A290375 -id:A290375 - 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.

A120817 10-adic integer x=...07839804103263499879186432 satisfying x^5 = x; also x^3 = -x = A120818; (x^2)^3 = x^2 = A091664; (x^4)^2 = x^4 = A018248.

Original entry on oeis.org

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

Offset: 0

Paul D. Hanna, Jul 06 2006

			x equals the limit of the (n+1) trailing digits of 2^(5^n):
2^(5^0)=(2), 2^(5^1)=(32), 2^(5^2)=33554(432),
2^(5^3)=4253529586511730793292182592897102(6432), ...
x=...93839649523223304553032451441224165530407839804103263499879186432.
x^2=...0557423423230896109004106619977392256259918212890624 (A091664).
x^3=...6695446967548558775834469592160195896736500120813568 (A120818).
x^4=...9442576576769103890995893380022607743740081787109376 (A018248).
x^5=...3304553032451441224165530407839804103263499879186432 = x.

Seiichi Manyama, Table of n, a(n) for n = 0..9999 (terms 0..999 from Paul D. Hanna)
Patrick A. Thomas, Solutions to x^5=x up to base 100

x^5 = x: this sequence (...6432), A120818 (...3568), A290372 (...5807), A290373 (...2943), A290374 (...7057), A290375 (...4193).

Cf. A091664, A018248.

{a(n)=local(b=2,v=[]);for(k=1,n+1,b=b^5%10^k;v=concat(v,(10*b\10^k)));v[n+1]}

{a(n)=if(n<0, 0, lift(chinese(Mod(truncate( teichmuller(2+O(5^(n+1)))), 5^(n+1)), Mod(0, 2^(n+1))))\10^n)} /* Michael Somos, Oct 03 2006 */

x = 10-adic limit_{n->infinity} 2^(5^n).

A290372 10-adic integer x = ...5807 satisfying x^5 = x.

Original entry on oeis.org

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

Offset: 0

Seiichi Manyama, Jul 28 2017

Also x^2 = A091661.

			     7^5 -    7 == 0 mod 10,
     7^5 -    7 == 0 mod 10^2,
   807^5 -  807 == 0 mod 10^3,
  5807^5 - 5807 == 0 mod 10^4.
From _Seiichi Manyama_, Aug 01 2019: (Start)
  2^(5^0) - 5^(2^0) ==    7 mod 10,
  2^(5^1) - 5^(2^1) ==    7 mod 10^2,
  2^(5^2) - 5^(2^2) ==  807 mod 10^3,
  2^(5^3) - 5^(2^3) == 5807 mod 10^4. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..9999

Cf. A120817, A120818, A290373, A290374, A290375.

Cf. A091661, A018247.

def P(n)
  s1, s2 = 2, 8
  n.times{|i|
    m = 10 ** (i + 1)
    (0..9).each{|j|
      k1, k2 = j * m + s1, (9 - j) * m + s2
      if (k1 ** 5 - k1) % (m * 10) == 0 && (k2 ** 5 - k2) % (m * 10) == 0
        s1, s2 = k1, k2
        break
      end
    }
  }
  s1
end
def Q(s, n)
  n.times{|i|
    m = 10 ** (i + 1)
    (0..9).each{|j|
      k = j * m + s
      if (k ** 2 - k) % (m * 10) == 0
        s = k
        break
      end
    }
  }
  s
end
def A290372(n)
  str = (10 ** (n + 1) + P(n) - Q(5, n)).to_s.reverse
  (0..n).map{|i| str[i].to_i}
end
p A290372(100)

p = A120817 = ...186432, q = A018247 = ...890625, x = p - q = ...295807.

A290373 10-adic integer x = ...2943 satisfying x^5 = x.

Original entry on oeis.org

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

Offset: 0

Seiichi Manyama, Jul 28 2017

Also x^2 = A091661.

			     3^5 -    3 == 0 mod 10,
    43^5 -   43 == 0 mod 10^2,
   943^5 -  943 == 0 mod 10^3,
  2943^5 - 2943 == 0 mod 10^4.
From _Seiichi Manyama_, Aug 01 2019: (Start)
  8^(5^0) - 5^(2^0) ==    3 mod 10,
  8^(5^1) - 5^(2^1) ==   43 mod 10^2,
  8^(5^2) - 5^(2^2) ==  943 mod 10^3,
  8^(5^3) - 5^(2^3) == 2943 mod 10^4. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..9999

Cf. A120817, A120818, A290372, A290374, A290375.

