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

1,3

Comments

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

Links

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

Example

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.

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 *)

Prog

(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

Crossrefs

Cf. A020639, A061228.

Sequence in context: A128538 A115588 A240471 * A105220 A083654 A164878

Adjacent sequences: A263566 A263567 A263568 * A263570 A263571 A263572

Keyword

nonn

Author

Gionata Neri, Oct 21 2015

Status

approved