A220886 -id:A220886 - cp's OEIS frontend

A060690 a(n) = binomial(2^n + n - 1, n).

1, 2, 10, 120, 3876, 376992, 119877472, 131254487936, 509850594887712, 7145544812472168960, 364974894538906616240640, 68409601066028072105113098240, 47312269462735023248040155132636160, 121317088003402776955124829814219234385920

Offset: 0

Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 19 2001

nonn

Also the number of n X n (0,1) matrices modulo rows permutation (by symmetry this is the same as the number of (0,1) matrices modulo columns permutation), i.e., the number of equivalence classes where two matrices A and B are equivalent if one of them is the result of permuting the rows of the other. The total number of (0,1) matrices is in sequence A002416.

Row sums of A220886. - Geoffrey Critzer, Nov 20 2014

Harry J. Smith, Table of n, a(n) for n = 0..59

Sequences of the form binomial(2^n +p*n +q, n): A136556 (0,-1), A014070 (0,0), A136505 (0,1), A136506 (0,2), this sequence (1,-1), A132683 (1,0), A132684 (1,1), A132685 (2,0), A132686 (2,1), A132687 (3,-1), A132688 (3,0), A132689 (3,1).
Cf. A002416, A060336, A088309, A163767.
Cf. A136555, A220886.
Main diagonal of A092056.
Central terms of A137153.

[Binomial(2^n +n-1, n): n in [0..20]]; // G. C. Greubel, Mar 14 2021

with(combinat): for n from 0 to 20 do printf(`%d,`,binomial(2^n+n-1, n)) od:

Table[Binomial[2^n+n-1,n],{n,0,20}] (* Harvey P. Dale, Apr 19 2012 *)

PARI
```
a(n)=binomial(2^n+n-1,n)
```

{a(n)=polcoeff(sum(k=0,n,(-log(1-2^k*x +x*O(x^n)))^k/k!),n)} \\ Paul D. Hanna, Dec 29 2007

a(n) = sum(k=0, n, stirling(n,k,1)*(2^n+n-1)^k/n!); \\ Paul D. Hanna, Nov 20 2014

from math import comb
def A060690(n): return comb((1<Chai Wah Wu, Jul 05 2024

[binomial(2^n +n-1, n) for n in (0..20)] # G. C. Greubel, Mar 14 2021

a(n) = [x^n] 1/(1-x)^(2^n).
a(n) = (1/n!)*Sum_{k=0..n} ( (-1)^(n-k)*Stirling1(n, k)*2^(k*n) ). - Vladeta Jovovic, May 28 2004
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(2^n+n,k) - Vladeta Jovovic, Jan 21 2008
a(n) = Sum_{k=0..n} Stirling1(n,k)*(2^n+n-1)^k/n!. - Vladeta Jovovic, Jan 21 2008
G.f.: A(x) = Sum_{n>=0} [ -log(1 - 2^n*x)]^n / n!. More generally, Sum_{n>=0} [ -log(1 - q^n*x)]^n/n! = Sum_{n>=0} C(q^n+n-1,n)*x^n ; also Sum_{n>=0} log(1 + q^n*x)^n/n! = Sum_{n>=0} C(q^n,n)*x^n. - Paul D. Hanna, Dec 29 2007
a(n) ~ 2^(n^2) / n!. - Vaclav Kotesovec, Jul 02 2016
a(n) = A163767(2^n). - Alois P. Heinz, Jun 12 2024

More terms from James Sellers, Apr 20 2001

Edited by N. J. A. Sloane, Mar 17 2008

A360693 Number T(n,k) of sets of n words of length n over binary alphabet where the first letter occurs k times; triangle T(n,k), n>=0, n-signum(n)<=k<=n*(n-1)+signum(n), read by rows.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 10, 15, 15, 10, 3, 4, 37, 108, 228, 336, 394, 336, 228, 108, 37, 4, 5, 101, 600, 2150, 5645, 11680, 19752, 27820, 32935, 32935, 27820, 19752, 11680, 5645, 2150, 600, 101, 5, 6, 226, 2490, 14745, 61770, 200529, 535674, 1211485, 2368200

Offset: 0

Alois P. Heinz, Feb 16 2023

T(n,k) is defined for all n >= 0 and k >= 0. The triangle contains only the positive elements.