Cf. A091661, A120818.

def P(n)
  s1, s2 = 2, 8
  n.times{|i|
    m = 10 ** (i + 1)
    (0..9).each{|j|
      k1, k2 = j * m + s1, (9 - j) * m + s2
      if (k1 ** 5 - k1) % (m * 10) == 0 && (k2 ** 5 - k2) % (m * 10) == 0
        s1, s2 = k1, k2
        break
      end
    }
  }
  s2
end
def Q(s, n)
  n.times{|i|
    m = 10 ** (i + 1)
    (0..9).each{|j|
      k = j * m + s
      if (k ** 2 - k) % (m * 10) == 0
        s = k
        break
      end
    }
  }
  s
end
def A290373(n)
  str = (10 ** (n + 1) + P(n) - Q(5, n)).to_s.reverse
  (0..n).map{|i| str[i].to_i}
end
p A290373(100)

p = A120818 = ...813568, q = A018247 = ...890625, x = p - q = ...922943.

A290374 10-adic integer x = ...7057 satisfying x^5 = x.

Original entry on oeis.org

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

Offset: 0

Seiichi Manyama, Jul 28 2017

Also x^2 = A091661.

			     7^5 -    7 == 0 mod 10,
    57^5 -   57 == 0 mod 10^2,
    57^5 -   57 == 0 mod 10^3,
  7057^5 - 7057 == 0 mod 10^4.
From _Seiichi Manyama_, Aug 01 2019: (Start)
  2^(5^0) + 5^(2^0) ==    7 mod 10,
  2^(5^1) + 5^(2^1) ==   57 mod 10^2,
  2^(5^2) + 5^(2^2) ==   57 mod 10^3,
  2^(5^3) + 5^(2^3) == 7057 mod 10^4. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..9999

x^5 = x: A120817 (...6432), A120818 (...3568), A290372 (...5807), A290373 (...2943), this sequence (...7057), A290375 (...4193).

Cf. A091661, A018247.

def P(n)
  s1, s2 = 2, 8
  n.times{|i|
    m = 10 ** (i + 1)
    (0..9).each{|j|
      k1, k2 = j * m + s1, (9 - j) * m + s2
      if (k1 ** 5 - k1) % (m * 10) == 0 && (k2 ** 5 - k2) % (m * 10) == 0
        s1, s2 = k1, k2
        break
      end
    }
  }
  s1
end
def Q(s, n)
  n.times{|i|
    m = 10 ** (i + 1)
    (0..9).each{|j|
      k = j * m + s
      if (k ** 2 - k) % (m * 10) == 0
        s = k
        break
      end
    }
  }
  s
end
def A290374(n)
  str = (P(n) + Q(5, n)).to_s.reverse
  (0..n).map{|i| str[i].to_i}
end
p A290374(100)

p = A120817 = ...186432, q = A018247 = ...890625, x = p + q = ...077057.

A376842 Asymptotic phase shift of the tetration base n written by juxtaposing its representative congruence classes modulo 10 (i.e., the 1, 2, or 4, distinct sfasamenti taken by starting from the minimum height such that the congruence speed of n is constant), and a(n) = -1 if n is a multiple of 10.

Original entry on oeis.org

8, 46, 6248, 5, 4268, 2684, 6842, 2, -1, 4, 46, 6248, 8, 5, 64, 4, 28, 4862, -1, 6248, 6248, 2486, 8, 5, 46, 2684, 4862, 6842, -1, 8426, 8426, 28, 28, 5, 4, 2684, 28, 82, -1, 64, 4268, 46, 4268, 5, 4862, 4, 2, 6842, -1, 5, 6, 8, 6248, 5, 8426, 2684, 4268, 2

Offset: 2

Marco Ripà, Oct 06 2024

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).

			a(3) = 46 since the congruence speed of 3^^b becomes constant starting from height 2 and its value is 0 for b = 1 and 1 for any b >= 2, then (3^3 - 3^(3^3))/10 == 4 (mod 10) while (3^(3^3) - 3^(3^(3^3)))/10^2 == 6 (mod 10).

