A316571 a(1) = 1; for n > 1: a(n) = smallest number such that (Sum_{k=1..n} a(k)) is divisible by n - 1.

1, 2, 3, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124

Offset: 1

Author

Jaroslav Krizek, Aug 20 2018

Keywords

• nonn

Comments

This is the lexicographically earliest increasing sequence such that n-1 divides the sum of the first n terms.

Sequence b(n) of the sums of the first n+1 terms, Sum_{k=1..n+1} a(k): 3, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, ...

Sequence c(n) of quotients when a(n) is calculated, (Sum_{k=1..n+1} a(k) ) / n is 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...

Examples

			a(1) = 1 because 1 divides the sum of the first 2 (i.e., 1 + 1) terms (a(1) + a(2)) for whatever term a(2) > a(1).
a(2) = 2 because 2 is the smallest number > a(1) and 2 divides the sum of the first 3 (i.e., 2 + 1) terms (a(1) + a(2) + a(3)) for whatever term a(3) > a(2) such that 2 divides the sum a(1) + a(2) + a(3); the smallest number > a(2) with this property for a(3) is 3.
a(3) = 3.
a(4) = 6 because 6 is the smallest number > a(3) such that 3 divides the sum of the first 4 (i.e., 3 + 1) terms.
a(5) = 8 because 8 is the smallest number > a(4) such that 4 divides the sum of the first 5 (i.e., 4 + 1) terms.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (2, -1).

Crossrefs

Cf. A005843, A020739, A316156.

Programs

Mathematica

CoefficientList[Series[(1 + 2 x^3 - x^4)/(-1 + x)^2 , {x, 0, 50}], x] (* Stefano Spezia, Sep 16 2018 *)
Join[{1, 2, 3}, LinearRecurrence[{2, -1}, {6, 8}, 60]] (* Jean-François Alcover, Dec 30 2018 *)

Formula

a(1)=1, a(2)=2, a(3)=3, a(n)=2(n-1) for n >= 4.

a(n) = A005843(n-1) for n >= 4. - Antti Karttunen, Sep 16 2018

G.f.: (1 + 2*x^3 - x^4)/(-1 + x)^2. - Stefano Spezia, Sep 16 2018

a(n) = A020739(n-4) = 2*(n - 4) + 6 for n >= 4. - Georg Fischer, Jan 19 2019

Extensions

Definition clarified by Georg Fischer and Alois P. Heinz, Jan 19 2019