			T(2,3) = 2: {aa,ab}, {aa,ba}.
T(3,3) = 10: {aab,abb,bbb}, {aab,bab,bbb}, {aab,bba,bbb}, {aba,abb,bbb}, {aba,bab,bbb}, {aba,bba,bbb}, {abb,baa,bbb}, {abb,bab,bba}, {baa,bab,bbb}, {baa,bba,bbb}.
T(4,3) = 4: {abbb,babb,bbab,bbbb}, {abbb,babb,bbba,bbbb}, {abbb,bbab,bbba,bbbb}, {babb,bbab,bbba,bbbb}.
Triangle T(n,k) begins:
  1;
  1, 1;
  .  2, 2,  2;
  .  .  3, 10, 15,  15,  10,    3;
  .  .  .   4, 37, 108, 228,  336,  394,   336,   228,   108,    37,     4;
  .  .  .   .   5, 101, 600, 2150, 5645, 11680, 19752, 27820, 32935, 32935, ...;
  ...

Alois P. Heinz, Rows n = 0..33, flattened

Row sums give A014070.
Column sums give A360695.
Main diagonal T(n,n) gives A154323(n-1) for n>=1.
T(n,n-1) gives A000027(n) for n>=1.
T(2n,2n^2) gives A360702.
Cf. A000290, A057427, A220886 (similar triangle for multisets).

g:= proc(n, i, j) option remember; expand(`if`(j=0, 1, `if`(i<0, 0, add(
      g(n, i-1, j-k)*x^(i*k)*binomial(binomial(n, i), k), k=0..j))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=n-signum(n)..n*(n-1)+signum(n)))(g(n$3)):
seq(T(n), n=0..6);

g[n_, i_, j_] := g[n, i, j] = Expand[If[j == 0, 1, If[i < 0, 0, Sum[g[n, i - 1, j - k]*x^(i*k)*Binomial[Binomial[n, i], k], {k, 0, j}]]]];
T[n_] := Table[Coefficient[#, x, i], {i, n - Sign[n], n(n - 1) + Sign[n]}]&[g[n, n, n]];
Table[T[n], {n, 0, 6}] // Flatten (* Jean-François Alcover, May 26 2023, after Alois P. Heinz *)

T(n,k) = T(n,n^2-k).

A360660 Number of inequivalent n X n {0,1} matrices modulo permutation of the rows, with exactly n 1's.

Original entry on oeis.org

1, 1, 4, 20, 133, 1027, 9259, 94033, 1062814, 13176444, 177427145, 2573224238, 39924120823, 658921572675, 11513293227271, 212109149134617, 4105637511110979, 83238756058333277, 1762856698153603049, 38905470655863251479, 892840913430059075405

Offset: 0

Alois P. Heinz, Feb 15 2023

nonn

Also the number of multisets of n words of length n over binary alphabet where the first letter occurs n times. E.g., a(2) = 4: {aa,bb}, {ab,ab}, {ab,ba}, {ba,ba}.

			a(3) = 20: [111/000/000], [110/100/000], [110/010/000], [110/001/000], [101/100/000], [101/010/000], [101/001/000], [100/100/100], [100/100/010], [100/100/001], [100/011/000], [100/010/010], [100/010/001], [100/001/001], [011/010/000], [011/001/000], [010/010/010], [010/010/001], [010/001/001], [001/001/001].

Alois P. Heinz, Table of n, a(n) for n = 0..200

Main diagonal of A220886.

Cf. A091058, A091059, A154323, A246107, A256070, A360664, A383073.

g:= proc(n, i, j) option remember; expand(`if`(j=0, 1, `if`(i<0, 0, add(
      g(n, i-1, j-k)*x^(i*k)*binomial(binomial(n, i)+k-1, k), k=0..j))))
    end:
a:= n-> coeff(g(n$3), x, n):
seq(a(n), n=0..20);

g[n_, i_, j_] := g[n, i, j] = Expand[If[j == 0, 1, If[i < 0, 0, Sum[g[n, i - 1, j - k]*x^(i*k)*Binomial[Binomial[n, i] + k - 1, k], {k, 0, j}]]]];
a[n_] := SeriesCoefficient[g[n, n, n], {x, 0, n}];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 28 2023, after Alois P. Heinz *)
Table[SeriesCoefficient[Product[1/(1 - x^k)^Binomial[n, k], {k, 1, n}], {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Apr 15 2025 *)

a(n) = A220886(n,n).

a(n) = [x^n] Product_{k=1..n} 1/(1 - x^k)^binomial(n,k). - Vaclav Kotesovec, Apr 15 2025

cp's OEIS Frontend

A060690 a(n) = binomial(2^n + n - 1, n).

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A360693 Number T(n,k) of sets of n words of length n over binary alphabet where the first letter occurs k times; triangle T(n,k), n>=0, n-signum(n)<=k<=n*(n-1)+signum(n), read by rows.

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A360660 Number of inequivalent n X n {0,1} matrices modulo permutation of the rows, with exactly n 1's.

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Maple

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