A379243 a(n) = (10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1)^n, where v_p(n) indicates the p-adic valuation of n.

111, 1212201, 1331363033001, 146415324072600440001, 1610517320513310012100005500001, 1771561966306219615026620001815000066000001, 194871722400927338207105124350046585000254100000770000001, 2143588825589736849603708090188560102487000074536000033880000008800000001

Offset

1,1

Comments

For any n, a(n) == 1 (mod 10^n), while it is not congruent to 1 modulo 10^(n + 1).

If n is not a multiple of 10, 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 has a total of n + 2 digits and they are n - 1 0s and 3 1s. Conversely, if there is a pair of positive integers (m, k) not ending with 0 and such that n := m*10^k, then 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 has n - k + 2 digits (three 1s and the rest are all 0s) and a(n) = (10^(n + 1) + 10^(n - k) + 1)^n.

Since a(n) == 1 (mod 5) for any n, the constant congruence speed of a(n) (i.e., V(a(n))) is guaranteed to be constant starting from height v_5(a(n) - 1) + 2 (for this sufficient condition, see “Number of stable digits of any integer tetration” in Links).

Then, for any positive integer n, a(n) is (exactly) a n-th perfect power (since, for any given n, 10^(n + 1) + 10^(n - min{v_2(n), v_5(n)}) + 1 is divisible by 3 only once) and is also characterized by a constant congruence speed of n (for a strict proof of the general formula V((10^(t + k) + 10^(t - min{v_2(n), v_5(n)}) + 1)^n) = t, holding for any chosen positive integer k as long as t is an integer above min{v_2(n), v_5(n)} + 1, see Section 3 of “On the relation between perfect powers and tetration frozen digits” in Links).

Links

Table of n, a(n) for n=1..8.

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, On the relation between perfect powers and tetration frozen digits, Journal of AppliedMath, 2024, 2(5), 1771.

Wikipedia, Tetration.

Formula

If n <> 0 (mod 10), then a(n) = (11...[n - 1 trailing 0s]...1)^n.

Example

a(3) = (10^4 + 10^3 + 1)^3 = 11001^3  = 1331363033001 is a perfect cube whose constant congruence speed is 3.

Mathematica

pAdicValuation[n_, p_] := Module[{v = 0, k = n}, While[Mod[k, p] == 0 && k > 0, k = k/p; v++; ]; v];
a[n_] := Module[{v2, v5, minVal}, v2 = pAdicValuation[n, 2]; v5 = pAdicValuation[n, 5];
minVal = Min[v2, v5]; (10^(n + 1) + 10^(n - minVal) + 1)^n]; sequence = Table[a[n], {n, 1, 20}]; sequence

Crossrefs

Cf. A000533, A001597, A063006, A121520, A199691, A317905, A372490, A373387.

Sequence in context: A262631 A259747 A176194 * A261820 A262661 A156407

Adjacent sequences: A379240 A379241 A379242 * A379244 A379245 A379246

Keyword

nonn,base,new

Author

Marco Ripà, Dec 18 2024

Status

approved