%I #62 May 23 2024 04:30:50

%S 1,3,2,5,9,7,8,13,15,11,14,16,26,24,41,29,18,28,20,30,22,32,25,33,43,

%T 45,31,37,50,52,54,56,58,35,87,38,55,67,40,60,72,44,63,77,79,47,70,49,

%U 121,88,53,129,94,96,98,100,59,89,105,107,62,158,113,65,102,68,103,189

%N The lexicographically earliest sequence of distinct terms with a(1) = 1 such that a(n) divides the sum of the first a(n) terms.

%C Once there is a k such that k > n and a(k) > n, n can no longer appear in the sequence, otherwise a(k) would be n. - _Franklin T. Adams-Watters_, Jun 11 2014

%C If the sum a(1) + a(2) + ... + a(m) is not divisible by m, then m does not belong to this sequence. Sequence A019444 gives a variant of this sequence, where every positive integer is a term. - _Max Alekseyev_, Jun 11 2014

%C Positive integers that do not appear in this sequence form A243864.

%C Is there any index n > 3 such that a(n) <= n? - _Max Alekseyev_, Jun 13 2014

%C From _Bill McEachen_, May 21 2024: (Start)

%C Conjecture: For n > 1000, a(n) falls within 1% of one of the following six values. a(n) = n, 1.576385*n, 1.788185*n, 2.576385*n, 2.788185*n, or 3.576285*n, using floor at the low bound and ceiling at the high bound, inclusive.

%C For example, a(1153) = 1836. This is between floor(1.576385 * 1153 * 0.99) and ceiling(1.576385 * 1153 * 1.01). About 90% of values fall in the three lower slopes. (End)

%H Max Alekseyev, <a href="/A243700/b243700.txt">Table of n, a(n) for n = 1..100000</a> (first 1100 terms from Jean-Marc Falcoz)

%H Éric Angelini, <a href="http://list.seqfan.eu/pipermail/seqfan/2014-June/027781.html">a(n) divides the sum of the first a(n) terms of T</a>, posting to the Sequence Fans Mailing List, Jun 11 2014

%H Hugo Pfoertner, <a href="https://www.randomwalk.de/sequences/b243700.7z">1.73*10^6 terms,</a> 7z compressed b-file.

%e 1 divides the sum of the first 1 term (yes: 1/1=1)

%e 3 divides the sum of the first 3 terms (yes: 6/3=2)

%e 2 divides the sum of the first 2 terms (yes: 4/2=2)

%e 5 divides the sum of the first 5 terms (yes: 20/5=4)

%e 9 divides the sum of the first 9 terms (yes: 63/9=7)

%e 7 divides the sum of the first 7 terms (yes: 35/7=5)

%e 8 divides the sum of the first 8 terms (yes: 48/8=6)

%e ...

%o (PARI) { printA243700() = my( S=Set(), T=[], s=0, m=1, k); for(n=1,10^5, k=m; while( ((k==n || setsearch(S,n)) && Mod(s+k,n)) || if(k<n,sum(i=1,k,T[i])%k) || setsearch(S,k), k++); S=setunion(S,[k]); T=concat(T,[k]); s+=k; if(s%n,S=setunion(S,[n]);); while(setsearch(S,m),m++); print1(k,", "); ) } \\ _Max Alekseyev_, Jun 13 2014

%Y Cf. A019444, A243864 (complement), A244010 (partial sums), A244011 (the quotients), A244016 (sorted), A244017, A244018.

%K nonn,nice

%O 1,2

%A _N. J. A. Sloane_, Jun 12 2014

%E First 1100 terms were computed by _Jean-Marc Falcoz_.