A244671 - OEIS

%I #9 Jul 07 2014 00:19:39

%S 1,3,5,6,10,11,12,13,14,15,20,22,24,26,28,29,30,31,32,48,49,55,56,60,

%T 61,67,68,72,89,93,97,101,102,103,104,105,106,107,108,109,110,111,112,

%U 113,114,115,116,141,161,162,163,164,165,166,175,188,189,190,191,222,269

%N The lexicographically earliest increasing sequence such that a(n) divides the sum of the first a(n) terms.

%C See A244672(n) - partial sums of a(n).

%F A244672(a(n)) / a(n) = integer.

%e a(1) = 1 because 1 divides the first term (1/1=1); a(2) cannot be 2 because 2 does not divide the sum of the first 2 terms (3/2 is not integer), a(2) must be 3; if a(2) = 3 then a(3) must be 5 (5 is the smallest number > a(2) such that the sum of the first 3 terms (i.e. 9) is divisible by a(2) = 3; if a(4) = 6 (holds 6 > a(3)), a(5) must be 10 (10 is the smallest number > a(4) such that the sum of first 5 terms (i.e. 25) is divisible by a(3) = 5; etc…

%p N:= 1000: # to get the first N terms

%p A:= {1,3}: s:= 4:

%p for n from 3 to N do

%p   if member(n,A,'p') then

%p      r:= A[n-1]+1 + (-s-A[n-1]-1 mod A[p])

%p   else

%p      r:= A[n-1]+1

%p   fi;

%p   A:= A union {r};

%p   s:= s + r;

%p od:

%p A; # _Robert Israel_, Jul 06 2014

%Y Cf. A243700, A244672.

%K nonn

%O 1,2

%A _Jaroslav Krizek_, Jul 04 2014