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 and Luca Onnis, Number of stable digits of any integer tetration, Notes on Number Theory and Discrete Mathematics, 2022, 28(3), 441-457.
Marco Ripà, Graham's number stable digits: an exact solution, arXiv:2411.00015 [math.GM], 2024.
Marco Ripà, Twelve Python Programs to Help Readers Test Peculiar Properties of Integer Tetration, ResearchGate, 2024. See p. 22.
Wikipedia, Graham's Number.
Wikipedia, Tetration.

Cf. A317905, A370211, A372490, A373387, A376446.

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

A376446 Modular phase shift of the tetration base n written by juxtaposing its representative congruence classes modulo 10 taken by starting from height 2 + v(n), and a(n) = -1 if n is a multiple of 10.

Original entry on oeis.org

8, 64, 2486, 5, 4268, 8426, 8426, 2, -1, 4, 64, 4862, 8, 5, 46, 4, 82, 6248, -1, 4862, 6248, 4862, 8, 5, 64, 6842, 8624, 4268, -1, 4268, 2684, 28, 82, 5, 4, 8426, 28, 82, -1, 46, 4268, 46, 2684, 5, 4862, 4, 2, 2684, -1, 5, 6, 8, 2486, 5, 4268, 2684, 4268, 2

Offset: 2

Marco Ripà, Sep 23 2024

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).

			a(3) = 64 since the congruence speed of 3 at height 3^^(2 + u_5(3^2 + 1)) is constant and its value is 1, so (3^(3^3) - 3^(3^(3^3)))/10^2 == 6 (mod 10) while (3^(3^(3^3)) - 3^(3^(3^(3^3))))/10^3 == 4 (mod 10).

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. 7, 9, 12, 14, 27.
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. A317905, A370211, A372490, A373387.

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

# Function to calculate the p-adic valuation
def p_adic_valuation(n, p):
    count = 0
    while n % p == 0 and n != 0:
        n //= p
        count += 1
    return count
# Function to calculate tetration (tower of powers)
def tetration(base, height, last_digits=500):
    results = [base]
    for n in range(1, height):
        result = pow(base, results[-1], 10**last_digits)  # Only the last last_digits digits
        results.append(result)
    return results
# Function to find the first non-zero difference and compute modulo 10
def find_difference_mod_10(tetrations):
    differences = []
    for n in range(len(tetrations) - 1):
        string_n = str(tetrations[n]).zfill(500)  # Pad with zeros if needed
        string_n_plus_1 = str(tetrations[n+1]).zfill(500)
        # Find the first difference starting from the rightmost digit
        for i in range(499, -1, -1):  # From right to left
            if string_n[i] != string_n_plus_1[i]:
                difference = (int(string_n[i]) - int(string_n_plus_1[i])) % 10
                differences.append(difference)
                break
    return differences
# Function to determine the first hyperexponent based on modulo 5 congruences
def calculate_initial_exponent(a):
    mod_5 = a % 5
    if mod_5 == 1:
        valuation = p_adic_valuation(a - 1, 5)
        initial_exponent = valuation + 2
    elif mod_5 in [2, 3]:
        valuation = p_adic_valuation(a**2 + 1, 5)
        initial_exponent = valuation + 2
    elif mod_5 == 4:
        valuation = p_adic_valuation(a + 1, 5)
        initial_exponent = valuation + 2
    else:
        valuation = p_adic_valuation(a**2 - 1, 2)
        initial_exponent = valuation + 1
    return initial_exponent
# Main logic
try:
    a = int(input("Enter a positive integer greater than 1: "))
    # Check if the number ends with 0
    if a % 10 == 0:
        print(-1)
    elif a <= 1:
        raise ValueError("The number must be greater than 1.")
    else:
        # Calculate the initial exponent based on modulo 5 congruence
        initial_exponent = calculate_initial_exponent(a)
        # Generate tetrations for 30 iterations and the last 500 digits
        tetrations = tetration(a, 30, last_digits=500)
        # Find the modulo 10 differences for the 4 required iterations
        mod_10_differences = find_difference_mod_10(tetrations[initial_exponent-1:initial_exponent+4])
        # Optimization of the output
        if mod_10_differences[:2] == mod_10_differences[2:]:
            mod_10_differences = mod_10_differences[:2]
        if len(set(mod_10_differences)) == 1:
            mod_10_differences = [mod_10_differences[0]]
        # Convert the list of differences into a string without brackets or commas
        result_str = ''.join(map(str, mod_10_differences))
        # Print the optimized result
        print(f"a({a}) = {result_str}")
except Exception as e:
    print(f"ERROR!\n{e}")

A379906 Smallest integer greater than 1 and not ending in 0 whose congruence speed is not constant at height n (see A373387).

