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A048138
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a(n) = number of m such that sum of proper divisors of m (A001065(m)) is n.
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25
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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
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2,5
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COMMENTS
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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
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LINKS
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FORMULA
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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)
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EXAMPLE
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a(6) = 2 since 6 is the sum of the proper divisors of 6 and 25.
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MAPLE
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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:
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PROG
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(PARI) list(n)=my(v=vector(n-1), k); for(m=4, n^2, k=sigma(m)-m; if(k>1 & k<=n, v[k-1]++)); v \\ Charles R Greathouse IV, Apr 21 2011
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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