A319737 The lexicographically earliest increasing sequence such that n divides the sum of the first a(n) + 1 terms.

1, 2, 3, 6, 7, 8, 9, 14, 16, 18, 19, 20, 21, 22, 26, 27, 33, 34, 44, 55, 59, 63, 67, 68, 69, 70, 74, 89, 90, 91, 92, 93, 94, 109, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 150, 151, 152, 153, 169

Offset: 1

Author

Jaroslav Krizek, Sep 26 2018

Keywords

• nonn

Comments

Sequence b(n) of the sums of the first a(n)+1 terms = Sum_{k=1..a(n)+1} a(k): 3, 6, 12, 36, 50, 66, 84, 192, 252, 330, 385, 444, 507, 574, 855, 944, 1513, ...

Sequence c(n) of quotients when a(n) is calculated = (Sum_{k=1..a(n)+1} a(k) ) / n: 3, 3, 4, 9, 10, 11, 12, 24, 28, 33, 35, 37, 39, 41, 57, 59, 89, ...

Is there a lexicographically earliest bijective sequence such that n divides the sum of the first a(n)+1 terms?

Examples

			a(1) = 1.
a(2) = 2 because 2 is the smallest number > a(1) and n = 1 divides the sum of the first a(1) + 1 = 2 terms for all any term > 1.
a(3) = 3 because 3 is the smallest number > a(2) such that n = 2 divides the sum of the first a(2) + 1 = 3 terms.
a(4) = 6 because 6 is the smallest number > a(3) such that n = 3 divides the sum of the first a(3) + 1 = 4 terms.
a(5) = 7 and a(6) = 8; a(4) < a(5) < a(6).
a(7) = 9 because 9 is the smallest number > a(6) such that n = 4 divides the sum of the first a(4) + 1 = 7 terms.
a(8) = 14 because 14 is the smallest number > a(7) such that n = 5 divides the sum of the first a(5) + 1 = 8 terms.
		

Crossrefs

Cf. A316571 (similar sequence for n divides the sum of the first n+1 terms).

Cf. A319736.