A100036 a(n) = smallest m such that A100035(m) = n.

1, 2, 3, 5, 7, 12, 14, 23, 25, 38, 40, 57, 59, 80, 82, 107, 109, 138, 140, 173, 175, 212, 214, 255, 257, 302, 304, 353, 355, 408, 410, 467, 469, 530, 532, 597, 599, 668, 670, 743, 745, 822, 824, 905, 907, 992, 994, 1083, 1085, 1178, 1180, 1277, 1279, 1380

Offset: 1

Author

Reinhard Zumkeller, Oct 31 2004

Keywords

• nonn

Comments

Smallest positions of occurrences of the natural numbers as subsequence in A100035;

A100035(a(n)) = n and A100035(m) <> n for m < a(n);

a(n) < A100037(n) < A100038(n) < A100039(n).

Examples

			First terms (10=A,11=B,12=C) of A100035(a(n)):
123.4.5....6.7........8.9............A.B................C.
1231435425165764736271879869584938291A9BA8B7A6B5A4B3A2B1CBD;
a(1) = A084849(1) = 1, A100035(1) = 1;
a(2) = A014107(1) = 2, A100035(2) = 2;
a(3) = A033537(1) = 3, A100035(3) = 3;
a(4) = A100040(1) = 5, A100035(5) = 4;
a(5) = A100041(1) = 7, A100035(7) = 5.
		

Formula

Conjecture: a(n) = partial sums of sequence [1,1,1,2,2,5,2,9,2,13,2,17,2,21,2,25,2,29,2,33,...2,n/2-7,2,...]. In other words, a(n) consists of the numbers 1,2,3 and the sequences A096376 and A096376+2 interspersed. - Ralf Stephan, May 15 2007