A127943 a(n) = 2^binomial(n+1,2)/A046161(n).

1, 1, 1, 4, 8, 128, 2048, 131072, 2097152, 536870912, 137438953472, 140737488355328, 72057594037927936, 295147905179352825856, 1208925819614629174706176, 19807040628566084398385987584, 40564819207303340847894502572032, 2658455991569831745807614120560689152

Offset

0,4

Comments

Apparently, also numerator of 2^(n*(n-1)/2)/n!. - N. J. A. Sloane, Dec 31 2010

Links

G. C. Greubel, Table of n, a(n) for n = 0..82

Formula

a(n) = 2^binomial(n+1,2)/denominator(binomial(2*n,n)/4^n).

a(n) = 2^A127944(n).

Maple

a:=n->2^(binomial(n+1, 2))/denom(binomial(2*n, n)/4^n); seq(a(n), n=0..17); # Muniru A Asiru, Dec 10 2018

Mathematica

Table[2^Binomial[n+1, 2]/Denominator[Binomial[2*n, n]/4^n], {n, 0, 25}] (* G. C. Greubel, May 01 2018 *)

Prog

(PARI) for(n=0, 25, print1(2^(binomial(n+1, 2))/denominator(binomial(2*n, n)/4^n), ", ")) \\ G. C. Greubel, May 01 2018
(PARI) a(n) = numerator(2^(n*(n-1)/2)/n!); \\ Altug Alkan, May 02 2018
(Magma) [2^(Binomial(n+1, 2))/Denominator(Binomial(2*n, n)/4^n): n in [0..25]]; // G. C. Greubel, May 01 2018
(Sage) [2^binomial(n+1, 2)/denominator(binomial(2*n, n)/4^n) for n in range(30)] # G. C. Greubel, Dec 09 2018
(GAP) List([0..30], n-> 2^(Binomial(n+1, 2))/DenominatorRat(Binomial(2*n, n)/4^n)); # G. C. Greubel, Dec 09 2018

Crossrefs

Sequence in context: A013049 A193155 A060239 * A012498 A180745 A330860

Adjacent sequences: A127940 A127941 A127942 * A127944 A127945 A127946

Keyword

easy,nonn

Author

Paul Barry, Feb 08 2007

Status

approved