A136465 Row 0 of square array A136462: a(n) = C(2^(n-1), n) for n>=0.

1, 1, 1, 4, 70, 4368, 906192, 621216192, 1429702652400, 11288510714272000, 312268282598377321216, 30813235422145714150738944, 11005261717918037175659349191168, 14391972654784168932973746746691440640, 69538271351155829150354851003285125277716480, 1250303357941919088313448625534941836891635347865600

Offset: 0

Author

Paul D. Hanna, Dec 31 2007

Keywords

• nonn
• easy

Comments

a(n) is found in row n, column 0, of triangle A136467 for n>=0.

For n > 0, number of increasing integer sequences 1 <= a_1 < ... < a_n <= 2^(n-1). - Charles R Greathouse IV, Aug 08 2010

The (n-1)-dimensional hypercube has 2^(n-1) corners. There are binomial(2^(n-1),n) ways of selecting a set of n corners. So a(n) is the number of simplices (hyper-tetrahedra) with vertices defined by a corner subset of a (n-1)-dimensional hypercube. (This count includes degenerate polytopes with zero volume.) - R. J. Mathar, Jan 16 2016

Examples

			From _Paul D. Hanna_, Sep 26 2010: (Start)
G.f.: A(x) = 1 + x + x^2 + 4*x^3 + 70*x^4 + 4368*x^5 +...
A(x) = 1 + log(1+2*x)/2 + log(1 + 2^2*x)^2/(2!*2^2) + log(1 + 2^3*x)^3/(3!*2^3) + log(1 + 2^4*x)^4/(4!*2^4) +... (End)
		

Links

• J. Brandts, S. Dijkhuis, V. de Haan, M. Krizek, There are only two nonobtuse binary triangulations of the unit n-cube, Comp. Geom. 46 (2013) 286-297, Table 1.

Crossrefs

Cf. A136462; other rows: A014070, A136466, A101346; A136467.

Programs

Mathematica

Table[Binomial[2^(n-1),n], {n,0,15}] (* Vaclav Kotesovec, Jul 02 2016 *)

PARI

{a(n)=binomial(2^(n-1),n)}
for(n=0,20,print1(a(n),", "))

PARI

/* a(n) = Coefficient of x^k in series: */
{a(n)=polcoeff(sum(i=0,n,(1/2)^i*log(1+2^i*x +x*O(x^n))^i/i!),n)}
for(n=0,20,print1(a(n),", "))

PARI

{a(n)=polcoeff(sum(m=0,n,log(1+2^m*x+x*O(x^n))^m/(m!*2^m)),n)}
for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Sep 26 2010

Formula

a(n) = [x^n] Sum_{i>=0} (1/2)^i * log(1 + 2^i*x)^i/i!.

O.g.f.: Sum_{n>=0} log(1 + 2^n*x)^n / (n!*2^n). - Paul D. Hanna, Sep 26 2010

a(n) ~ 2^(n*(n-1)) / n!. - Vaclav Kotesovec, Jul 02 2016