A317824 -id:A317824 - cp's OEIS frontend

A317905 Convergence speed of m^^m, where m = A067251(n) and n >= 2. a(n) = f(m, m) - f(m, m - 1), where f(x, y) corresponds to the maximum value of k, such that x^^y == x^^(y + 1) (mod 10^k).

0, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 2, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 6, 1, 1, 3, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 5, 1, 1

Offset: 2

Marco Ripà, Aug 10 2018

It is possible to anticipate the convergence speed of m^^m, where ^^ indicates tetration or hyper-4 (e.g., 3^^4 = 3^(3^(3^3))), simply looking at the congruence (mod 25) of m. In fact, assuming m > 2, a(n) = 1 for any m == 2, 3, 4, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 21, 22, 23 (mod 25), and a(n) >= 2 otherwise.
It follows that 32/45 = 71.11% of the a(n) assume unitary value.
You can also obtain an arbitrary high convergence speed, such as taking the beautiful base b = 999...99 (9_9_9... n times), which gives a(n) = len(b), for any len(b) > 1. Thus, 99...9^^m == 99...9^^(m + 1) (mod m*10^len(b)), as proved by Ripà in "La strana coda della serie n^n^...^n", pages 25-26. In fact, m = 99...9 == 24 (mod 25) and a(m=24) > 1.
From Marco Ripà, Dec 19 2021: (Start)
Knowing the "constant congruence speed" of a given base (a.k.a. the convergence speed of the base m, assuming m > 2) is very useful in order to calculate the exact number of stable digits of all its tetrations of height b > 1. As an example, let us consider all the a(n) such that n is congruent to 4 (mod 9) (i.e., all the tetration bases belonging to the congruence class 5 (mod 10)). Then, the exact number of stable digits (#S(m, b)) of any tetration m^^b (i.e., the number of its last "frozen" digits) such that m is congruent to 5 (mod 10), for any b >= 3, can automatically be calculated by simply knowing that (under the stated constraint) the congruence speed of the m corresponds to the 2-adic valuation of (m^2 - 1) minus 1. Thus, let k = 1, 2, 3, ..., and we have that
If m = 20*k - 5, then #S(m, b > 2) = b*(v_2(m^2 - 1) - 1) + 1 = b*(v_2(m + 1) + 1);
If m = 20*k + 5, then #S(m, b > 2) = (b + 1)*(v_2(m^2 - 1) - 1) = (b + 1)*(v_2(m - 1));
If m = 5, then #S(m, 1) = 1, #S(m, 2) = 4, #S(m, b > 2) = 8 + 2*(b - 3).
(End)
For any n > 2, the value of a(n) depends on the congruence modulo 18 of n, since the constant congruence speed of m arises from the 14 nontrivial solutions of the fundamental equation y^5 = y in the (commutative) ring of decadic integers (e.g., y = -1 = ...9999 is a solution of y^5 = y, so it originates the law a(n) = min(v_2(m + 1), v_5(m + 1)) concerning every n belonging to the congruence class 0 modulo 18, as stated in the "Formula" section of the present sequence). - Marco Ripà, Feb 17 2022
a(n) satisfies the following multiplicative constraint: for each pair (m_1, m_2) of terms of A067251, a(m_1*m_2) is necessarily greater than or equal to the minimum between a(m_1) and a(m_2) (see Equation 2.4 and Appendix of "A Compact Notation for Peculiar Properties Characterizing Integer Tetration" in Links). - Gabriele Di Pietro, Apr 29 2025

			For m = 25, a(23) = 3 implies that 25^^(25 + i) freezes 3*i "new" rightmost digits (i >= 0).

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

Michel Marcus, reducetower.gp (from Math StackExchange).
Math StackExchange, Calculating a^n (mod m) in the general case
Marco Ripà, On the Convergence Speed of Tetration, ResearchGate (2018).
Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, Volume 26, 2020, Number 3, Pages 245—260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.
Marco Ripà and Gabriele Di Pietro, A Compact Notation for Peculiar Properties Characterizing Integer Tetration, Zenodo, 2025.
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

Cf. A067251, A317824, A317903, A349425, A370211, A370775, A371129, A371671, A371720, A372490, A373387.

