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A343294 a(n+1) is the smallest preimage k such that A008477(k) = a(n) with a(1) = 100.

%I A343294 #10 Mar 31 2023 09:17:45
%S A343294 100,1024,625,33554432,2116,70368744177664
%N A343294 a(n+1) is the smallest preimage k such that A008477(k) = a(n) with a(1) = 100.
%C A343294 Equivalently, when g is the reciprocal map of f = A008477 as defined in the Name, the terms of this sequence are the successive terms of the infinite iterated sequence {m, g(m), g(g(m)), g(g(g(m))), ...} that begins with m = a(1) = 100, hence f(a(n)) = a(n-1).
%C A343294 Why choose 100? Because it is the second integer, after 36, for which there exists a new infinite iterated sequence that begins with g(100) = 1024; then f(100) = 128 with the periodic sequence (128, 49, 128, 49, ...) (see A062307). Explanation: 100 is the 4th nonsquarefree number in A342973 that is also squareful, but the 3 previous such first integers 36, 64, 81 are yet terms of the infinite iterated sequence A343293. Remember that the nonsquarefree terms in A342973 that are not squareful (A332785) have no preimage by f.
%C A343294 When a(n-1) has several preimages by f, as a(n) is the smallest preimage, this sequence is well defined (see examples).
%C A343294 All the terms are nonsquarefree but also powerful, hence they are in A001694.
%C A343294 a(n) < a(n+2) (last comment in A008477) but a(n) < a(n+1) or a(n) > a(n+1).
%e A343294 a(1) = 100; 1024 = 2^10 so f(1024) = 10^2 = 100: also 5120 = 2^10*5^1 and f(5120) = 10^2*1^5 = 100; we have f(1024) = f(5120) = 100, but as 1024 < 5120, hence g(100) = 1024 and a(2) = 1024.
%e A343294 a(2) = 1024 = f(625) = f(1250), but as 625 < 1250, g(1024) = 625 and a(3) = 625.
%Y A343294 Cf. A001694, A008477, A062307, A332785, A342973, A343293.
%K A343294 nonn,more
%O A343294 1,1
%A A343294 _Bernard Schott_, Apr 12 2021