cp's OEIS Frontend

3^500 ends in 7610001

3^5000 ends in 276100001

3^50000 ends in 32761000001

3^500000 ends in 9327610000001

The last digits before the zeros are converging to the present sequence.

7^500 ends in 64030001

7^5000 ends in 5640300001

7^50000 ends in 756403000001

7^500000 ends in 77564030000001

The last digits before the zeros tend to converge to A216105.

7^500 ends in 64030001

7^5000 ends in 5640300001

7^50000 ends in 756403000001

7^500000 ends in 77564030000001

The last digits before the zeros tend to converge to the present sequence.

11^50 ends in 23001

11^500 ends in 2230001

11^5000 ends in 422300001

11^50000 ends in 34223000001

The last digits before the zeros tend to converge.

11^50 ends in 23001

11^500 ends in 2230001

11^5000 ends in 422300001

11^50000 ends in 34223000001

The last digits before the zeros converge to this sequence.

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).

Since a(28) = A376446(700001) = 9317, the present sequence has at least 28 terms.

If we merge A376446(n) and A377124(n*10), taking A376446(n) if and only if n is not a multiple of 10 and A376446(n*10) otherwise, we should get the sequence: 8, 64, 2486, 5, 4268, 8426, 2, 1, 4, 4862, 46, 82, 6248, 6, 6842, 8624, 2684, 28, 9, 7139, 3179, 19, 1397, 1793, 91, 3971, 7931, 9713, 9317 (which the author conjectures to be complete, as the present one).

Moreover, by construction, each term of this sequence is necessarily a circular permutation of the digits of one term of A376842 (e.g., a(4) = 2486 since A376842(4) = 6248).

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

For n = 1, 2, ..., 50, A381460(n) equals a(n)^n with the sole exception of n = 3 (i.e., A381460(3) = (5*5)^3 <> 55^3 = a(3)^3, and this is the only known value of n such that A381460(n) <> a(n)^n).

It is conjectured that a(n) is a multiple of 5 for any n > 2 (probabilistic argument).