Cf. A000007, A018247, A018248, A063006, A091661, A091663, A091664, A120817, A120818, A290372, A290373, A290374, A290375.

\\ uses reducetower.gp from links
f2(x,y) = my(k=0); while(reducetower(x, 10^k, y) == reducetower(x, 10^k, y+1), k++); k;
f1(n) = polcoef(x*(x+1)*(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1) / ((x-1)^2*(x^2+x+1)*(x^6+x^3+1)) + O(x^(n+1)), n, x); \\ A067251
a(n) =  my(m=f1(n)); f2(m, m) - f2(m, m-1);
lista(nn) = {for (n=2, nn, print1(a(n), ", "););} \\ Michel Marcus, Jan 27 2021

Let n > 2. For any integer c >= 0, if n is an element of the set {5, 7, 14, 17, 22, 23, 24, 29, 32, 39, 41, 45, 46}, then a(n + 45*c) >= 2; whereas a(n) = 1 otherwise. - Marco Ripà, Sep 28 2018
If n == 5 (mod 9), then a(n) = v_2(a(n)^2 - 1) - 1, where v_2(x) indicates the 2-adic valuation of x. - Marco Ripà, Dec 19 2021
If n == 1 (mod 18) and n <> 1, then a(n) = min(v_2(m - 1), v_5(m - 1)) (i.e., 1 plus the number of trailing zeros, if any, next to the rightmost digit of m);
if n == 10 (mod 18), then a(n) = min(v_2(m + 1), v_5(m - 1));
if n == {2,8}(mod 9) and n <> 2, then a(n) = v_5(m^2 + 1);
if n == {3,7}(mod 18), then a(n) = min(v_2(m + 1), v_5(n^2 + 1));
if n == {12,16}(mod 18), then a(n) = min(v_2(m - 1), v_5(n^2 + 1));
if n == 4 (mod 9), then a(n) = v_5(m + 1);
if n == 5 (mod 18), then a(n) = v_2(m - 1);
if n == 14 (mod 18), then a(n) = v_2(m + 1);
if n == 6 (mod 9), then a(n) = v_5(m - 1);
if n == 9 (mod 18), then a(n) = min(v_2(m - 1), v_5(m + 1));
if n == 0 (mod 18), then a(n) = min(v_2(m + 1), v_5(m + 1)) (i.e., number of digits of the rightmost repunit "9's" of m); where v_2(x) and v_5(x) indicates the 2-adic valuation of (x) and the 5-adic valuation of (x), respectively. - Marco Ripà, Feb 17 2022

Edited by Jinyuan Wang, Aug 30 2020

A373387 Constant congruence speed of the tetration base n (in radix-10), or -1 if n is a multiple of 10.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 1, 2, 1, 1, -1, 1, 1, 1, 1, 4, 1, 1, 2, 1, -1, 1, 1, 1, 2, 3, 2, 1, 1, 1, -1, 1, 2, 1, 1, 2, 1, 1, 1, 1, -1, 1, 1, 2, 1, 2, 1, 1, 1, 2, -1, 2, 1, 1, 1, 3, 1, 3, 1, 1, -1, 1, 1, 1, 1, 6, 1, 1, 3, 1, -1, 1, 1, 1, 2, 2, 2, 1, 1, 1, -1, 1, 2, 1

