A134057 a(n) = binomial(2^n-1,2).
0, 0, 3, 21, 105, 465, 1953, 8001, 32385, 130305, 522753, 2094081, 8382465, 33542145, 134193153, 536821761, 2147385345, 8589737985, 34359345153, 137438167041, 549754241025, 2199020109825, 8796086730753, 35184359505921
Offset: 0
Examples
a(2) = 3 because for P(A) = {{},{1},{2},{1,2}} we have for case 0 {{1},{2}} and we have for case 2 {{1},{1,2}}, {{2},{1,2}}. There are 0 {x,y} of P(A) in this example that fall under case 1.
Links
- Robert Israel, Table of n, a(n) for n = 0..1600
- Ross La Haye, Binary Relations on the Power Set of an n-Element Set, Journal of Integer Sequences, Vol. 12 (2009), Article 09.2.6. - _Ross La Haye_, Feb 22 2009
- Tilman Piesk, 3-bit Walsh permutation; conjugacy class 2+2 (Wikiversity)
- Index entries for linear recurrences with constant coefficients, signature (7,-14,8).
Programs
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Magma
[Binomial(2^n-1, 2): n in [0..30]]; // Vincenzo Librandi, Nov 30 2015
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Maple
seq((2^n-1)*(2^(n-1)-1), n=0..100); # Robert Israel, Nov 30 2015
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Mathematica
Table[Binomial[2^n - 1, 2], {n, 0, 30}] (* Vincenzo Librandi, Nov 30 2015 *)
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PARI
a(n) = binomial(2^n-1, 2); \\ Michel Marcus, Nov 30 2015
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Python
print([(2**n-1)*(2**(n-1)-1) for n in range(23)]) # Alex Ratushnyak, Jun 02 2013
Formula
a(n) = (1/2)*(4^n - 3*2^n + 2) = 3*(Stirling2(n+1,4) + Stirling2(n+1,3)).
a(n) = 3 *A006095(n).
a(n) = (2^n-1)*(2^(n-1)-1). - Alex Ratushnyak, Jun 02 2013
a(n) = Stirling2(2^n - 1,2^n - 2).
G.f.: 3*x^2/(1-x)/(1-2*x)/(1-4*x). - Colin Barker, Feb 22 2012
a(n) = A000217(2^n-2). - Michel Marcus, Nov 30 2015
a(n) = 7*a(n-1) - 14*a(n-2) + 8*a(n-3). - Wesley Ivan Hurt, May 17 2021
E.g.f.: exp(x)*(exp(x) - 1)^2*(exp(x) + 2)/2. - Stefano Spezia, Apr 06 2022
Comments