A167485 - OEIS

%I #24 Sep 27 2024 10:19:28

%S 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,

%T 30,12,102,29,16,21,42,62,84,22,36,37,18,27,63,20,50,43,66,52,129,33,

%U 75,40,78,48,220,34,36,28,49,60,265,24,132,61,32,56,117,54,100,67,90,84

%N 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.

%C 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.

%C Is a(n) = o(n)? - _Arkadiusz Wesolowski_, Nov 09 2013

%C The above question has been posed by Erdős. See Guy. - _Stefano Spezia_, Sep 25 2024

%C 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

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

%H Robert Israel, <a href="/A167485/b167485.txt">Table of n, a(n) for n = 0..20000</a> (n = 0 .. 1000 from Michel Marcus)

%H Robert Israel, <a href="/A167485/a167485.png">Plot of a(n)/n for n = 1 .. 10^6</a>

%e 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.

%p N:= 100: # for a(0) .. a(N)

%p count:= 1: V:= Array(0..N): V[0]:= 1:

%p for m from 1 while count < N-1 do

%p   L:= ListTools:-PartialSums(sort(convert(numtheory:-divisors(m),list)));

%p   for x in L do

%p     if x > N then break fi;

%p     if V[x] = 0 then V[x]:= m; count:= count+1 fi;

%p od od:

%p convert(V,list); # _Robert Israel_, Sep 26 2024

%o (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}

%Y Cf. A000203, A001065, A078587, A051444.

%K nonn,look

%O 0,4

%A _Franklin T. Adams-Watters_, Nov 04 2009