A120817 -id:A120817 - cp's OEIS frontend

A120818 10-adic integer x=...92160195896736500120813568 satisfying x^5 = x; also x^3 = -x = A120817; (x^2)^3 = x^2 = A091664; (x^4)^2 = x^4 = A018248.

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

Offset: 0

Paul D. Hanna, Jul 06 2006

			x equals the limit of the (n+1) trailing digits of 8^(5^n):
8^(5^0)=(8), 8^(5^1)=327(68), 8^(5^2)=37778931862957161709(568), ...
x=...06160350476776695446967548558775834469592160195896736500120813568.
x^2=...0557423423230896109004106619977392256259918212890624 (A091664).
x^3=...3304553032451441224165530407839804103263499879186432 (A120817).
x^4=...9442576576769103890995893380022607743740081787109376 (A018248).
x^5=...6695446967548558775834469592160195896736500120813568 = x.

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

Cf. A120817, A091664, A018248.

{a(n)=local(b=8,v=[]);for(k=1,n+1,b=b^5%10^k;v=concat(v,(10*b\10^k)));v[n+1]}

x = 10-adic lim_{n->oo} 8^(5^n).

A009003 Hypotenuse numbers (squares are sums of 2 nonzero squares).

Original entry on oeis.org

5, 10, 13, 15, 17, 20, 25, 26, 29, 30, 34, 35, 37, 39, 40, 41, 45, 50, 51, 52, 53, 55, 58, 60, 61, 65, 68, 70, 73, 74, 75, 78, 80, 82, 85, 87, 89, 90, 91, 95, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, 113, 115, 116, 117, 119, 120, 122, 123, 125, 130, 135, 136, 137, 140

Offset: 1

David W. Wilson

nonn

Multiples of Pythagorean primes A002144 or of primitive Pythagorean triangles' hypotenuses A008846. - Lekraj Beedassy, Nov 12 2003
This is exactly the sequence of positive integers with at least one prime divisor of the form 4k + 1. Compare A072592. - John W. Layman, Mar 12 2008 and Franklin T. Adams-Watters, Apr 26 2009
Circumradius R of the triangles such that the area, the sides and R are integers. - Michel Lagneau, Mar 03 2012
The 2 squares summing to a(n)^2 cannot be equal because sqrt(2) is not rational. - Jean-Christophe Hervé, Nov 10 2013
Closed under multiplication. The primitive elements are those with exactly one prime divisor of the form 4k + 1 with multiplicity one, which are also those for which there exists a unique integer triangle = A084645. - Jean-Christophe Hervé, Nov 11 2013
a(n) are numbers whose square is the mean of two distinct nonzero squares. This creates 1-to-1 mapping between a Pythagorean triple and a "Mean" triple. If the Pythagorean triple is written, abnormally, as {j, k, h} where j^2 +(j+k)^2 = h^2, and h = a(n), then the corresponding "Mean" triple with the same h is {k, 2j, h} where (k^2 + (k+2j)^2)/2 = h^2. For example for h = 5, the Pythagorean triple is {3, 1, 5} and the Mean triple is {1, 6, 5}. - Richard R. Forberg, Mar 01 2015
Integral side lengths of rhombuses with integral diagonals p and q (therefore also with integral areas A because A = pq/2 is some multiple of 24). No such rhombuses are squares. - Rick L. Shepherd, Apr 09 2017
Conjecture: these are bases n in which exists an n-adic integer x satisfying x^5 = x, and 5 is the smallest k>1 such that x^k =x (so x^2, x^3 and x^4 are not x). Example: the 10-adic integer x = ...499879186432 (A120817) satisfies x^5 = x, and x^2, x^3, and x^4 are not x, so 10 is in this sequence. See also A120817, A210850 and A331548. - Patrick A. Thomas, Mar 01 2020
Didactic comment: When students solve a quadratic equation a*x^2 + b*x + c = 0 (a, b, c: integers) with the solution formula, they often make the mistake of calculating b^2 + 4*a*c instead of b^2 - 4*a*c (especially if a or c is negative). If the root then turns out to be an integer, they feel safe. This sequence lists the absolute values of b for which this error can happen. Reasoning: With p^2 = b^2 - 4*a*c and q^2 = b^2 + 4*a*c it follows by addition immediately that p^2 + q^2 = 2*b^2. If 4*a*c < 0, let p = x + y and q = x - y. If 4*a*c > 0, let p = x - y and q = x + y. In both cases follows that y^2 + x^2 = b^2. So every Pythagorean triple gives an absolute value of b for which this error can occur. Example: From (y, x, b) = (3, 4, 5) follows (q^2, b^2, p^2) = (1, 25, 49) or (p^2, b^2, q^2) = (1, 25, 49) with abs(4*a*c) = 24. - Felix Huber, Jul 22 2023
Conjecture: Numbers m such that the limit: Limit_{s->1} zeta(s)*Sum_{k=1..m} [k|m]*A008683(k)*(i^k)/(k^(s - 1)) exists, which is equivalent to numbers m such that abs(Sum_{k=1..m} [k|m]*A008683(k)*(i^k)) = 0. - Mats Granvik, Jul 06 2024

Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 98-104.

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
R. Chapman, Pythagorean triples and sums of squares
Steven R. Finch, Landau-Ramanujan Constant [Broken link]
Steven R. Finch, Landau-Ramanujan Constant [From the Wayback machine]
Ron Knott, Pythagorean Triples and Online Calculators
J. Pahikkala, On contraharmonic mean and Pythagorean triples, Elemente der Mathematik, 65:2 (2010), 62-67.
Patrick A. Thomas, Solutions of x^5=x up to base 100
Index entries for sequences related to sums of squares

Cf. A000404 (sums of 2 squares), A004431 (sums of 2 distinct squares), A009000 (hypotenuse numbers with repetition), A072592, A004613, A187811.
Complement of A004144. Primes in this sequence give A002144. Same as A146984 (integer contraharmonic means) as sets - see Pahikkala 2010, Theorem 5.
Cf. A083025, A084645 (primitive elements), A084646, A084647, A084648, A084649, A006339.

import Data.List (findIndices)
a009003 n = a009003_list !! (n-1)
a009003_list = map (+ 1) $ findIndices (> 0) a005089_list
-- Reinhard Zumkeller, Jan 07 2013

isA009003 := proc(n)
    local p;
    for p in numtheory[factorset](n) do
        if modp(p,4) = 1 then
            return true;
        end if;
    end do:
    false;
end proc:
for n from 1 to 200 do
    if isA009003(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Nov 17 2014

f[n_] := Module[{k = 1}, While[(n - k^2)^(1/2) != IntegerPart[(n - k^2)^(1/2)], k++; If[2 * k^2 >= n, k = 0; Break[]]]; k]; A009003 = {}; Do[If[f[n^2] > 0, AppendTo[A009003, n]], {n, 3, 100}]; A009003 (* Vladimir Joseph Stephan Orlovsky, Jun 15 2009 *)
Select[Range[200], Length[PowersRepresentations[#^2, 2, 2]] > 1 &] (* Alonso del Arte, Feb 11 2014 *)

is_A009003(n)=setsearch(Set(factor(n)[,1]%4),1)  \\ M. F. Hasler, May 27 2012

list(lim)=my(v=List(),u=vectorsmall(lim\=1)); forprimestep(p=5,lim,4, forstep(n=p,lim,p, u[n]=1)); for(i=5,lim, if(u[i], listput(v,i))); u=0; Vec(v) \\ Charles R Greathouse IV, Jan 13 2022

from itertools import count, islice
from sympy import primefactors
def A009003_gen(): # generator of terms
    return filter(lambda n:any(map(lambda p: p % 4 == 1,primefactors(n))),count(1))
A009003_list = list(islice(A009003_gen(),20)) # Chai Wah Wu, Jun 22 2022

A005089(a(n)) > 0. - Reinhard Zumkeller, Jan 07 2013
a(n) ~ n. - Charles R Greathouse IV, Jan 13 2022
a(n) = sqrt(n-th square in A000404), where A000404 lists the sums of two nonzero squares. - M. F. Hasler, Jun 20 2025

Definition edited by Jean-Christophe Hervé, Nov 10 2013

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).

