A343293 -id:A343293 - cp's OEIS frontend

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

Bernard Schott, Apr 12 2021

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).

			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.

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

A343295 a(n) is the smallest k such that A008477(k) = a(n-1) with a(1) = 144.

Original entry on oeis.org

144, 4096, 1225, 34359738368, 549081

Offset: 1

Bernard Schott, May 10 2021

nonn

The next term is a(6) = 2^741 with 224 digits.
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) = 144, hence f(a(n)) = a(n-1).
Why choose 144? Because it is the third integer, after 36 and 100, for which there exists a new infinite iterated sequence that begins with g(144) = 4096; then f(144) = 128 with the periodic sequence (128, 49, 128, 49, ...) (see A062307). Explanation: 144 is the 5th nonsquarefree number in A342973 that is also squareful; the 3 such first integers 36, 64, 81 are terms of the infinite iterated sequence A343293, while 100 is a term of the infinite iterated sequence A343294.
Remember that the nonsquarefree terms in A342973 that are not squareful (A332785) have no preimage by f.
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(6): 2^4*3^2, 2^12, 5^2*7^2, 2^35, 3^2*13^2*19^2, 2^741.
It appears that a(2m) = 2^q for some q > 1, and a(2m+1) = r^2 for some r > 1.

			a(1) = 144; 4096 = 2^12 so f(4096) = 12^2 = 144: also 12288 = 2^12*3^1 and f(12288) = 12^2*1^3 = 144; we have f(4096) = f(12288) = 144, but as 4096 < 12288, hence g(144) = 4096 and a(2) = 4096.
a(2) = 4096 = f(1225) = f(2450), but as 1225 < 2450, g(4096) = 1225 and a(3) = 1225.

Cf. A001694, A008477, A062307, A332785, A342511, A342973, A343293, A343294.

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

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A343295 a(n) is the smallest k such that A008477(k) = a(n-1) with a(1) = 144.

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