A157163 Product_{n>=1} (1 + 2*a(n)*x^n) = Sum_{k>=0} binomial(2*k, k)*x^k = 1/sqrt(1 - 4*x), with the central binomial numbers A000984(n).

1, 3, 4, 27, 48, 156, 576, 2955, 7168, 27792, 95232, 352188, 1290240, 5105856, 17743872, 77010795, 252641280, 1000224768, 3616800768, 14484040464, 52102692864, 208963943424, 764877471744, 3025006038012, 11258183024640, 44968060784640, 166308918329344

Offset

1,2

Comments

In the original problem 2*a(n) = [2, 6, 8, 54, 96, 312, 1152, 5910, 14336, 55584, 190464, 704376, ...] appears.

Links

Robert Israel, Table of n, a(n) for n = 1..1665

Giedrius Alkauskas, One curious proof of Fermat's little theorem, arXiv:0801.0805 [math.NT], 2008.

Giedrius Alkauskas, A curious proof of Fermat's little theorem, Amer. Math. Monthly 116(4) (2009), 362-364.

Giedrius Alkauskas, Algebraic functions with Fermat property, eigenvalues of transfer operator and Riemann zeros, and other open problems, arXiv:1609.09842 [math.NT], 2016.

Art of Problem Solving, Product formula for a generating function

H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.

H. Gingold and A. Knopfmacher, Analytic properties of power product expansions, Canad. J. Math. 47 (1995), 1219-1239.

W. Lang, Recurrences for the general problem.

Formula

Recurrence I: With FP(n,m) the set of partitions of n with m distinct parts (which could be called fermionic partitions fp):

a(n) = binomial(2*n, n)/2 - Sum_{m=2..maxm(n)} 2^(m-1)*(Sum_{fp from FP(n,m)} (Product_{j=1..m} a(k[j]))), with maxm(n) = A003056(n) and the distinct parts k[j], j = 1..m, of the partition fp of n, n >= 3. Inputs a(1) = 1, a(2) = 3. See the array A008289(n,m) for the cardinality of the set FP(n,m).

Recurrence II: With P(n,m) the set of all partitions of n with m parts, and the multinomial numbers M0 (given for every partition under A048996):

a(n) = (1/2)*Sum_{d|n, 1<d<n} (d/n)*(-2*a(d))^(n/d) + (1/2)*Sum_{m=1..n-1} (-1)^(m-1)*(1/m)*(Sum{p=(1^e(1),...,n^e(n)) from P(n,m)} M0(p)*cbi(1)^e(1)*...*cbi(n)^e(n)), n >= 2; a(1) = 1, with cbi(n) = binomial(2*n, n) = A000984(n). The exponents e(j) >= 0 satisfy Sum_{j=1..n} j*e(j) =n and Sum_{j=1..n} e(j) = m. The M0 numbers are m!/(Product_{j=1..n} (e(j))!).

Recurrence II (rewritten, due to email from V. Jovovic, Mar 10 2009):

a(n) = ((-2*a(1))^n + Sum_{d|n, 1<d<n} (d*(-2*a(d))^(n/d) + B(n))/(2*n), with the o.g.f. x*(log(1/sqrt(1 - 4*x)))' = 2*x/(1 - 4*x) for the sequence {B(n)}; hence B(n) = (1/2)*4^n.

Example

Recurrence I: a(4) = binomial(8, 4)/2 - 2*a(1)*a(3) = 35 - 8 = 27.
Recurrence II: a(4) = (1/2)*(1/2)*(-2*a(2))^2 + (1/2)*(1*cbi(4) - (1/2)*(2*cbi(1)*cbi(3) + 1*cbi(2)^2) + (1/3)*3*cbi(1)^2*cbi(2)) = 27.
Recurrence II (rewritten): a(4)= (1/8)*((-2)^4 + 2*(-2*a(2))^2 + (1/2)*4^4) = 27.

Maple

N:= 100: # for a(1)..a(N)
S:= convert(series(-ln(1-4*x)/2, x, N+1), polynom):
for n from 1 to N do
  a[n]:= coeff(S, x, n)/2;
  S:= S - add((-1)^(k-1)*(2*a[n])^k*x^(k*n)/k, k=1..N/n)
od:
seq(a[n], n=1..N); # Robert Israel, Jan 03 2019

Prog

(PARI) a(n) = if (n==1, 1, (1/(2*n))*((-2*a(1))^n + sumdiv(n, d, if ((d!=1) && (d!=n), d*(-2*a(d))^(n/d), 0)) + 4^n/2)); \\ after 2nd Recurrence II; Michel Marcus, Jul 06 2015

Crossrefs

Cf. A147542 (Fibonacci), A157161 (Catalan).

Sequence in context: A032832 A041021 A041022 * A042225 A094084 A362887

Adjacent sequences: A157160 A157161 A157162 * A157164 A157165 A157166

Keyword

nonn,easy

Author

Wolfdieter Lang, Aug 10 2009

Status

approved