A194260 A194259(n) - n, where A194259(n) is the number of distinct prime factors of p(1)*p(2)*...*p(n) and p(n) is the n-th partition number.

-1, -1, -1, -1, -1, -1, -2, -3, -4, -5, -6, -7, -7, -8, -9, -10, -11, -12, -13, -13, -14, -14, -14, -15, -15, -15, -15, -15, -15, -15, -15, -15, -16, -16, -16, -16, -16, -17, -18, -18, -18, -18, -18, -17, -17, -16, -16, -16, -16, -16, -16, -16, -16, -15, -15, -14, -14, -14, -14, -13, -13, -13, -12, -12, -12, -12, -11, -11, -10, -10, -10, -10, -9, -9, -9, -9, -9, -8, -7, -7, -7, -8, -8, -8, -8, -7, -7, -7, -7, -6, -5, -4, -4, -4, -3, -3, -4, -4, -4, -4, -4, -3, -3, -3, -3, -3, -3, -3, -3, -2, -2, -2, -2, -2, 0, 1

Offset: 1

Author

Jonathan Sondow, Aug 20 2011

Keywords

• sign

Comments

Schinzel and Wirsing proved that a(n) > C*log n - n, for any positive constant C < 1/log 2 and all large n. In fact, it appears that a(n) > 0 for all n > 115.

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).

Examples

			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 - 8 = -3.
		

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.

Crossrefs

Cf. A000041, A001221, A087175, A194259.

Programs

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)) -n:
seq(a(n), n=1..116); # Alois P. Heinz, Aug 20 2011

Mathematica

a[n_] := PrimeNu[Product[PartitionsP[k], {k, 1, n}]] - n; Table[a[n], {n, 1, 116}] (* Jean-François Alcover, Jan 28 2014 *)

Formula

a(n) = A001221(product(k=1..n, A000041(k))) - n.