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
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Aug 10 2009
STATUS
approved