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

1,2

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?

Links

Table of n, a(n) for n=1..62.

Example

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).

Sequence in context: A175054 A154366 A226227 * A100452 A004201 A109054

Adjacent sequences: A319733 A319734 A319735 * A319737 A319738 A319739

Keyword

nonn

Author

Jaroslav Krizek, Sep 26 2018

Status

approved