%I #16 Jan 25 2017 15:15:05
%S 1,3,2,5,6,4,8,9,10,7,12,13,14,15,11,17,18,19,20,21,16,23,24,25,26,27,
%T 28,22,30,31,32,33,34,35,36,29,38,39,40,41,42,43,44,45,37,47,48,49,50,
%U 51,52,53,54,55,46,57,58,59,60,61,62,63,64,65,66,56,68,69,70,71,72,73
%N Permutation of the integers with cycle form {1}, {2, 3}, {4, 5, 6}, {7, 8, 9, 10}, ...
%C Arrange natural numbers 1,2,3,4,5,... as a triangle like A000027, then rotate each row of triangle one left. - _Antti Karttunen_, May 07 2002
%C For a group of n terms a^(n)(k)= k where n(n-1)/2 < k <= n(n+1)/2. E.g. for the three terms 5, 6 and 4, a(5)= 6, a(6) = 4, a(4) = 5; a(a(a(5))) = 5 = a^(3)(5). - _Amarnath Murthy_, May 31 2003
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F a(n) = 1 + n + binomial(round(sqrt(2*n)),2) - binomial(round(sqrt(2*n+2)),2). - _Brian Tenneson_, Jan 23 2017
%p a := proc(n) floor(sqrt(2*n));n+1;`if`(2*n=%%*(%%+1),%-%%,%) end: # _Peter Luschny_, Jan 25 2017
%t FromCycles[Table[n(n-1)/2+Range[n], {n, 13}]]
%Y Inverse permutation: A066182.
%K easy,nonn,tabl
%O 1,2
%A _Wouter Meeussen_, Dec 15 2001