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A284465
Number of compositions (ordered partitions) of n into prime power divisors of n (not including 1).
4
1, 0, 1, 1, 2, 1, 2, 1, 6, 2, 2, 1, 36, 1, 2, 2, 56, 1, 90, 1, 201, 2, 2, 1, 4725, 2, 2, 20, 1085, 1, 15778, 1, 5272, 2, 2, 2, 476355, 1, 2, 2, 270084, 1, 302265, 1, 35324, 3910, 2, 1, 67279595, 2, 14047, 2, 219528, 1, 5863044, 2, 14362998, 2, 2, 1, 47466605656, 1, 2, 35662, 47350056, 2, 119762253, 1, 9479643
OFFSET
0,5
LINKS
FORMULA
a(n) = [x^n] 1/(1 - Sum_{p^k|n, p prime, k>=1} x^(p^k)).
a(n) = 1 if n is a prime.
a(n) = 2 if n is a semiprime.
EXAMPLE
a(8) = 6 because 8 has 4 divisors {1, 2, 4, 8} among which 3 are prime powers > 1 {2, 4, 8} therefore we have [8], [4, 4], [4, 2, 2], [2, 4, 2], [2, 2, 4] and [2, 2, 2, 2].
MAPLE
F:= proc(n) local f, G;
G:= 1/(1 - add(add(x^(f[1]^j), j=1..f[2]), f = ifactors(n)[2]));
coeff(series(G, x, n+1), x, n);
end proc:
map(F, [$0..100]); # Robert Israel, Mar 29 2017
MATHEMATICA
Table[d = Divisors[n]; Coefficient[Series[1/(1 - Sum[Boole[PrimePowerQ[d[[k]]]] x^d[[k]], {k, Length[d]}]), {x, 0, n}], x, n], {n, 0, 68}]
PROG
(Python)
from sympy import divisors, primefactors
from sympy.core.cache import cacheit
@cacheit
def a(n):
l=[x for x in divisors(n) if len(primefactors(x))==1]
@cacheit
def b(m): return 1 if m==0 else sum(b(m - j) for j in l if j <= m)
return b(n)
print([a(n) for n in range(71)]) # Indranil Ghosh, Aug 01 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Mar 27 2017
STATUS
approved