G.f.: Product_{k>0} 1/(1-x^k) = Sum_{k>= 0} x^k Product_{i = 1..k} 1/(1-x^i) = 1 + Sum_{k>0} x^(k^2)/(Product_{i = 1..k} (1-x^i))^2.
G.f.: 1 + Sum_{n>=1} x^n/(Product_{k>=n} 1-x^k). -
Joerg Arndt, Jan 29 2011
a(n) - a(n-1) - a(n-2) + a(n-5) + a(n-7) - a(n-12) - a(n-15) + ... = 0, where the sum is over n-k and k is a generalized pentagonal number (
A001318) <= n and the sign of the k-th term is (-1)^([(k+1)/2]). See
A001318 for a good way to remember this!
a(n) = (1/n) * Sum_{k=0..n-1} sigma(n-k)*a(k), where sigma(k) is the sum of divisors of k (
A000203).
a(n) ~ 1/(4*n*sqrt(3)) * e^(Pi * sqrt(2n/3)) as n -> infinity (Hardy and Ramanujan). See
A050811.
a(n) = a(0)*b(n) + a(1)*b(n-2) + a(2)*b(n-4) + ... where b =
A000009.
It appears that the above approximation from Hardy and Ramanujan can be refined as
a(n) ~ 1/(4*n*sqrt(3)) * e^(Pi * sqrt(2n/3 + c0 + c1/n^(1/2) + c2/n + c3/n^(3/2) + c4/n^2 + ...)), where the coefficients c0 through c4 are approximately
c0 = -0.230420145062453320665537
c1 = -0.0178416569128570889793
c2 = 0.0051329911273
c3 = -0.0011129404
c4 = 0.0009573,
as n -> infinity. (End)
c0 = -0.230420145062453320665536704197233... = -1/36 - 2/Pi^2
c1 = -0.017841656912857088979502135349949... = 1/(6*sqrt(6)*Pi) - sqrt(3/2)/Pi^3
c2 = 0.005132991127342167594576391633559... = 1/(2*Pi^4)
c3 = -0.001112940489559760908236602843497... = 3*sqrt(3/2)/(4*Pi^5) - 5/(16*sqrt(6)*Pi^3)
c4 = 0.000957343284806972958968694349196... = 1/(576*Pi^2) - 1/(24*Pi^4) + 93/(80*Pi^6)
a(n) ~ exp(Pi*sqrt(2*n/3))/(4*sqrt(3)*n) * (1 - (sqrt(3/2)/Pi + Pi/(24*sqrt(6)))/sqrt(n) + (1/16 + Pi^2/6912)/n).
a(n) ~ exp(Pi*sqrt(2*n/3) - (sqrt(3/2)/Pi + Pi/(24*sqrt(6)))/sqrt(n) + (1/24 - 3/(4*Pi^2))/n) / (4*sqrt(3)*n).
(End)
a(n) < exp( (2/3)^(1/2) Pi sqrt(n) ) (Ayoub, p. 197).
a(n) = Sum_{i=0..n-1} P(i, n-i), where P(x, y) is the number of partitions of x into at most y parts and P(0, y)=1. -
Jon Perry, Jun 16 2003
G.f.: Product_{i>=1} Product_{j>=0} (1+x^((2i-1)*2^j))^(j+1). -
Jon Perry, Jun 06 2004
a(n) = determinant of the n X n Toeplitz matrix:
1 -1
1 1 -1
0 1 1 -1
0 0 1 1 -1
-1 0 0 1 1 -1
. . .
d_n d_(n-1) d_(n-2)...1
where d_q = (-1)^(m+1) if q = m(3m-1)/2 = p_m, the m-th generalized pentagonal number (
A001318), otherwise d_q = 0. Note that the 1's run along the diagonal and the -1's are on the superdiagonal. The (n-1) row (not written) would end with ... 1 -1. (End)
Empirical: let F*(x) = Sum_{n=0..infinity} p(n)*exp(-Pi*x*(n+1)), then F*(2/5) = 1/sqrt(5) to a precision of 13 digits.
F*(4/5) = 1/2+3/2/sqrt(5)-sqrt(1/2*(1+3/sqrt(5))) to a precision of 28 digits. These are the only values found for a/b when a/b is from F60, Farey fractions up to 60. The number for F*(4/5) is one of the real roots of 25*x^4 - 50*x^3 - 10*x^2 - 10*x + 1. Note here the exponent (n+1) compared to the standard notation with n starting at 0. -
Simon Plouffe, Feb 23 2011
The constant (2^(7/8)*GAMMA(3/4))/(exp(Pi/6)*Pi^(1/4)) = 1.0000034873... when expanded in base exp(4*Pi) will give the first 52 terms of a(n), n>0, the precision needed is 300 decimal digits. -
Simon Plouffe, Mar 02 2011
G.f.: A(x)=1+x/(G(0)-x); G(k) = 1 + x - x^(k+1) - x*(1-x^(k+1))/G(k+1); (continued fraction Euler's kind, 1-step ). -
Sergei N. Gladkovskii, Jan 25 2012
G.f.: 1 + x*(1 - G(0))/(1-x) where G(k) = 1 - 1/(1-x^(k+1))/(1-x/(x-1/G(k+1) )); (continued fraction). -
Sergei N. Gladkovskii, Jan 22 2013
G.f.: Q(0) where Q(k) = 1 + x^(4*k+1)/( (x^(2*k+1)-1)^2 - x^(4*k+3)*(x^(2*k+1)-1)^2/( x^(4*k+3) + (x^(2*k+2)-1)^2/Q(k+1) )); (continued fraction). -
Sergei N. Gladkovskii, Feb 16 2013
a(n-1) = Sum_{parts k in all partitions of n} mu(k), where mu(k) is the arithmetical Möbius function (see
A008683).
