This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A343293 #40 Mar 31 2023 09:17:49
%S A343293 36,64,81,512,196,16384,1089,8589934592,3844,4611686018427387904,
%T A343293 31329,191561942608236107294793378393788647952342390272950272,478864
%N A343293 a(n+1) is the smallest preimage k such that A008477(k) = a(n) with a(1) = 36.
%C A343293 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) = 36, hence f(a(n)) = a(n-1).
%C A343293 Why choose 36? Because it is the smallest integer for which there exists such an infinite iterated sequence, with g(36) = 64; then f(36) = 32 with the periodic sequence (32, 25, 32, 25, ...) (see A062307). Explanation: 36 is the first nonsquarefree number in A342973 that is also squareful. The nonsquarefree terms < 36: 12, 18, 20, 24, 28 in A342973 are not squareful (A332785), so they have no preimage by f.
%C A343293 When a(n-1) has several preimages by f, as a(n) is the smallest preimage, this sequence is well defined (see examples).
%C A343293 All the terms are nonsquarefree but also powerful, hence they are in A001694.
%C A343293 a(n) < a(n+2) (last comment in A008477) but a(n) < a(n+1) or a(n) > a(n+1).
%C A343293 Prime factorizations from a(1) to a(13): 2^2*3^2, 2^6, 3^4, 2^9, 2^2*7^2, 2^14, 3^2*11^2, 2^33, 2^2*31^2, 2^62, 3^2*59^2, 2^177, 2^4*173^2.
%C A343293 It appears that a(2m) = 2^q for some q>1 and a(2m+1) = r^2 for some r>1.
%C A343293 a(14) <= 2^692.
%H A343293 Annales Concours Général, <a href="https://www.freemaths.fr/annales-composition-mathematiques-concours-general/concours-general-mathematiques-sujet-serie-s-2012.pdf">Sujet Concours Général 2012</a> (in French, problems).
%H A343293 Annales Concours Général, <a href="https://www.freemaths.fr/annales-composition-mathematiques-concours-general/concours-general-mathematiques-corrige-serie-s-2012.pdf">Corrigé Concours Général 2012</a> (in French, solutions).
%e A343293 a(1) = 36; 64 = 2^6 so f(64) = 6^2 = 36, also 192 = 2^6*3^1 and f(192) = 6^2*1^3 = 36 we have f(64) = f(192) = 36; but as 64 < 192, hence g(36) = 64 and a(2) = 64.
%e A343293 a(2) = 64 = f(81) = f(256), but as 81 < 256, g(64) = 81 and a(3) = 81.
%e A343293 a(4) = 512 = f(196) = f(400), but as 196 < 400, g(512) = 196 and a(5) = 196.
%Y A343293 Cf. A001694, A008477, A062307, A332785, A342973.
%K A343293 nonn,more
%O A343293 1,1
%A A343293 _Bernard Schott_, Apr 11 2021
%E A343293 a(10)-a(13) from _Bert Dobbelaere_, Apr 13 2021