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

Author

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

Keywords

• nonn

Comments

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

Links

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

Crossrefs

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.

Programs

Magma

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

Maple

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

Mathematica

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)

PARI

{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

PARI

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

Python

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

Sage

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

Formula

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

Extensions

More terms from James Sellers, Apr 20 2001

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