A018248 -id:A018248 - cp's OEIS frontend

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

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.

A091661 Coefficients in a 10-adic square root of 1.

Original entry on oeis.org

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

Offset: 0

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

10-adic integer x=.....239954784512519836425781249 satisfying x^3 = x.

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

Another 10-adic root of 1 is given by A063006.

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.

def A(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 A091661(n)
  str = (10 ** (n + 1) + A(5, n) - A(6, n)).to_s.reverse
  (0..n).map{|i| str[i].to_i}
end
p A091661(100) # Seiichi Manyama, Jul 31 2017

For n>0, a(n) = 9 - A063006(n).

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

A055620 Digits of an idempotent 6-adic number.

Original entry on oeis.org

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

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), Jun 04 2000

( a(0) + a(1)*6 + a(2)*6^2 + ... )^k = a(0) + a(1)*6 + a(2)*6^2 + ... for each k. Apart from 0 and 1 in base 6 there are only 2 numbers with this property. For the other see A054869.

			(a(0) + a(1)*6 + a(2)*6^2 + a(3)*6^3)^2 == (a(0) + a(1)*6 + a(2)*6^2 + a(3)*6^3) mod 6^4 because 1478656 == 1216 (mod 1296).

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Kenny Lau, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

first(p)=Vecrev(digits(lift(Mod(4,6^p)^3^p), 6)) \\ Charles R Greathouse IV, Nov 01 2022

n=10000;res=pow((3**n+1)//2,n,3**n)*2**n
for i in range(n):print(i,res%6);res//=6
# Kenny Lau, Jun 09 2018

If A is the 6-adic number, A == 4^(3^n) mod 6^n. - Robert Dawson, Oct 28 2022

A216093 a(n) = 10^n - (5^(2^n) mod 10^n).

Original entry on oeis.org

5, 75, 375, 9375, 9375, 109375, 7109375, 87109375, 787109375, 1787109375, 81787109375, 81787109375, 81787109375, 40081787109375, 740081787109375, 3740081787109375, 43740081787109375, 743740081787109375

Offset: 1

V. Raman, Sep 01 2012

nonn

a(n)^3 mod 10^n = a(n).
a(n) is the unique positive integer less than 10^n such that a(n) is divisible by 5^n and a(n) + 1 is divisible by 2^n. - Eric M. Schmidt, Sep 01 2012
a(n+1) + a(n)^2 == 0 (mod 10^(n+1)). - Robert Israel, Apr 24 2017

Robert Israel, Table of n, a(n) for n = 1..999

Cf. A007185, A016090, A216092, A091663, A018248.

f:= n -> (-5 &^(2^n) mod 10^n):
map(f, [$1..30]); # Robert Israel, Apr 24 2017

def A216093(n) : return crt(-1, 0, 2^n, 5^n) # Eric M. Schmidt, Sep 01 2012

2^(4*5^(n-1)) mod 10^n - 1.

A259468 Digits of an idempotent 12-adic number.

Original entry on oeis.org

4, 5, 8, 3, 11, 1, 6, 11, 1, 2, 11, 2, 5, 4, 3, 11, 7, 6, 2, 11, 6, 8, 9, 6, 1, 10, 3, 10, 1, 8, 10, 3, 3, 1, 10, 8, 6, 4, 6, 11, 2, 0, 4, 0, 3, 8, 9, 9, 5, 3, 1, 0, 9, 7, 11, 8, 7, 6, 5, 2, 5, 1, 10, 2, 11, 0, 2, 7, 6, 6, 5, 4, 7, 2, 0, 5, 11, 7, 8, 7, 2, 6

Offset: 0

N. J. A. Sloane, Jul 02 2015

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Eric M. Schmidt, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 No. 3, (1968), 173-179.

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

def a_list(n) : return crt(1, 0, 3^n, 4^n).digits(12) # Eric M. Schmidt, Jul 09 2015

More terms from Eric M. Schmidt, Jul 09 2015

A259469 Digits of an idempotent 12-adic number.

Original entry on oeis.org

9, 6, 3, 8, 0, 10, 5, 0, 10, 9, 0, 9, 6, 7, 8, 0, 4, 5, 9, 0, 5, 3, 2, 5, 10, 1, 8, 1, 10, 3, 1, 8, 8, 10, 1, 3, 5, 7, 5, 0, 9, 11, 7, 11, 8, 3, 2, 2, 6, 8, 10, 11, 2, 4, 0, 3, 4, 5, 6, 9, 6, 10, 1, 9, 0, 11, 9, 4, 5, 5, 6, 7, 4, 9, 11, 6, 0, 4, 3, 4, 9, 5, 11

Offset: 0

N. J. A. Sloane, Jul 02 2015

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Eric M. Schmidt, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179.

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

def a_list(n) : return crt(0, 1, 3^n, 4^n).digits(12) # Eric M. Schmidt, Jul 09 2015

More terms from Eric M. Schmidt, Jul 09 2015

A054869 Digits of an idempotent 6-adic number.

Original entry on oeis.org

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

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

( a(0) + a(1)*6 + a(2)*6^2 + ... )^k = a(0) + a(1)*6 + a(2)*6^2 + ... for each k. Apart from 0 and 1, in base 6 there are only 2 numbers with this property. For the other see A055620.

V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.

Kenny Lau, Table of n, a(n) for n = 0..9999
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179

The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

n=10000;res=1-pow((3**n+1)//2,n,3**n)*2**n
for i in range(n):print(i,res%6);res//=6
# Kenny Lau, Jun 09 2018

a(n) == 3^(2^n) (mod 6^n). - Robert Dawson, Oct 28 2022

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}")

cp's OEIS Frontend

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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A091661 Coefficients in a 10-adic square root of 1.

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

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A055620 Digits of an idempotent 6-adic number.

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A216093 a(n) = 10^n - (5^(2^n) mod 10^n).

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Maple

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A259468 Digits of an idempotent 12-adic number.

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A259469 Digits of an idempotent 12-adic number.

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Sage