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

Author

Franklin T. Adams-Watters, Nov 04 2009

Keywords

• nonn
• look

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

The above question has been posed by Erdős. See Guy. - Stefano Spezia, Sep 25 2024

a(A000203(n)) <= n. Since A000203(n)/n can be arbitrarily large, that shows that lim inf_{n -> oo} a(n)/n = 0. - Robert Israel, Sep 26 2024

Examples

			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.
		

References

• R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B2.

Links

• Robert Israel, Table of n, a(n) for n = 0..20000 (n = 0 .. 1000 from Michel Marcus)
• Robert Israel, Plot of a(n)/n for n = 1 .. 10^6

Crossrefs

Cf. A000203, A001065, A078587, A051444.

Programs

Maple

N:= 100: # for a(0) .. a(N)
count:= 1: V:= Array(0..N): V[0]:= 1:
for m from 1 while count < N-1 do
  L:= ListTools:-PartialSums(sort(convert(numtheory:-divisors(m),list)));
  for x in L do
    if x > N then break fi;
    if V[x] = 0 then V[x]:= m; count:= count+1 fi;
od od:
convert(V,list); # Robert Israel, Sep 26 2024

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}