%I #6 Dec 18 2018 00:41:55

%S 1,3,4,7,8,9,12,16,18,19,20,23,24,25,26,33,34,42,46,48,49,50,59,61,63,

%T 65,66,67,68,69,70,71,72,78,79,80,81,82,83,84,85,98,99,100,101,115,

%U 116,131,133,155,156,157,158,159,160,161,162,163,169,170,189,190

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

%C Sequence b(n) of the sums of the first a(n) terms = Sum_{k=1..a(n)} a(k): 1, 8, 15, 44, 60, 78, 140, 248, 324, 370, 418, 576, 637, 700, 765, 1248, ...

%C Sequence c(n) of quotients when a(n) is calculated = (Sum_{k=1..a(n)} a(k) ) / n: 1, 4, 5, 11, 12, 13, 20, 31, 36, 37, 38, 48, 49, 50, 51, 78, 78, 111, ...

%C Is there a lexicographically earliest bijective sequence such that n divides the sum of the first a(n) terms?

%e a(1) = 1 because n = 1 divides the sum of the first 1 term.

%e a(2) is not 2 because 2 not divide the sum of the first a(2)= 2 terms (i.e., 1 + 2).

%e a(2) = 3 because 3 is the smallest number > a(1) such that 3 divides the sum of the first a(2)= 3 terms if a(3) = 4 whereas a(3) > a(2).

%e a(3) = 4.

%e a(4) = 7 because 7 is the smallest number > a(3) such that n = 3 divides the sum of the first 4 (i.e., a(3)) terms.

%e a(5) = 8 and a(6) = 9; a(4) < a(5) < a(6).

%e a(7) = 12 because 12 is the smallest number > a(6) such that n = 4 divides the sum of the first 7 (i.e., a(4)) terms.

%e a(8) = 16 because 16 is the smallest number > a(7) such that n = 5 divides the sum of the first 8 (i.e., a(5)) terms.

%Y Cf. A005408 (similar sequence for n divides the sum of first n terms).

%K nonn

%O 1,2

%A _Jaroslav Krizek_, Sep 26 2018