A370532 Integers m > 0 such that m^m and m^(m^m) have the same rightmost m digits.

1, 5, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 490, 500, 510, 520, 530

Offset

1,2

Comments

This sequence consists of the positive integers m such that m^m == m^(m^m) (mod 10^m).

All multiples of 10 are terms, since then m^m == m^(m^m) == 0 (mod 10^m).

5 is the only term m > 1 not a multiple of 10 (5^5 = 3125 and 5^(5^5) = ...03125). A compact proof of this fact has been published on Mathematics Stack Exchange by John Omelian (see Links).

Links

Table of n, a(n) for n=1..55.

Marco Ripà, Congruence speed of tetration bases ending with 0, arXiv:2402.07929 [math.NT], 2024.

Mathematics Stack Exchange, Easy proof that n = 5 is the only solution of n^n == n^n^n (mod 10^(n - 1)) if n > 1 is not a multiple of 10 .

Index entries for linear recurrences with constant coefficients, signature (2,-1).

Formula

a(n) = 10*(n - 2) for any n > 2.

From Stefano Spezia, Mar 03 2024: (Start)

G.f.: x*(1 + 3*x + x^2 + 5*x^3)/(1 - x)^2.

E.g.f.: 20 + 10*exp(x)*(x - 2) + 11*x + 5*x^2/2. (End)

Example

20 is a term since 20^20 == 0 (mod 10^20) and also 20^(20^20) == 0 (mod 10^20).

Mathematica

Join[{1, 5}, 10*Range[100]] (* Paolo Xausa, Mar 15 2024 *)

Prog

(Python)
def A370532(n): return 10*(n-2) if n>2 else 4*n-3 # Chai Wah Wu, Mar 11 2024

Crossrefs

Cf. A002489, A317905, A370211, A369826.

Essentially A008592.

Sequence in context: A065969 A306775 A027884 * A331141 A236391 A026357

Adjacent sequences: A370529 A370530 A370531 * A370533 A370534 A370535

Keyword

nonn,easy,base

Author

Marco Ripà, Feb 21 2024

Status

approved