, 112916830, 10062563610, 1061196371665, 129369938790070, 17909387604206371, 2776290021986848588, 476539253976442601735, 89736215305419802692184, 18395742890606906720656524, 4078527943680251523126851306, 972490249766494185823234587681

Offset: 0

Ilya Gutkovskiy, Sep 26 2023

nonn

Cf. A006318, A064063, A135861, A291699, A292798, A365816, A366012, A366016, A366037.

A366038 := proc(n)
    add(binomial(n+k,k)*binomial(n*(n+1),n-k)*n^k,k=0..n) ;
    %/(n+1) ;
end proc:
seq(A366038(n),n=0..80) ; # R. J. Mathar, Oct 24 2024

Unprotect[Power]; 0^0 = 1; Table[1/(n + 1) Sum[Binomial[n + k, k] Binomial[n (n + 1) , n - k] n^k, {k, 0, n}], {n, 0, 16}]
Table[Binomial[n (n + 1), n] Hypergeometric2F1[-n, n + 1, n^2 + 1, -n]/(n + 1), {n, 0, 16}]
Table[SeriesCoefficient[(1/x) InverseSeries[Series[x (1 - n x)/(1 + x)^n, {x, 0, n + 1}], x], {x, 0, n}], {n, 0, 16}]

a(n) = [x^n] (1/x) * Series_Reversion( x * (1 - n * x) / (1 + x)^n ).

a(n) ~ phi^(3*n + 3/2) * exp(n/phi^2 + 1/(2*phi)) * n^(n - 3/2) / (5^(1/4) * sqrt(2*Pi)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Sep 27 2023

cp's OEIS Frontend

A107111 Number array whose rows are the series reversions of x(1-x)/(1+x)^k, read by antidiagonals.

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Formula

A366038 a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(n(n+1),n-k) n^k.

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cp's OEIS Frontend

A107111 Number array whose rows are the series reversions of x(1-x)/(1+x)^k, read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A366038 a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(n*(n+1),n-k) * n^k.

Views

Author

Keywords

Crossrefs

Programs

Maple

Mathematica

Formula

A366038 a(n) = (1/(n+1)) * Sum_{k=0..n} binomial(n+k,k) * binomial(n(n+1),n-k) n^k.