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A188795

a(n) counts all integers k in [2,floor(sqrt(n))] such that the number of divisors d>1 of n-k with k|(n-d) equals A188550(n).

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 3, 1, 2, 2, 3, 2, 1, 1, 1, 1, 1, 1, 4, 2, 2, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 1, 1

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OFFSET

4,8

LINKS

Alois P. Heinz, Table of n, a(n) for n = 4..10000

MAPLE

with(numtheory):

a:= proc(n) option remember; local c, h, k, m;

m, c:= 0, 0;

for k from 2 to floor(sqrt(n)) do

h:= nops(select(x-> irem(x, k)=0,

[seq (n-d, d=divisors(n-k) minus{1})]));