A315615 -id:A315615

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A269501

Subsequence immediately following the instances of n in the sequence is n, n-1, ..., 1, n+1, n+2, ....

+0
4

1, 1, 2, 2, 1, 3, 3, 2, 3, 1, 4, 4, 3, 4, 2, 4, 1, 5, 5, 4, 5, 3, 5, 2, 5, 1, 6, 6, 5, 6, 4, 6, 3, 6, 2, 6, 1, 7, 7, 6, 7, 5, 7, 4, 7, 3, 7, 2, 7, 1, 8, 8, 7, 8, 6, 8, 5, 8, 4, 8, 3, 8, 2, 8, 1, 9, 9, 8, 9, 7, 9, 6, 9, 5, 9, 4, 9, 3, 9, 2, 9, 1, 10, 10, 9, 10, 8, 10, 7, 10, 6, 10, 5, 10, 4, 10, 3, 10, 2, 10, 1

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OFFSET

0,3

COMMENTS

The sequence includes every ordered pair of positive integers exactly once as consecutive terms of the sequence. Through n = k^2, it has every pair i,j with 0 < i,j <= k.

Can be regarded as an irregular triangle where row k contains 1, k, k, k-1, k, k-2, ..., 2, k, with 2n-1 terms.

See A305615 for an essentially identical sequence: a(n) = A305615(n)+1. - N. J. A. Sloane, Jul 03 2018

LINKS

Table of n, a(n) for n=0..100.

FORMULA

Let r = ceiling(sqrt(n)) = A003059(n). If n and r have the same parity, a(n) = (r^2-n)/2 + 1; otherwise a(n) = r.

EXAMPLE

The first 3 occurs as a(5), so a(6) = 3, the first term of 3, 2, 1, 4, 5, 6, .... The second 3 is thus a(6), so a(7) = 2. The third 3 is a(8), so a(9) = 1. The fourth 3 is a(12), now we start incrementing, and a(13) = 4.

The triangle starts:

1, 2, 2

1, 3, 3, 2, 3

1, 4, 4, 3, 4, 2, 4

1, 5, 5, 4, 5, 3, 5, 2, 5

PROG

(PARI) a(n) = my(r = if(n<=0, 0, sqrtint(n-1)+1); if((n-r)%2, r, (r^2-n)/2 + 1)

CROSSREFS

Cf. A003059, A060747 (row lengths), A000326 (row sums), A097291, A269780.