

A048138


a(n) = number of m such that sum of proper divisors of m (A001065(m)) is n.


26



0, 1, 1, 0, 2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 2, 3, 2, 2, 1, 3, 1, 2, 1, 2, 1, 5, 2, 3, 1, 3, 1, 4, 1, 1, 3, 4, 2, 5, 2, 3, 2, 3, 1, 6, 2, 4, 0, 3, 2, 6, 1, 5, 1, 3, 1, 6, 2, 3, 3, 6, 1, 6, 1, 2, 1, 5, 1, 8, 3, 4, 3, 5, 1, 7, 1, 6, 1, 4, 1, 8, 1, 5, 0, 5, 2, 9, 2, 4, 1, 4, 0, 9, 1, 3, 2, 6, 1, 8, 2, 7, 4
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OFFSET

2,5


COMMENTS

The offset is 2 since there are infinitely many numbers (all the primes) for which A001065 = 1.
The graph of this sequence, shifted by 1, looks similar to that of A061358, which counts Goldbach partitions of n.  T. D. Noe, Dec 05 2008
The smallest k > 0 such that there are exactly n numbers whose sum of proper divisors is k is A125601(n).  Bernard Schott, Mar 23 2023


LINKS



FORMULA

a(n) = 0 iff n is in A005114 (untouchable numbers).
a(n) = 1 iff n is in A057709 ("hermit" numbers).
a(n) > 1 iff n is in A160133. (End)


EXAMPLE

a(6) = 2 since 6 is the sum of the proper divisors of 6 and 25.


MAPLE

with(numtheory): for n from 2 to 150 do count := 0: for m from 1 to n^2 do if sigma(m)  m = n then count := count+1 fi: od: printf(`%d, `, count): od:


PROG

(PARI) list(n)=my(v=vector(n1), k); for(m=4, n^2, k=sigma(m)m; if(k>1 & k<=n, v[k1]++)); v \\ Charles R Greathouse IV, Apr 21 2011


CROSSREFS



KEYWORD

easy,nonn


AUTHOR



EXTENSIONS



STATUS

approved



