

A269501


Subsequence immediately following the instances of n in the sequence is n, n1, ..., 1, n+1, n+2, ....


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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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, k1, k, k2, ..., 2, k, with 2n1 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^2n)/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
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(n1)+1); if((nr)%2, r, (r^2n)/2 + 1)


CROSSREFS

Cf. A003059, A060747 (row lengths), A000326 (row sums), A097291, A269780.
See also A315615.
Sequence in context: A182535 A181186 A097291 * A254044 A274515 A291564
Adjacent sequences: A269498 A269499 A269500 * A269502 A269503 A269504


KEYWORD

nonn,tabf


AUTHOR

Franklin T. AdamsWatters, Mar 04 2016


STATUS

approved



