cp's OEIS Frontend

This sequence has infinitely many terms since a (trivial) upper bound for a(n) is given by (10^(n - v_{10}(n)) + 1)^n, where v_{10} corresponds to the number of trailing 0's of n (if any), and each of these terms is mandatorily a perfect power of degree n or a multiple of n (e.g., a(3) = 25^3 = 5^6).

All the a(n) are generated by the 13 nontrivial 10-adic solutions of the fundamental equation y^5 = y: ...480163574218751 (see A063006), ...263499879186432 (see A120817), ...996418333704193 (see A290375), ...476581907922943 (see A290373), ...259918212890624 (see A091664), ...259918212890625 (see A018247), ...740081787109375 (see A091663), ...740081787109376 (see A018248), ...003581666295807 (see A290372), ...523418092077057 (see A290374), ...736500120813568 (see A120818), ...519836425781249 (see A091661), and ...999999999999999.

The bases of a(11), a(12), ..., a(50) have been provided by Max Alekseyev on February 14, 2025 (see "Closed form for the general term of 2, 49, 15625, 625, ..." in Links).

It is conjectured that if n > 2 is given, then a(n) is generated by {5^2^k}_oo or -{5^2^k}_oo (this probabilistic argument is based on the study of a(n) up to n = 50 since a(50) is a 762 digit number generated by {5^2^k}_oo = ...6259918212890625).

Constant congruence speed of (10^n + 1)^n, i.e., a(n) = A373387((10^n + 1)^n).

Since 10^n + 1 is never a perfect power by Catalan's conjecture (Mihăilescu's theorem), it follows that if 10 does not divide n, then (10^n + 1)^n is exactly an n-th perfect power with a constant congruence speed of a(n) = n.

Moreover, for any positive integer n, the congruence speed of (10^n + 1)^n equals 2*a(n) at height 1 and then becomes stable at height 2.