Let P(2,n) denote the set of partitions of n into parts k >= 2. Then a(n-2) = -Sum_{parts k in all partitions in P(2,n)} mu(k).
n*( a(n) - a(n-1) ) = Sum_{parts k in all partitions in P(2,n)} k (see
A138880).
Let P(3,n) denote the set of partitions of n into parts k >= 3. Then
a(n-3) = (1/2)*Sum_{parts k in all partitions in P(3,n)} phi(k), where phi(k) is the Euler totient function (see
A000010). Using this result and Mertens's theorem on the average order of the phi function, we can find an approximate 3-term recurrence for the partition function: a(n) ~ a(n-1) + a(n-2) + (Pi^2/(3*n) - 1)*a(n-3). For example, substituting the values a(47) = 124754, a(48) = 147273 and a(49) = 173525 into the recurrence gives the approximation a(50) ~ 204252.48... compared with the true value a(50) = 204226. (End)
A production matrix for the sequence with offset 1 is M, an infinite n x n matrix of the following form:
a, 1, 0, 0, 0, 0, ...
b, 0, 1, 0, 0, 0, ...
c, 0, 0, 1, 0, 0, ...
d, 0, 0, 0, 1, 0, ...
.
.
... such that (a, b, c, d, ...) is the signed version of
A080995 with offset 1: (1,1,0,0,-1,0,-1,...)
and a(n) is the upper left term of M^n.
This operation is equivalent to the g.f. (1 + x + 2x^2 + 3x^3 + 5x^4 + ...) = 1/(1 - x - x^2 + x^5 + x^7 - x^12 - x^15 + x^22 + ...). (End)
a(n) = Sum_{k=-inf..+inf} (-1)^k a(n-k(3k-1)/2) with a(0)=1 and a(negative)=0. The sum can be restricted to the (finite) range from k = (1-sqrt(1-24n))/6 to (1+sqrt(1-24n))/6, since all terms outside this range are zero. -
Jos Koot, Jun 01 2016
G.f.: (conjecture) (r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) is
A000009: (1, 1, 1, 2, 2, 3, 4, ...). -
Gary W. Adamson, Sep 18 2016;
Doron Zeilberger observed today that "This follows immediately from Euler's formula 1/(1-z) = (1+z)*(1+z^2)*(1+z^4)*(1+z^8)*..."
Gary W. Adamson, Sep 20 2016
a(n) ~ 2*Pi * BesselI(3/2, sqrt(24*n-1)*Pi/6) / (24*n-1)^(3/4). -
Vaclav Kotesovec, Jan 11 2017
G.f.: Product_{k>=1} (1 + x^k)/(1 - x^(2*k)). -
Ilya Gutkovskiy, Jan 23 2018
a(n) = p(1, n) where p(k, n) = p(k+1, n) + p(k, n-k) if k < n, 1 if k = n, and 0 if k > n. p(k, n) is the number of partitions of n into parts >= k. -
Lorraine Lee, Jan 28 2020
Sum_{n>=0} a(n)/exp(Pi*n) = 2^(3/8)*Gamma(3/4)/(Pi^(1/4)*exp(Pi/24)).
Sum_{n>=0} a(n)/exp(2*Pi*n) = 2^(1/2)*Gamma(3/4)/(Pi^(1/4)*exp(Pi/12)).
[These are the reciprocals of phi(exp(-Pi)) (
A259148) and phi(exp(-2*Pi)) (
A259149), where phi(q) is the Euler modular function. See B. C. Berndt (RLN, Vol. V, p. 326), and formulas (13) and (14) in I. Mező, 2013. -
Peter Luschny, Mar 13 2021]
a(n) =
A008284(2*n,n) is also the number of partitions of 2n into n parts. -
Ryan Brooks, Jun 11 2022
a(n) ~ exp(Pi*sqrt(2*n/3)) / (4*n*sqrt(3)) * (1 + Sum_{r>=1} w(r)/n^(r/2)), where w(r) = 1/(-4*sqrt(6))^r * Sum_{k=0..(r+1)/2} binomial(r+1,k) * (r+1-k) / (r+1-2*k)! * (Pi/6)^(r-2*k) [Cormac O'Sullivan, 2023, pp. 2-3]. -
Vaclav Kotesovec, Mar 15 2023
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