Offset: 0

Marco Ripà, Jun 02 2024

It has been proved that this sequence contains arbitrarily large entries, while a(0) = a(1) = 0 by definition (given the fact that 0^0 = 1 is a reasonable choice and then 0^^b is 1 if b is even, whereas 0^^b is 0 if b is even). For any nonnegative integer n which is not a multiple of 10, a(n) is given by Equation (16) of the paper "Number of stable digits of any integer tetration" (see Links).
Moreover, a sufficient condition for having a constant congruence speed of any tetration base n, greater than 1 and not a multiple of 10, is that 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)
  (u_5 and u_2 indicate the 5-adic and the 2-adic valuation of the argument, respectively).
Therefore b >= n + 1 is always a sufficient condition for the constancy of the congruence speed (as long as n > 1 and n <> 0 (mod 10)).
As a trivial application of this property, we note that the constant congruence speed of the tetration 3^^b is 1 for any b > 1, while 3^3 is not congruent to 3 modulo 10. Thus, we can easily calculate the exact number of the rightmost digits of Graham’s number, G(64) (see A133613), that are the same of the homologous rightmost digits of 3^3^3^... since 3^3 is not congruent to 3 modulo 10, while the congruence speed of n = 3 is constant from height 2 (see A372490). This means that the last slog_3(G(64))-1 digits of G(64) are the same slog_3(G(64))-1 final digits of 3^3^3^..., whereas the difference between the slog_3(G(64))-th digit of G(64) and the slog_3(G(64))-th digit of 3^3^3^... is congruent to 6 modulo 10.
The constant congruence speed of tetration satisfies the following multiplicative constraint: for each pair (n_1, n_2) of nonnegative integers whose product is not divisible by 10, a(n_1*n_2) is necessarily greater than or equal to the minimum between a(n_1) and a(n_2) (see Equation 2.4 and Appendix of "A Compact Notation for Peculiar Properties Characterizing Integer Tetration" in Links). - Marco Ripà, Apr 26 2025

			a(3) = 1 since 3^^b := 3^3^3^... freezes 1 more rightmost digit for each unit increment of b, starting from b = 2.

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

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, Volume 26, 2020, Number 3, Pages 245—260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43—61.
Marco Ripà, Twelve Python Programs to Help Readers Test Peculiar Properties of Integer Tetration, ResearchGate, 2024. See pp. 22-23, 27.
Marco Ripà and Gabriele Di Pietro, A Compact Notation for Peculiar Properties Characterizing Integer Tetration, Zenodo, 2025.
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, Graham's Number.
Wikipedia, Tetration.

Cf. A067251, A133613, A317824, A317903, A317905, A349425, A370211, A370775, A371129, A371671, A372490.

Cf. A000007, A018247, A018248, A063006, A091661, A091663, A091664, A120817, A120818, A290372, A290373, A290374, A290375.

def v2(n):
    count = 0
    while n % 2 == 0 and n > 0:
        n //= 2
        count += 1
    return count
def v5(n):
    count = 0
    while n % 5 == 0 and n > 0:
        n //= 5
        count += 1
    return count
def V(a):
    mod_20 = a % 20
    mod_10 = a % 10
    if mod_20 == 1:
        return min(v2(a - 1), v5(a - 1))
    elif mod_20 == 11:
        return min(v2(a + 1), v5(a - 1))
    elif mod_10 in {2, 8}:
        return v5(a ** 2 + 1)
    elif mod_20 in {3, 7}:
        return min(v2(a + 1), v5(a ** 2 + 1))
    elif mod_20 in {13, 17}:
        return min(v2(a - 1), v5(a ** 2 + 1))
    elif mod_10 == 4:
        return v5(a + 1)
    elif mod_20 == 5:
        return v2(a - 1)
    elif mod_20 == 15:
        return v2(a + 1)
    elif mod_10 == 6:
        return v5(a - 1)
    elif mod_20 == 9:
        return min(v2(a - 1), v5(a + 1))
    elif mod_20 == 19:
        return min(v2(a + 1), v5(a + 1))
def generate_sequence():
    sequence = []
    for a in range(1026):
        if a == 0 or a == 1:
            sequence.append(0)
        elif a % 10 == 0:
            sequence.append(-1)
        else:
            sequence.append(V(a))
    return sequence
sequence = generate_sequence()
print("a(0), a(1), a(2), ..., a(1025) =", ", ".join(map(str, sequence)))

a(n) = -1 iff n == 0 (mod 10), a(n) = 0 iff n = 1 or 2. Otherwise, a(n) >= 1 and it is given by Equation (16) from Ripà and Onnis.

