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A194259
Number of distinct prime factors of p(1)*p(2)*...*p(n), where p(n) is the n-th partition number.
5
0, 1, 2, 3, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 8, 9, 9, 10, 11, 12, 13, 14, 15, 16, 17, 17, 18, 19, 20, 21, 21, 21, 22, 23, 24, 25, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 39, 40, 42, 43, 44, 45, 47, 48, 49, 51, 52, 53, 54, 56, 57, 59, 60, 61
OFFSET
1,3
COMMENTS
Schinzel and Wirsing proved that a(n) > C*log n, for any positive constant C < 1/log 2 and all large n. In fact, it appears that a(n) > n for all n > 115 (see A194260).
It also appears that a(n) > a(n-1), for all n > 97, so that some prime factor of p(n) does not divide p(1)*p(2)*...*p(n-1). See A194261, A194262.
LINKS
Alois P. Heinz and Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 2000 terms from Alois P. Heinz)
A. Schinzel and E. Wirsing, Multiplicative properties of the partition function, Proc. Indian Acad. Sci., Math. Sci. (Ramanujan Birth Centenary Volume), 97 (1987), 297-303; alternative link.
Eric Weisstein's World of Mathematics, Partition Function
FORMULA
a(n) = A001221(product(k=1..n, A000041(k))).
EXAMPLE
p(1)*p(2)*...*p(8) = 1*2*3*5*7*11*15*22 = 2^2 * 3^2 * 5^2 * 7 * 11^2, so a(8) = 5.
MAPLE
with(combinat): with(numtheory):
b:= proc(n) option remember;
`if`(n=1, {}, b(n-1) union factorset(numbpart(n)))
end:
a:= n-> nops(b(n)):
seq(a(n), n=1..100); # Alois P. Heinz, Aug 20 2011
MATHEMATICA
a[n_] := Product[PartitionsP[k], {k, 1, n}] // PrimeNu; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Jan 28 2014 *)
PROG
(PARI) a(n)=my(v=[]); for(k=2, n, v=concat(v, factor(numbpart(k))[, 1])); #vecsort(v, , 8) \\ Charles R Greathouse IV, Feb 01 2013
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Aug 20 2011
STATUS
approved