cp's OEIS Frontend

a(n) > 10^floor(n/2).

All terms have last digit 1 or 9.

Squares of terms are listed in A085877.

Decimal representation of each term is formed by the reverse concatenation of initial terms of either A063006 or A091661.

Except for 3, there are no solutions for n>1 and m^2 == -1 (mod 10^n). See comment in A063006 under extensions. - Robert G. Wilson v, Jan 26 2013

If a(n)<(10^n)/2 then (10^n-a(n))^2 is also congruent (modulo 10^n), it is just not the least. - Robert G. Wilson v, Jan 26 2013

For any n, a(n) == 1 (mod 10^n), while it is not congruent to 1 modulo 10^(n + 1).

If n is not a multiple of 10, 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 has a total of n + 2 digits and they are n - 1 0s and 3 1s. Conversely, if there is a pair of positive integers (m, k) not ending with 0 and such that n := m*10^k, then 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 has n - k + 2 digits (three 1s and the rest are all 0s) and a(n) = (10^(n + 1) + 10^(n - k) + 1)^n.

Since a(n) == 1 (mod 5) for any n, the constant congruence speed of a(n) (i.e., V(a(n))) is guaranteed to be constant starting from height v_5(a(n) - 1) + 2 (for this sufficient condition, see “Number of stable digits of any integer tetration” in Links).

Then, for any positive integer n, a(n) is (exactly) a n-th perfect power (since, for any given n, 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 is divisible by 3 only once) and is also characterized by a constant congruence speed of n (for a strict proof of the general formula V((10^(t + k) + 10^(t - min{v_2(n), v_5(n)}) + 1)^n) = t, holding for any chosen positive integer k as long as t is an integer above min{v_2(n), v_5(n)} + 1, see Section 3 of “On the relation between perfect powers and tetration frozen digits” in Links).

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).

It is possible to prove that for any integer n >= 1 there are infinitely many prime numbers with a convergence speed equal to n (invoking Dirichlet's theorem on arithmetic progressions and considering the bases of the form 10^j - 1 + (2*k)*10^j = (2*k + 1)*10^j - 1, since their convergence speed is always equal to j and 10 never divides (2*k + 1)).

Since the only base with a convergence speed of 0 is a = 1 (and 1 is not a prime number), this sequence starts from a(1) = 2, while the convergence speed of 2 has been assumed to be 1 because the tetration 2^^b "freezes" one more rightmost digit for any unitary increment of b for any b >= 3 (the "constant" convergence speed of 2 is 1, even if V(2) = 0 according to the definition used in A317905). In general, a sufficient but not necessary condition to find the constant convergence speed of the base a, is to assume b >= a + 1 (e.g., V(2) corresponds to the new rightmost frozen digit going from 2^^(b >= 3) to 2^^(b + 1)).

This is not a strictly increasing sequence, since 3640476581907922943 = a(20) < a(19) = 23640476581907922943 (while a(19) < a(21) = 803640476581907922943).

For any n >= 3, a(n) == {1,3,7,9}(mod 10), since any prime above 5 is coprime to 10.

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).