A349425 n-th stable digit (in decimal system) of n^(n^(...^n)).

Original entry on oeis.org

1, 3, 3, 8, 0, 2, 5, 9, 3, 0, 7, 4, 5, 8, 6, 7, 8, 3, 1, 0, 1, 4, 4, 8, 7, 8, 0, 6, 4, 0, 9, 8, 0, 1, 0, 3, 5, 8, 7, 0, 6, 8, 2, 4, 1, 0, 3, 3, 3, 0, 2, 3, 5, 2, 3, 6, 8, 7, 0, 0, 7, 3, 3, 5, 0, 8, 0, 0, 8, 0, 7, 8, 8, 0, 3, 0, 6, 9, 1, 0, 7, 7, 6, 6, 5, 7, 3

Offset: 1

Marco Ripà, Nov 17 2021

The integer tetration (or hyper-4) n^^b is characterized by a well-known property involving its rightmost digits (as b grows an increasing number of the rightmost digits of n^^b are frozen - following the general rule described by Equation (16) of the linked paper "Number of stable digits of any integer tetration", p. 454).
In 2011 Ripà conjectured that, for any n >= 1, if b >= n + 2, then the n rightmost digits of n^^b are stable.
The above-mentioned paper, published in 2022, proved that this conjecture is true and also stated the stronger sufficient condition that the height of the hyperexponent is greater than or equal to tilde(v(a)) + 2, where tilde(v(a)) := v_5(a - 1) iff a == 1 (mod 5), v_5(a^2 + 1) iff a == {2, 3} (mod 5), v_5(a + 1) iff a == 4 (mod 5), v_2(a^2 - 1) - 1 iff a == 5 (mod 10), where v_2(x) = A007814(x) and v_5(x) = A112765(x) are the 2-adic and 5-adic valuations of x, respectively. - Marco Ripà, Jul 24 2024

			For n = 3, a(3) = 3 since 3^^5 == 387(mod 10^3). Thus, (387(mod 10^3) - 387(mod 10^2))/10^2 = 3.

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

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, 2020, 26(3), 245-260.
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.

Cf. A007814, A112765, A120962, A317824, A317903, A317905, A369624, A371048, A371078, A373387.

b:= proc(n) option remember; local m, v, w; m, w:= 10^n, n;
      do v:= n&^w mod m; if w=v then return v else w:=v fi od
    end:
a:= n-> `if`(irem(n, 10)=0, 0, iquo(b(n), 10^(n-1))):
seq(a(n), n=1..100);  # Alois P. Heinz, Nov 17 2021

def A349425(n):
    if n % 10 == 0: return 0
    m, n1, n2 = n, 10**n, 10**(n-1)
    while (k := pow(n,m,n1)) != m: m = k
    return k//n2 # Chai Wah Wu, Dec 19 2021

a(n) = (n^^(n + 2)(mod 10^n) - n^^(n + 2)(mod 10^(n - 1)))/10^(n - 1).

A317903 a(n) = A038394(n)^^A038394(n) (mod 10^len(A038394(n))), where ^^ indicates tetration or hyper-4 (e.g., 3^^4=3^(3^(3^3))).

Original entry on oeis.org

4, 76, 176, 4176, 314176, 91314176, 891314176, 80891314176, 88080891314176, 5288080891314176, 705288080891314176, 10705288080891314176, 2410705288080891314176, 912410705288080891314176, 42912410705288080891314176, 9242912410705288080891314176, 989242912410705288080891314176

Offset: 1

Marco Ripà, Aug 10 2018

For any n >= 2, a(n) (mod 10^len(A038394(n))) == a(n + 1) (mod 10^len(A038394(n))), where len(k) := number of digits in k. Assuming len(a(n))>1, this is a general property of every concatenated sequence with fixed rightmost digits (such as A014925 or A092447), as shown in Ripà's book "La strana coda della serie n^n^...^n".