Original entry on oeis.org

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

A018248 The 10-adic integer x = ...1787109376 satisfies x^2 = x.

Original entry on oeis.org

6, 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, 9, 0, 6, 3, 7, 5, 3, 2

Offset: 0

Yoshihide Tamori (yo(AT)salk.edu)

The 10-adic numbers a and b defined in A018247 and this sequence satisfy a^2=a, b^2=b, a+b=1, ab=0. - Michael Somos

			x equals the limit of the (n+1) trailing digits of 6^(5^n):
6^(5^0)=(6), 6^(5^1)=77(76), 6^(5^2)=28430288029929701(376), ...
x = ...9442576576769103890995893380022607743740081787109376.
From _Peter Bala_, Nov 05 2022: (Start)
Trailing digits of 2^(10^n), 4^(10^n) and 6^(10^n) for n = 5:
2^(10^5) = ...9883(109376);
4^(10^5) = ...7979(109376);
6^(10^5) = ...4155(109376). (End)

W. W. R. Ball, Mathematical Recreations & Essays, N.Y. Macmillan Co, 1947.
R. Cuculière, Jeux Mathématiques, in Pour la Science, No. 6 (1986), 10-15.
V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.
M. Kraitchik, Sphinx, 1935, p. 1.
A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see pp. 63, 419.

Seiichi Manyama, Table of n, a(n) for n = 0..9999 (terms 0..999 from Paul D. Hanna).
Anonymous, Automorphic numbers (2)
Peter Bala, A note on A018248
V. deGuerre and R. A. Fairbairn, Automorphic numbers, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179.
MathOverflow, Distribution of digits of pq-adic idempotents (a.k.a. "automorphic numbers"), 2014.
Eric Weisstein's World of Mathematics, Automorphic numbers (1)
Index entries for sequences related to automorphic numbers

A016090 gives associated automorphic numbers.
Cf. A018247, A033819.
The difference between this sequence & A018247 is A075693 and their product is A075693.
Cf. A120817, A120818, A091664.
The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.

a := proc (n) option remember; if n = 1 then 2 else irem(a(n-1)^10, 10^n) end if; end proc:
# display the digits of a(100) from right to left
S := convert(a(100), string):
with(ListTools):
the_List := [seq(parse(S[i]), i = 1..length(S))]:
Reverse(the_List); # Peter Bala, Nov 04 2022

b = {6}; g[n_] := Block[{k = 0, c}, While[c = FromDigits[Prepend[b, k]]; Mod[c^2, 10^n] != c, k++ ]; b = Prepend[b, k]]; Do[ g[n], {n, 2, 105}]; Reverse[b]
With[{n = 150}, Reverse[IntegerDigits[PowerMod[16, 5^n, 10^n]]]] (* IWABUCHI Yu(u)ki, Feb 16 2024 *)

{a(n)=local(b=6,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

Vecrev(digits(lift(chinese(Mod(0, 2^100), Mod(1, 5^100))))) \\ Seiichi Manyama, Aug 07 2019

x = r^4 where r=...1441224165530407839804103263499879186432 (A120817). x = 10-adic limit_{n->oo} 6^(5^n). - Paul D. Hanna, Jul 06 2006

For n >= 2, the final n+1 digits of either 2^(10^n), 4^(10^n) or 6^(10^n), when read from right to left, give the first n+1 entries in the sequence. - Peter Bala, Nov 05 2022

More terms from David W. Wilson

Edited by David W. Wilson, Sep 26 2002

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

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.

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

A120818 10-adic integer x=...92160195896736500120813568 satisfying x^5 = x; also x^3 = -x = A120817; (x^2)^3 = x^2 = A091664; (x^4)^2 = x^4 = A018248.

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A009003 Hypotenuse numbers (squares are sums of 2 nonzero squares).

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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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A018248 The 10-adic integer x = ...1787109376 satisfies x^2 = x.

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

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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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