Original entry on oeis.org

2, 2, 5, 307, 807, 72943, 795807, 1295807, 16295807, 166295807, 16666295807, 31666295807, 81666295807, 8581666295807, 26581907922943, 503581666295807, 2003581666295807, 90476581907922943, 140476581907922943, 6847003581666295807, 61847003581666295807, 911847003581666295807

Offset: 1

Marco Ripà, Jan 05 2025

The present sequence is a subsequence of A068407.
Although the congruence speed of any integer m > 1 not divisible by 10 is certainly stable at height m + 1 (for a tighter upper bound see "Number of stable digits of any integer tetration" in Links), this sequence contains infinitely many terms, implying the existence of infinitely many tetration bases whose congruence speed does not stabilize in less than b + 1 iterations, for any chosen positive integer b.
As a nontrivial example, the congruence speed of m := 45115161423787862411847003581666295807 becomes stable at height 41, which exactly matches the mentioned tight bound, for the numbers ending in 2, 3, 7, or 8, of v_5(45115161423787862411847003581666295807^2 + 1) + 2, where v_5(...) indicates the 5-adic valuation of the argument.

			a(5) = 807 since the congruence speed of 807 is 0 at height 1, 4 at heights 2, 3, 4, and 5, finally matching the value of the constant congruence speed of 807 at height 6 (and it is the smallest integer whose congruence speed stabilizes at height 6 or above).

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.
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. A068407, A290372, A290373, A290374, A290375, A317905, A370211, A370775, A371129, A371671, A372490, A373387.

As long as ceiling(log_10(a(n))) < n, for any n > 3, the least significant floor(log_10(a(n))) digits of a(n) (from right to left) are given by the first floor(log_10(a(n))) entries of A290372(n), or A290373(n), or A290374(n), or A290375(n) (i.e., all but the first digit of each a(n) are described by ({5^2^k}_oo + {2^5^k}_oo) := ...17196359523418092077057, ({5^2^k}_oo - {2^5^k}_oo) := ...37588152996418333704193, (- {5^2^k}_oo + {2^5^k}_oo) := ...2411847003581666295807, and (- {5^2^k}_oo - {2^5^k}_oo) := ...2803640476581907922943).

A381460 Smallest n-th perfect power greater than 1 satisfying A373387(a(n)) = n.

Original entry on oeis.org

2, 49, 15625, 625, 7737809375, 735091890625, 1253790880222890625, 6634204312890625, 47312447868976594992787109375, 72624607478879073313928212890625, 471781339858152691906169456697218212890625, 1344888824246298437178134918212890625

Offset: 1

Marco Ripà, Feb 24 2025

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).

			a(3) = 15625 since 15625 = 25^3 and 15625 is the smallest perfect cube whose constant congruence speed equals 3.

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.
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.
Marco Ripà, On the relation between perfect powers and tetration frozen digits, Journal of AppliedMath, 2024, 2(5), 1771.
Math Overflow, Closed form for the general term of 2, 49, 15625, 625, ....
Wikipedia, Tetration.

Cf. A018247, A018248, A063006, A091661, A091663, A091664, A120817, A120818, A290372, A290373, A290374, A290375, A317905, A373387, A379243.

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).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

PARI

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Python

Formula

A120817 10-adic integer x=...07839804103263499879186432 satisfying x^5 = x; also x^3 = -x = A120818; (x^2)^3 = x^2 = A091664; (x^4)^2 = x^4 = A018248.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

PARI

PARI

Formula

A290372 10-adic integer x = ...5807 satisfying x^5 = x.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Ruby

Formula

A290373 10-adic integer x = ...2943 satisfying x^5 = x.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Ruby

Formula

A290374 10-adic integer x = ...7057 satisfying x^5 = x.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Ruby

Formula

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

A376446 Modular phase shift of the tetration base n written by juxtaposing its representative congruence classes modulo 10 taken by starting from height 2 + v(n), and a(n) = -1 if n is a multiple of 10.

Views