A263569 Number of distinct prime divisors p of 2*n such that lpf(2*n - p) = p, where lpf = least prime factor (A020639).

0, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 3, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 1, 1, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 2, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 3, 1, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 4

Offset: 1

Author

Gionata Neri, Oct 21 2015

Keywords

• nonn

Comments

a(n) gives the number of times 2*n occurs in A061228.

Examples

			a(10) = 1 since the distinct prime divisors of 2*10 = 20 are 2 and 5, A020639(20 - 2) = 2 and A020639(20 - 5) = 3, so only prime 2 is to be considered.
a(15) = 3 since the distinct prime divisors of 2*15 = 30 are 2, 3 and 5, A020639(30 - 2) = 2 and A020639(30 - 3) = 3 and A020639(30 - 5) = 5, so all three prime 2, 3 and 5 are to be considered.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A020639, A061228.

Programs

Mathematica

f[n_] := Select[First /@ FactorInteger[2 n], FactorInteger[2 n - #][[1, 1]] == # &]; Length /@ Table[f@ n, {n, 2, 105}] (* Michael De Vlieger, Oct 22 2015 *)

PARI

a(n) = {my(f=factor(2*n)); sum(k=1, #f~, p=f[k,1]; p == factor(2*n-p)[1,1]);} \\ Michel Marcus, Oct 31 2015