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

144, 4096, 1225, 34359738368, 549081

Offset: 1

Author

Bernard Schott, May 10 2021

Keywords

• nonn

Comments

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.

Examples

			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.
		

Crossrefs

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