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
Examples
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.
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.
Programs
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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
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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
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= 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
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