cp's OEIS Frontend

Numbers in A018247 and A018248 are known as automorphic numbers in base 10, meaning that the infinite integers a=(...256259918212890625) or b=(...743740081787109376) provides a nontrivial solution to x*x == x (mod any power of 10).

Read backwards so as to match their counterparts (A007185 & A016090), A018247(0)+A018248(0) = 11 & A018247(n)+A018248(n) = 9 for all n's > 0 and their product is A076308.

All entries must be odd.

Is the accumulative sum equally positive and negative, i.e. does the sum equal 0 infinitely often?

Also called curious numbers.

For entries after the second, two successive terms sum up to a total having the form 10^n + 1. - Lekraj Beedassy, Apr 29 2005 [This comment is clearly wrong as stated. The sums of two consecutive terms are 1, 6, 11, 31, 101, 452, 1001, 10001, 100001, 200001, 1000001, 3781250, .... - T. D. Noe, Nov 14 2010]

If a d-digit number n is in the sequence, then so is 10^d+1-n. However, the same number can be 10^d+1-n for different n in the sequence (e.g., 10^3+1-376 = 10^4+1-9376 = 625), which spoils Beedassy's comment. - Robert Israel, Jun 19 2015

Substring of both its square and its cube not congruent to 0 (mod 10). See A029943. - Robert G. Wilson v, Jul 16 2005

a(n)^k ends with a(n) for k > 0; see also A029943. - Reinhard Zumkeller, Nov 26 2011

Apart from initial term, a subsequence of A046831. - M. F. Hasler, Dec 05 2012

This is also the sequence of numbers such that the n-th m-gonal number ends in n for any m == 0,4,8,16 (mod 20). - Robert Dawson, Jul 09 2018

Apart from 6, a subsequence of A301912. - Robert Dawson, Aug 01 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

Conjecture: For any m coprime to 10 and for any k, the density of n such that a(n) == k (mod m) is 1/m. - Eric M. Schmidt, Aug 01 2012

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, Aug 18 2012

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

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.

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.

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

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.