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A319736 The lexicographically earliest increasing sequence such that n divides the sum of the first a(n) terms. 1
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified November 20 14:54 EST 2019. Contains 329337 sequences. (Running on oeis4.)