A263717 Number of partitions of n into perfect odd powers (1 being excluded).

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 1, 1, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 2, 0, 3, 0, 0, 0, 1, 1, 0, 2, 0, 3, 0, 1, 1, 2, 2, 0, 3, 0, 5, 0, 1, 1, 2, 2, 0, 3, 0, 5, 0, 1, 1, 2, 2, 0, 3, 1, 6

Offset

1,27

Links

Martin Y. Champel, Table of n, a(n) for n = 1..999

Example

a(97) = #{8*9+25, 5*9+25+27, 2*9+25+2*27} = 3.

Mathematica

Needs["Combinatorica`"]; Length@ Select[Combinatorica`Partitions@ #, AllTrue[#, And[PrimePowerQ@ #, ! PrimeQ@ #, OddQ@ #] &, 1] &] & /@ Range[52] (* Michael De Vlieger, Nov 05 2015, Version 10 *)

Prog

(Python)
from math import log
def a(n):
    base = sorted(list(set([a**b for b in range(2, int(log(n)/log(2))) for a in range(3, 1+int(n**(1./b)), 2)])))
    lb = len(base)
    if lb == 0:
        return 0
    sol = 0
    s = [n // base[0]]
    if lb == 1:
        if n % base[0] == 0: return 1
        return 0
    while True:
        k = s.pop()
        while k < 0:
            if s ==(lb-1)*[0]:
                return sol
            k = s.pop() - 1
        s.append(k)
        x = n - sum([s[i]*base[i] for i in range(len(s))])
        ls = len(s)
        if ls == lb:
            continue
        a = x // base[ls]
        b = x % base[ls]
        if b == 0:
            s.append(a)
            sol +=1
            if len(s) == lb:
                s.pop()
                s.append(-1)
            r = s.pop() - 2
            s.append(r)
        else:
            s.append(a-1)
            if a!=0:
                if len(s) == lb: s[lb-1]=-1

Crossrefs

Cf. A075109, A112344.

Sequence in context: A216512 A078359 A107329 * A230279 A376366 A085859

Adjacent sequences: A263714 A263715 A263716 * A263718 A263719 A263720

Keyword

nonn

Author

Martin Y. Champel, Oct 24 2015

Extensions

Missing a(1) = 0 prepended by Andrei Zabolotskii, Jan 01 2026

Status

approved