A167485 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.

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

Offset

0,4

Comments

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.

Is a(n) = o(n)? - Arkadiusz Wesolowski, Nov 09 2013

Links

Michel Marcus, Table of n, a(n) for n = 0..1000

Example

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.

Prog

(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}

Crossrefs

Cf. A000203, A001065, A078587, A051444.

Sequence in context: A299762 A057637 A258913 * A294141 A265513 A140508

Adjacent sequences: A167482 A167483 A167484 * A167486 A167487 A167488

Keyword

nonn

Author

Franklin T. Adams-Watters, Nov 04 2009

Status

approved