A319736 The lexicographically earliest increasing sequence such that n divides the sum of the first a(n) terms.

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, 65, 66, 67, 68, 69, 70, 71, 72, 78, 79, 80, 81, 82, 83, 84, 85, 98, 99, 100, 101, 115, 116, 131, 133, 155, 156, 157, 158, 159, 160, 161, 162, 163, 169, 170, 189, 190

Offset: 1

Author

Jaroslav Krizek, Sep 26 2018

Keywords

• nonn

Comments

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, ...

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, ...

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

Examples

			a(1) = 1 because n = 1 divides the sum of the first 1 term.
a(2) is not 2 because 2 not divide the sum of the first a(2)= 2 terms (i.e., 1 + 2).
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).
a(3) = 4.
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.
a(5) = 8 and a(6) = 9; a(4) < a(5) < a(6).
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.
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.
		

Crossrefs

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