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

100, 1024, 625, 33554432, 2116, 70368744177664

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) = 100, hence f(a(n)) = a(n-1).

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.

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).

Links

Table of n, a(n) for n=1..6.

Example

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.
a(2) = 1024 = f(625) = f(1250), but as 625 < 1250, g(1024) = 625 and a(3) = 625.

Crossrefs

Cf. A001694, A008477, A062307, A332785, A342973, A343293.

Sequence in context: A267127 A266893 A267153 * A025421 A300254 A124166

Adjacent sequences: A343291 A343292 A343293 * A343295 A343296 A343297

Keyword

nonn,more

Author

Bernard Schott, Apr 12 2021

Status

approved