A068049 The first term greater than one on each row of A068009. a(n) = A068009[n, A002024[n]].

2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 4, 3, 3, 2, 2, 2, 5, 4, 3, 3, 2, 2, 2, 6, 5, 4, 3, 3, 2, 2, 2, 7, 6, 5, 4, 3, 3, 2, 2, 2, 9, 7, 6, 5, 4, 3, 3, 2, 2, 2, 11, 9, 7, 6, 5, 4, 3, 3, 2, 2, 2, 13, 11, 9, 7, 6, 5, 4, 3, 3, 2, 2, 2, 16, 13, 11, 9, 7, 6, 5, 4, 3, 3, 2, 2, 2, 19, 16, 13, 11, 9, 7, 6, 5

Offset

1,1

Comments

In row 1 of A068009 the first term > 1 is found at position 1, for rows 2 & 3 at position 2, for rows 4,5,6 at position 3, for rows 7,8,9,10 at position 4 etc., thus it is natural to view this also as a triangular table.

Links

Alois P. Heinz, Table of n, a(n) for n = 1..1000

Maple

[seq(A000009(A025581(j-1))+1, j=1..99)];
A025581 := n-> binomial(1+floor(1/2+sqrt(2+2*n)), 2)-(n+1);
N := 100; t1 := series(mul(1+x^k, k=1..N), x, N); A000009 := proc(n) coeff(t1, x, n); end;

Mathematica

a[n_] := PartitionsQ[(1/2)(Floor[Sqrt[2n]+1/2]^2 + Floor[Sqrt[2n]+1/2] - 2n)] + 1; Array[a, 100] (* Jean-François Alcover, Mar 02 2016 *)

Crossrefs

a(n) = A000009(A025581(n-1))+1. Specifically, the left edge is equal to A000009[n]+1 (i.e. apart from the first term = A052839) and the right edge is all-2 sequence A007395.

Sequence in context: A335108 A359238 A320011 * A297850 A171092 A141256

Adjacent sequences: A068046 A068047 A068048 * A068050 A068051 A068052

Keyword

nonn,tabl

Author

Antti Karttunen, Feb 11 2002

Status

approved