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%I #21 Apr 26 2024 03:30:36
%S 1,2,4,5,8,7,11,9,17,16,22,13,23,15,28,29,37,19,34,21,44,36,40,25,54,
%T 35,46,47,51,31,67,33,79,53,58,56,82,39,64,60,89,43,87,45,97,88,76,49,
%U 120,65,112,77,102,55,104,80,145,84,94,61,142,63,100,114,174
%N Let P(n) of a sequence s(1),s(2),s(3),... be obtained by leaving s(1),...,s(n) fixed and forward-cyclically permuting every n consecutive terms thereafter; apply P(2) to 1,2,3,... to get PS(2), then apply P(3) to PS(2) to get PS(3), then apply P(4) to PS(3), etc. The limit of PS(n) is A057064.
%C It appears that this is not a permutation of the integers: 3, 6, 10, 12, 14, 18, 20, 24, ... are not terms. - _Michel Marcus_, Feb 19 2016
%C Indeed, see the first formula here and the first comment in A069829. - _Mikhail Kurkov_, Mar 08 2023
%F a(n) = A057032(n-1) + 1 for n > 1. - _Sean A. Irvine_, May 19 2022
%e PS(2) begins with 1,2,4,3,6,5,8; PS(3) with 1,2,4,5,3,6,10; PS(4) with 1,2,4,5,8,3,6.
%o (PARI) get(v, iv) = if (iv > #v, 0, v[iv]);
%o fcp(nbn, nbp, startv, v) = {w = vector(nbn); for (k=1, nbn, j = k % nbp; if (j == 1, jv = startv+k+nbp-2, jv = startv+k-2); w[k] = get(v, jv);); w;}
%o lista(nn) = {v = vector(nn, n, n); print1(v[1], ", ", v[2], ", "); startv = 3; for (n=3, nn, w = fcp(nn-n+1, n-1, startv, v); startv = 2; if (w[1] == 0, break); print1(w[1], ", "); v = w;);} \\ _Michel Marcus_, Feb 19 2016
%Y Cf. A007062, A057030, A057032, A057033, A057063, A069829.
%K nonn
%O 1,2
%A _Clark Kimberling_, Aug 01 2000
%E More terms from _Michel Marcus_, Feb 19 2016
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