			For n = 6, a(6) = 13117532^^13117532 (mod 10^8) == 91314176.

Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011, page 60. ISBN 978-88-6178-789-6

Marco Ripà, On the Convergence Speed of Tetration, ResearchGate (2018).

Cf. A038394, A068670, A171882 (tetration), A317824.

tmod(b, n) = {if (b % n == 0, return (0)); if (b % n == 1, return (1)); if (gcd(b, n)==1, return (lift(Mod(b, n)^tmod(b, lift(znorder(Mod(b, n))))))); lift(Mod(b, n)^(eulerphi(n) + tmod(b, eulerphi(n))));}
f(n) = fromdigits(concat([digits(p) | p<-Vecrev(primes(n))])); \\ A038394
a(n) = if (n==1, 4, my(x=f(n)); tmod(x, 10^#Str(x))); \\ Michel Marcus, Sep 12 2021

a(n) = (p(n)p(n-1)_p(n-2)...3_2)^^(p(n)_p(n-1)_p(n-2)...3_2) (mod 10^len(p(n)_p(n-1)_p(n-2)..._3_2)), where len(k) := number of digits in k.

More terms from Jinyuan Wang, Aug 30 2020

A321130 Values of m (mod 25) such that V(m) >= 2, where V(m) indicates the constant convergence speed of the tetration base m.

Original entry on oeis.org

0, 1, 5, 7, 15, 18, 24

Offset: 1

Marco Ripà, Oct 27 2018

This sequence represents the values of the base a such that a^^m, where ^^ indicates tetration or hyper-4 (e.g., 3^^4=3^(3^(3^3))), is characterized by a convergence speed at or above 2 (fast m-adic convergence). Only 26% of the positive integers belong to this list (see A317905).

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

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, Volume 26, 2020, Number 3, pp. 245—260.
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

Cf. A067251, A317824, A317903, A317905, A321131.

For m = 57, m (mod 25) == 7 and 7^^n has a convergence speed greater than 1, since A317905(m = 57) = 3 > 1 and also A317905(m = 7) = 2 > 1.

A321131 Values of m (mod 25), where A317905(m) = 1. Values of m (mod 25) such that V(m) = 1, where V(m) indicates the constant convergence speed of the tetration base m.

Original entry on oeis.org

2, 3, 4, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 21, 22, 23

Offset: 2

Marco Ripà, Oct 27 2018

This sequence represents the values of the base a such that a^^m, where ^^ indicates tetration or hyper-4 (e.g., 3^^4=3^(3^(3^3))), is characterized by a unitary convergence speed.

64% of the positive integers belong to this list (see A317905).

			For m = 47, m (mod 25) == 22 and 22^^n has a unitary convergence speed, since A317905(m = 47) = 1 = A317905(m = 22).

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

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, Volume 26, 2020, Number 3, pp. 245—260.
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

Cf. A067251, A317824, A317903, A317905, A321130.

A369624 Numbers formed by the rightmost decimal digits of n^(n^n) that are the same as those of n^(n^(n^n)).

Original entry on oeis.org

0, 1, 6, 87, 96, 8203125, 8656, 2343, 56, 289, 0, 6611, 16, 53, 36, 765380859375, 5616, 777, 76, 179, 0, 2421, 6, 47, 4976, 908447265625, 84203776, 83, 96, 669, 0, 6431, 4176, 713, 16, 8046875, 7136, 917, 6, 759, 0, 7641, 6, 1107, 56, 48828125, 9696, 23, 36

Offset: 0

Marco Ripà, Jan 27 2024

The common digits might include leading 0's (such as at n = 5) and they are discarded (in particular, a(0) = 0 indicates that the corresponding zero digit term results in a 0 integer entry).

a(k*10) = 0 for every positive integer k, since (k*10)^((k*10)^(k*10)) and (k*10)^((k*10)^((k*10)^(k*10))) have in common only their rightmost (k*10)^(k*10) digits.

