login
A263569
Number of distinct prime divisors p of 2*n such that lpf(2*n - p) = p, where lpf = least prime factor (A020639).
1
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
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
Sequence in context: A128538 A115588 A240471 * A105220 A083654 A164878
KEYWORD
nonn
AUTHOR
Gionata Neri, Oct 21 2015
STATUS
approved