%I #10 Sep 24 2015 01:41:08

%S 1,0,5,4,3,2,7,6,11,10,9,8,19,18,23,22,21,20,13,12,17,16,15,14,25,24,

%T 29,28,27,26,31,30,35,34,33,32,43,42,47,46,45,44,37,36,41,40,39,38,49,

%U 48,53,52,51,50,55,54,59,58,57,56,67,66,71,70,69,68,61,60,65,64,63,62,97,96,101,100,99,98,103,102,107,106,105,104,115,114,119,118,117,116,109,108,113,112,111,110,73

%N Row 1 of A261216.

%C Equally, column 1 of A261217.

%C Take the n-th (n>=0) permutation from the list A060117, change 1 to 2 and 2 to 1 to get another permutation, and note its rank in the same list to obtain a(n).

%C Equally, we can take the n-th (n>=0) permutation from the list A060118, swap the elements in its two leftmost positions, and note the rank of that permutation in A060118 to obtain a(n).

%C Self-inverse permutation of nonnegative integers.

%H Antti Karttunen, <a href="/A261218/b261218.txt">Table of n, a(n) for n = 0..5039</a>

%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>

%F a(n) = A261216(1,n).

%F By conjugating related permutations:

%F a(n) = A060126(A261098(A060119(n))).

%e In A060117 the permutation with rank 2 is [1,3,2], and swapping the elements 1 and 2 we get permutation [2,3,1], which is listed in A060117 as the permutation with rank 5, thus a(2) = 5.

%e Equally, in A060118 the permutation with rank 2 is [1,3,2], and swapping the elements in the first and the second position gives permutation [3,1,2], which is listed in A060118 as the permutation with rank 5, thus a(2) = 5.

%Y Row 1 of A261216, column 1 of A261217.

%Y Cf. A060117, A060118.

%Y Cf. also A004442.

%Y Related permutations: A060119, A060126, A261098.

%K nonn

%O 0,3

%A _Antti Karttunen_, Aug 26 2015