			For n = 3, 3^(3^3) = 7625597484987 and 3^(3^(3^3)) == 387 (mod 1000) so there are two common final digits a(3) = 87.

Jorge Jiménez Urroz and José Luis Andrés Yebra, On the Equation a^x == x (mod b^n), Journal of Integer Sequences, Article 09.8.8, 2009.
Marco Ripà, Congruence speed of tetration bases ending with 0, arXiv:2402.07929 [math.NT], 2024.
Eric Weisstein's World of Mathematics, Joyce Sequence.
Wikipedia, Knuth's up-arrow notation.

Cf. A002488, A317824, A317903, A317905, A349425.

a(n) = A002488(n) (mod 10^k), where k is such that n^(n^n) == n^(n^(n^n)) (mod 10^k) and n^(n^n) <> n^(n^(n^n)) (mod 10^(k+1)).

A337392 Minimum m such that the convergence speed of m^^m is equal to n >= 2, where A317905(n) represents the convergence speed of m^^m (and m = A067251(n), the n-th non-multiple of 10).

Original entry on oeis.org

5, 25, 15, 95, 65, 385, 255, 1535, 1025, 6145, 4095, 24575, 16385, 98305, 65535, 393215, 262145, 1572865, 1048575, 6291455, 4194305, 25165825, 16777215, 100663295, 67108865, 402653185, 268435455, 1610612735, 1073741825, 6442450945, 4294967295, 25769803775

Offset: 2

Marco Ripà, Aug 25 2020

This sequence has an unbounded number of terms, since it has been proved that the congruence speed (aka "convergence speed") of m^^m (an integer number by definition) covers any value from zero (iff m = 1) to infinity. In particular, for any n >= 2, a(n) == 5 (mod 10).
From Marco Ripà, Dec 19 2021: (Start)
Moreover, given any m which is congruent to 5 (mod 10), the congruence speed of m corresponds to the 2-adic valuation of (m^2 - 1) minus 1 (e.g., the congruence speed of 15 is equal to 4 since (15^2 - 1) is divisible by 2 exactly 5 times, so that 5 - 1 = 4 = congruence speed of the tetration base 15).
The aforementioned result, let us easily calculate the exact number of stable digits (#S(m, b)) of any tetration m^^b (i.e., the number of its last "frozen" digits) such that m is congruent to 5 (mod 10), for any b >= 3, as follows:
Let k = 1, 2, 3, ...
If m = 20*k - 5, then #S(m, b > 2) = b*(v_2(m^2 - 1) - 1) + 1;
If m = 20*k + 5, then #S(m, b > 2) = (b + 1)*(v_2(m^2 - 1) - 1);
If m = 5, then #S(m, 1) = 1, #S(m, 2) = 3, #S(m, 3) = 4, #S(m, b > 3) = 2.
(End)

			For n = 4, a(4) = 15 by Corollary 1 of "https://doi.org/10.7546/nntdm.2021.27.4.43-61" (see Equation 20). - _Marco Ripà_, Dec 19 2021

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

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, Volume 26, 2020, Number 3, Pages 245—260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,4).

