A386005 Number of stable digits of the integer tetration n^^n (i.e., maximum nonnegative integer m such that n^^n is congruent modulo 10^m to n^^(n + 1)), or -1 if n is a multiple of 10.

0, 2, 3, 12, 7, 12, 7, 9, -1, 12, 11, 13, 13, 61, 17, 17, 32, 19, -1, 22, 20, 22, 46, 78, 54, 26, 27, 29, -1, 32, 62, 33, 33, 71, 37, 37, 36, 39, -1, 42, 40, 84, 43, 92, 47, 46, 47, 98, -1, 103, 51, 53, 53, 166, 57, 171, 56, 59, -1, 62, 60, 62, 63, 396, 67, 66

Offset: 2

Author

Marco Ripà, Jul 14 2025

Keywords

• sign
• base
• hard

Comments

If n is divisible by 10, the corresponding a(n) would be too large to be include in the data section (it would be equal to the number of trailing zeros that appear at the end of n^^n).

This sequence directly follows from the constancy of the "congruence speed" property characterizing tetration (for an explicit formula to calculate a(n) for any n not a multiple of 10, see the linked paper entitled "Number of stable digits of any integer tetration").

Examples

			For n = 5, 5^5^5^5^5 is congruent to 5^5^5^5^5^5 (mod 10^12) and 5^5^5^5^5 is not congruent to 5^5^5^5^5^5 (mod 10^13). Thus, a(5) = 12.
		

References

• Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6.

Links

• Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.
• Marco Ripà and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441-457.
• Wikipedia, Tetration

Crossrefs

Cf. A317824, A317903, A317905, A349425, A356946.

Programs

Python

def last_k_digits_tetration(base, height, k):
    mod = 10 ** k
    result = base
    for _ in range(height - 1):
        result = pow(base, result, mod)
    return str(result).zfill(k)
def count_stable_digits(base, k=500):
    try:
        x = last_k_digits_tetration(base, base, k)
        y = last_k_digits_tetration(base, base + 1, k)
        count = 0
        for i in range(1, k + 1):
            if x[-i] == y[-i]:
                count += 1
            else:
                break
        return count
    except:
        return -1
for n in range(2, 101):
    if n % 10 == 0:
        print(f"n = {n}: -1")
    else:
        print(f"n = {n}: {count_stable_digits(n)}")

Formula

If n == 0 (mod 10), then a(n) = -1, and a(n) = A356946(n) otherwise.