A091661 -id:A091661 - 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

A091664 10-adic integer x=.....06619977392256259918212890624 satisfying x^3 = x.

Original entry on oeis.org

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

Offset: 0

Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004

Let a,b be integers defined in A018247, A018248 satisfying a^2=a, b^2=b, obviously a^3=a, b^3=b; let c,d,e,f be integers defined in A091661, A063006, A091663, A091664; then c^3=c, d^3=d, e^3=e, f^3=f, c+d=1, a+e=1, b+f=1, b+c=a, d+f=e, a+f=c, a=f+1, b=e+1, cd=-1, af=-1, gh=-1 where -1=.....999999999.

			x equals the limit of the (n+1) trailing digits of 4^(5^n):
4^(5^0) = (4), 4^(5^1) = 10(24), 4^(5^2) = 1125899906842(624), ...
x = ...0557423423230896109004106619977392256259918212890624.

Seiichi Manyama, Table of n, a(n) for n = 0..9999 (terms 0..999 from Paul D. Hanna)

Cf. A018247, A018248, A120817, A120818, A091661, A063006, A091663.

To calculate c, d, e, f use Mathematica algorithms for a, b and equations: c=a-b, d=1-c, e=b-1, f=a-1.

{a(n)=local(b=4,v=[]); for(k=1, n+1, b=b^5%10^k; v=concat(v,(10*b\10^k))); v[n+1]} \\ Paul D. Hanna, Jul 06 2006

(A091664_vec(n)=Vecrev(digits(lift(chinese(Mod(0,2^n),Mod(-1,5^n))))))(99) \\ M. F. Hasler, Jan 26 2020

x = r^2 where r = ...1441224165530407839804103263499879186432 (A120817). x = 10-adic lim_{n->oo} 4^(5^n). - Paul D. Hanna, Jul 06 2006

For n > 0, a(n) = 9 - A018248(n) = A018247(n). - Seiichi Manyama, Jul 28 2017

A091663 10-adic integer x=.....93380022607743740081787109375 satisfying x^3 = x.

Original entry on oeis.org

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

Offset: 0

Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004

Let a,b be integers defined in A018247, A018248 satisfying a^2=a, b^2=b, obviously a^3=a, b^3=b; let c,d,e,f be integers defined in A091661, A063006, A091663, A091664 then c^3=c, d^3=d, e^3=e, f^3=f, c+d=1, a+e=1, b+f=1, b+c=a, d+f=e, a+f=c, a=f+1, b=e+1, cd=-1, af=-1, gh=-1 where -1=.....999999999.

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

Cf. A018247, A018248.

To calculate c, d, e, f use Mathematica algorithms for a, b and equations: c=a-b, d=1-c, e=b-1, f=a-1.

For n > 0, a(n) = 9 - A018247(n) = A018248(n). - Seiichi Manyama, Jul 28 2017

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.

A063006 Coefficients in a 10-adic square root of 1.

Original entry on oeis.org

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

Offset: 0

Robert Lozyniak (11(AT)onna.com), Aug 03 2001

10-adic integer x=.....86760045215487480163574218751 satisfying x^3=x.
A "bug" in the decimal enumeration system: another square root of 1.
Let a,b be integers defined in A018247, A018248 satisfying a^2=a,b^2=b, obviously a^3=a,b^3=b; let c,d,e,f be integers defined in A091661, A063006, A091663, A091664 then c^3=c, d^3=d, e^3=e, f^3=f, c+d=1, a+e=1, b+f=1, b+c=a, d+f=e, a+f=c, a=f+1, b=e+1, cd=-1, af=-1, gh=-1 where -1=.....999999999. - Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004
What about the 10-adic square roots of -1, -2, -3, 2, 3, 4, ...? They do not exist, unless the square really is a square (+1, +4, +9, +16, ...). Then there's a nontrivial square root; for example, sqrt(4)=...44002229693692923584436016426479909569025039672851562498. - Don Reble, Apr 25 2006

			...4218751^2 = ...0000001

K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973.

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

Another 10-adic root of 1 is given by A091661.

Cf. A075693.

To calculate c, d, e, f use Mathematica algorithms for a, b and equations: c=a-b, d=1-c, e=b-1, f=a-1. - Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004

(a_0 + a_1*10 + a_2*10^2 + a_3*10^3 + ... )^2 = 1 + 0*10 + 0*10^2 + 0*10^3 + ...

For n > 0, a(n) = 9 - A091661(n).

More terms from Vladeta Jovovic, Aug 11 2001

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.

A290375 10-adic integer x = ...4193 satisfying x^5 = x.

Original entry on oeis.org

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

Offset: 0

Seiichi Manyama, Jul 28 2017

Also x^2 = A091661.

			     3^5 -    3 == 0 mod 10,
    93^5 -   93 == 0 mod 10^2,
   193^5 -  193 == 0 mod 10^3,
  4193^5 - 4193 == 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) ==   93 mod 10^2,
  8^(5^2) + 5^(2^2) ==  193 mod 10^3,
  8^(5^3) + 5^(2^3) == 4193 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), A290374 (...7057), this sequence (...4193).

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 A290375(n)
  str = (P(n) + Q(5, n)).to_s.reverse
  (0..n).map{|i| str[i].to_i}
end
p A290375(100)

p = A120818 = ...813568, q = A018247 = ...890625, x = p + q = ...704193.

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

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