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A167485
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Smallest positive integer m such that n can be expressed as the sum of an initial subsequence of the divisors of m, or 0 if no such m exists.
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2
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1, 1, 0, 2, 3, 0, 5, 4, 7, 15, 12, 21, 6, 9, 13, 8, 12, 30, 10, 42, 19, 18, 20, 57, 14, 36, 46, 30, 12, 102, 29, 16, 21, 42, 62, 84, 22, 36, 37, 18, 27, 63, 20, 50, 43, 66, 52, 129, 33, 75, 40, 78, 48, 220, 34, 36, 28, 49, 60, 265, 24, 132, 61, 32, 56, 117, 54, 100, 67, 90, 84
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OFFSET
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0,4
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COMMENTS
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It appears that 2 and 5 are the only zeros in this sequence. This would follow from a slightly stronger version of the Goldbach conjecture: every even integer > 22 can be expressed as the sum of two primes p and q, with 5 < p < q < 5p. Then odd numbers can be obtained for pq and even numbers for 5pq.
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LINKS
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EXAMPLE
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The divisors of 15 are 1,3,5,15, with cumulative sums 1,4,9,24. Since this is the smallest number where 9 occurs in the sums, a(9) = 15.
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PROG
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(PARI) {u=vector(100); for(n=1, 1000, ds=divisors(n); s=0; for(k=1, #ds, s+=ds[k]; if(s>#u, break); if(!u[s], u[s]=n))); u}
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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