A039824 Number of different coefficient values in expansion of Product (1+q^1+q^3...+q^(2i-1)), i=1 to n.

1, 2, 4, 11, 20, 31, 46, 61, 78, 97, 118, 141, 166, 193, 222, 253, 286, 321, 358, 397, 438, 481, 526, 573, 622, 673, 726, 781, 838, 897, 958, 1021, 1086, 1153, 1222, 1293, 1366, 1441, 1518, 1597, 1678, 1761, 1846, 1933, 2022, 2113, 2206, 2301, 2398, 2497

Offset: 1

Author

Olivier Gérard

Keywords

• nonn

Links

• J. Conrad, Table of n, a(n) for n = 1..180

Crossrefs

Cf. A028872.

Programs

Mathematica

p[1] = 1 + q; p[n_] := p[n] = p[n - 1] (1 + Sum[q^k, {k, 1, 2 n - 1, 2}]) // Expand; a[1] = 1; a[n_] := p[n] // CoefficientList[#, q]& // Union // Length; Array[a, 180] (* Jean-François Alcover, May 04 2017 *)

Python

def get(d, x): return d[x] if len(d) > x >= 0 else 0
def convolve(a, b):
    r = []
    for x in range(len(a) + len(b) - 1):
        n = 0
        for k in range(x + 1): n += get(a, k) * get(b, x - k)
        r.append(n)
    return r
def unique_in(d):
    out = list([])
    for elem in d:
        if elem not in out: out.append(elem)
    return len(out)
def A039824(x):
    seed = [0**k + k % 2 for k in range(2*(x+1))]
    product = seed[0:2]
    out = list([1])
    for k in range(2, x + 1):
        product = convolve(product, seed[0:2*k])
        out.append(unique_in(product))
    return out
# J. Conrad, May 02 2017

Formula

Conjecture: for n>6, a(n) = n^2 - 3. - Ralf Stephan, Mar 07 2004

Conjectures from Colin Barker, May 02 2017: (Start)

G.f.: x*(1 - x + x^2 + 4*x^3 - 3*x^4 + 2*x^6 - 4*x^7 + 2*x^8) / (1 - x)^3.

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>9.

(End)