cp's OEIS Frontend

By Definition 1.2 of “Number of stable digits of any integer tetration” (see Links), let bar_b be the smallest hyperexponent of the tetration m^^b such that the congruence speed of m is constant.

Then, assuming that n is not a multiple of 10, we define as asymptotic phase shift (original name "sfasamento asintotico” - see References, “La strana coda della serie n^n^...^n”, Chapter 7) the subsequence of the 4 (or 2 or even 1) congruence classes of the differences between the rightmost non-stable digits of n^^(bar_b) (i.e., the least significant digit of the tetration n^^(bar_b) that is not frozen as we move to n^^(b + 1)) and the corresponding digit of n^^(bar_b + 1), and ditto for (n^^(bar_b + 1) and n^^(bar_b + 2)), (n^^(bar_b + 2) and n^^(bar_b + 3)), (n^^(bar_b + 3) and n^^(bar_b + 4)).

Now, let us indicate this subsequence as [s_1, s_2, s_3, s_4] and then, if s_1 = s_3 and also s_2 = s_4, we transform [s_1, s_2, s_3, s_4] into [s_1, s_2], and lastly, if s_1 = s_2, we transform [s_1, s_2] into [s_1].

Finally, we get a(n) by juxtaposing all the 4 (or 2, or 1) digits inside [...] (e.g., if n = 3, then bar_b = 2 so that we get s_1 = 4 = s_3 and s_2 = 6 = s_4, and consequently a(3) = 46 instead of 4646 - see Definition 3.2 of "Graham's number stable digits: an exact solution" in Links).

Assuming that n is not a multiple of 10, in general, for any given pair of nonnegative integers c and k, the congruence class modulo 10 of the difference between the least significant digits of n^^(bar_b+c+4*k) and n^^(bar_b+c+1+4*k) does not depend on k. Moreover, if s_1 <> s_2, then s_1 + s_3 = s_2 + s_4 = 10.

In detail, if the last digit of n is 5, then a(n) = 5, while if n is coprime to 5, a(n) mandatorily belongs to the following set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 28, 37, 46, 64, 73, 82, 91, 1397, 1793, 2486, 2684, 3179, 3971, 4268, 4862, 6248, 6842, 7139, 7931, 8426, 8624, 9317, 9713}.

The author conjectures that if n is not a multiple of 10, then a(n) is necessarily an element of the subset {2, 4, 5, 6, 8, 9, 19, 28, 46, 64, 82, 2486, 2684, 3971, 4268, 4862, 6248, 6842, 7931, 8426, 8624}.

As a mere consequence of a(3) = 46 and the fact that congruence speed of 3 is 0 at height 1 and 1 at any height above 1, it follows that 4 is equal to the difference between the slog_3(G)-th least significant digit of Graham's number, G, and the slog_3(G)-th least significant digit of any power tower of the form 3^3^3^... whose height is above slog_3(G) (where slog(...) indicates the super-logarithm of the argument).

By definition, a(n) consists of a circular permutation of the digits of A376446(n).

The base-15 version of A120817. A, B, C, D, and E are the standard notations for the hexadecimal digits 10, 11, 12, 13, and 14, respectively.

Conjecture: If k is the number of prime factors congruent to 1 (mod 4) of an integer n, then there are exactly k n-adic integers x satisfying x^5 = x, while not satisfying x^h = x for h = 2, 3, or 4. This does not count -x, which also satisfies, in each case. - Patrick A. Thomas, Mar 31 2020

Let b := 2 + v(n), where v(n) is equal to u_5(n - 1) iff n == 1 (mod 5), u_5(n^2 + 1) iff n == 2,3 (mod 5), u_5(n + 1) iff n == 4 (mod 5), u_2(n^2 - 1) - 1 iff n == 5 (mod 10), while u_5 and u_2 indicate the 5-adic and the 2-adic valuation of the argument (respectively).

Assuming that n is not a multiple of 10, we define as modular phase shift (original name "sfasamento modulare" - see References, “La strana coda della serie n^n^...^n”, Chapter 7) the subsequence of the 4 (or 2 or even 1) congruence classes of the differences between the rightmost non-stable digits of n^^b (i.e., the least significant digit of the tetration n^^b that is not frozen as we move to n^^(b + 1)) and the corresponding digit of n^^(b + 1), and ditto for (n^^(b + 1) and n^^(b + 2)), (n^^(b + 2) and n^^(b + 3)), (n^^(b + 3) and n^^(b + 4)).

Now, let us indicate this subsequence as [s_1, s_2, s_3, s_4] and then, if s_1 = s_3 and also s_2 = s_4, we transform [s_1, s_2, s_3, s_4] into [s_1, s_2], and lastly, if s_1 = s_2, we transform [s_1, s_2] into [s_1].

Finally, we get a(n) by juxtaposing all the 4 (or 2, or 1) digits inside [...] (e.g., if n = 3, then b = 2 + 1 so that we get s_1 = 6 = s_3 and s_2 = 4 = s_4, and consequently a(3) = 64 instead of 6464).

Assuming that n is not a multiple of 10, in general, for any given pair of nonnegative integers c and k, the congruence class modulo 10 of the difference between the least significant digits of n^^(b+c+4*k) and n^^(b+c+1+4*k) does not depend on k. Moreover, if s_1 <> s_2, then s_1 + s_3 = s_2 + s_4 = 10.

In detail, if the last digit of n is 5, then a(n) = 5, while if n is coprime to 5, a(n) mandatorily belongs to the following set: {1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 28, 37, 46, 64, 73, 82, 91, 1397, 1793, 2486, 2684, 3179, 3971, 4268, 4862, 6248, 6842, 7139, 7931, 8426, 8624, 9317, 9713}.

The author conjectures that if n is not a multiple of 10, then a(n) is necessarily an element of the subset {2, 4, 5, 6, 8, 9, 19, 28, 46, 64, 82, 91, 1397, 1793, 2486, 2684, 3179, 3971, 4268, 4862, 6248, 6842, 7139, 7931, 8426, 8624, 9317, 9713}.

As a mere consequence of a(3) = 64 and the fact that congruence speed of 3 is 0 at height 1 and 1 at any height above 1, it follows that 4 is equal to the difference between the slog_3(G)-th least significant digit of Graham's number, G, and the slog_3(G)-th least significant digit of any power tower of the form 3^3^3^... whose height is above slog_3(G) (where slog(...) indicates the super-logarithm of the argument).

Letters A through J represent the base-20 digits 10 through 19, respectively.

Conjecture: There exists a nontrivial n-adic integer x satisfying x^5 = x, and x^2, x^3, and x^4 are not x, if and only if n has a prime factor of the form 4k+1. Further, there is one nontrivial pair (x and -x) for each different prime factor of the form 4k+1.

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).