A343295 - OEIS

%I #18 Mar 31 2023 09:17:39

%S 144,4096,1225,34359738368,549081

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

%C The next term is a(6) = 2^741 with 224 digits.

%C 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).

%C 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.

%C Remember that the nonsquarefree terms in A342973 that are not squareful (A332785) have no preimage by f.

%C All the terms are nonsquarefree but also powerful, hence they are in A001694.

%C a(n) < a(n+2) (last comment in A008477) but a(n) < a(n+1) or a(n) > a(n+1).

%C 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.

%C It appears that a(2m) = 2^q for some q > 1, and a(2m+1) = r^2 for some r > 1.

%e 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.

%e a(2) = 4096 = f(1225) = f(2450), but as 1225 < 2450, g(4096) = 1225 and a(3) = 1225.

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

%K nonn

%O 1,1

%A _Bernard Schott_, May 10 2021