OFFSET
1,1
COMMENTS
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).
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.
When a(n-1) has several preimages by f, as a(n) is the smallest preimage, this sequence is well defined (see examples).
All the terms are nonsquarefree but also powerful, hence they are in A001694.
a(n) < a(n+2) (last comment in A008477) but a(n) < a(n+1) or a(n) > a(n+1).
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.
It appears that a(2m) = 2^q for some q>1 and a(2m+1) = r^2 for some r>1.
a(14) <= 2^692.
LINKS
Annales Concours Général, Sujet Concours Général 2012 (in French, problems).
Annales Concours Général, Corrigé Concours Général 2012 (in French, solutions).
EXAMPLE
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.
a(2) = 64 = f(81) = f(256), but as 81 < 256, g(64) = 81 and a(3) = 81.
a(4) = 512 = f(196) = f(400), but as 196 < 400, g(512) = 196 and a(5) = 196.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Bernard Schott, Apr 11 2021
EXTENSIONS
a(10)-a(13) from Bert Dobbelaere, Apr 13 2021
STATUS
approved