%I #21 Mar 05 2023 20:51:02
%S 1,2,3,17,9,4,8,31,15,7,5,47,64,6,21,10,96,20,11,13,57,38,14,16,79,37,
%T 18,12,160,28,22,19,61,24,26,23,131,52,27,25,41,33,46,29,77,45,42,34,
%U 54,59,36,32,68,72,44,40,104,82,50,49,75,111,51,35,98,143,63,30,85
%N a(n) = k such that A361103(k1) = n, or 1 if n never appears in A361103.
%C Imagine the offset of A361103 is 1, and assume it really is a permutation of the natural numbers. In tabular form, it is
%C ..1..2..3..4..5..6..7..8..9.10.11...
%C ..1..2..3..6.11.14.10..7..5.16..19...
%C Then the inverse permutation would be
%C ..1..2..3.17..9..4..8.31.15..7..5.47...
%C which is the present sequence.
%H N. J. A. Sloane, <a href="/A361104/b361104.txt">Table of n, a(n) for n = 1..2721</a>
%e A361103(16) = 4, so a(4) = 17.
%Y Cf. A360519, A361102, A336957, A361103.
%K nonn
%O 1,2
%A _Scott R. Shannon_ and _N. J. A. Sloane_, Mar 02 2023
%E Edited by _N. J. A. Sloane_, Mar 04 2023
