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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.
2
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
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
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)
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.
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
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
KEYWORD
nonn,look
AUTHOR
STATUS
approved