Cf. A317824, A317903, A317905, A321130.

a(n) = 2^n*(2*cos(Pi*(n-1)/2) - 4*sin(Pi*(n-1)/2) + 5) + 1 iff n == {2,3} (mod 4), 2^n*(-2*cos(Pi*(n-1)/2) + 4*sin(Pi*(n-1)/2) + 5) - 1 iff n == {0,1} (mod 4), for n >= 2.
From Bruno Berselli, Sep 11 2020: (Start)
O.g.f.: 5*x^2*(1 + 5*x + 4*x^3)/((1 - 2*x)*(1 + 2*x)*(1 + x^2)).
a(n) = (2 - (-1)^n)*2^n + i^((n+1)*(n+2)), with i = sqrt(-1). (End)
From Marco Ripà, Dec 19 2021: (Start)
n = v_2(a(n)^2 - 1) - 1, where v_2(x) indicates the 2-adic valuation of x. (End)

A337833 Minimum m coprime to 5 such that the convergence speed of m^^m := m^(m^^(m-1)) is equal to n >= 0, where A317905(n) represents the convergence speed of m^^m (and m = A047201(n), the n-th non-multiple of 5).

Original entry on oeis.org

1, 2, 7, 57, 182, 3124, 1068, 32318, 390624, 280182, 3626068, 23157318, 120813568, 1220703124, 1097376068, 11109655182, 49925501068, 762939453124, 355101282318, 19073486328124, 15613890344818, 365855836217682, 2384185791015624

Offset: 0

Marco Ripà, Sep 24 2020

Let "s" denote the last digit of m, and V(m(s)) its convergence speed. For any n, the smallest bases that are not congruent to 5 modulo 10 (as in A337392) cannot be such that s = 6, since V(m(6)) = V(m(4)) + 2.

			For n = 19, a(19) = 19073486328124 is the smallest base (radix-10) of the tetration m^^m which is characterized by a congruence speed of 19.

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

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, 2020, 26(3), 245-260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.

Cf. A317824, A317903, A317905, A321130, A337392.

A356946 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)).

Original entry on oeis.org

1, 0, 2, 3, 12, 7, 12, 7, 9

Offset: 1

Marco Ripà, Sep 05 2022

a(10) = 10^^9 is too large to include. In general, if n is a multiple of 10, then a(n) is given by the number of trailing zeros that appear at the end of n^^n.

This follows from the constancy of the "congruence speed" (AKA "convergence speed" here on the OEIS) of hyper-3 for any exponentiation base which is a multiple of 10, otherwise the congruence speed is constant only for hyper-4 and it is strictly positive for any tetration base n >= 1 that is not a multiple of 10 (for an explicit formula to calculate a(n) for any n, see the linked paper entitled "Number of stable digits of any integer tetration").

			For n = 3, 3^3^3 is congruent to 3^3^3^3 (mod 10^2) and 3^3^3 is not congruent to 3^3^3^3 (mod 10^3). Thus, a(3) = 2.

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

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, 2020, 26(3), 245-260.
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

Cf. A317824, A317903, A317905, A349425.

cp's OEIS Frontend

A317905 Convergence speed of m^^m, where m = A067251(n) and n >= 2. a(n) = f(m, m) - f(m, m - 1), where f(x, y) corresponds to the maximum value of k, such that x^^y == x^^(y + 1) (mod 10^k).

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A373387 Constant congruence speed of the tetration base n (in radix-10), or -1 if n is a multiple of 10.

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A349425 n-th stable digit (in decimal system) of n^(n^(...^n)).

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A317903 a(n) = A038394(n)^^A038394(n) (mod 10^len(A038394(n))), where ^^ indicates tetration or hyper-4 (e.g., 3^^4=3^(3^(3^3))).

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A321130 Values of m (mod 25) such that V(m) >= 2, where V(m) indicates the constant convergence speed of the tetration base m.

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A321131 Values of m (mod 25), where A317905(m) = 1. Values of m (mod 25) such that V(m) = 1, where V(m) indicates the constant convergence speed of the tetration base m.

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A369624 Numbers formed by the rightmost decimal digits of n^(n^n) that are the same as those of n^(n^(n^n)).

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A337392 Minimum m such that the convergence speed of m^^m is equal to n >= 2, where A317905(n) represents the convergence speed of m^^m (and m = A067251(n), the n-th non-multiple of 10).