A381015 a(n) = n + (number of trailing 0's of n).

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 41, 42, 43, 44, 45, 46, 47, 48, 49, 51, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 71, 72, 73, 74, 75, 76, 77

Offset

1,2

Comments

Constant congruence speed of (10^n + 1)^n, i.e., a(n) = A373387((10^n + 1)^n).

Since 10^n + 1 is never a perfect power by Catalan's conjecture (Mihăilescu's theorem), it follows that if 10 does not divide n, then (10^n + 1)^n is exactly an n-th perfect power with a constant congruence speed of a(n) = n.

Moreover, for any positive integer n, the congruence speed of (10^n + 1)^n equals 2*a(n) at height 1 and then becomes stable at height 2.

Links

Table of n, a(n) for n=1..77.

Mathematics Stack Exchange, Non-existence of perfect powers of the form 10^n+1 or 2*10^n+1.

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

Wikipedia, Catalan's_conjecture.

Formula

a(n) = n + A122840(n).

a(n) = A373387(A121520(n)).

Example

a(10) = 11 since A373387((10^10 + 1)^10) = 11.

Mathematica

a[n_]:=n+IntegerExponent[n, 10]; Array[a, 77] (* Stefano Spezia, Feb 13 2025 *)

Prog

(PARI) a(n) = n + valuation(n, 10); \\ Michel Marcus, Feb 13 2025

Crossrefs

Cf. A121520, A122840, A317905, A372490, A373387, A379243.

Sequence in context: A354719 A340835 A123241 * A355222 A033862 A329201

Adjacent sequences: A381007 A381008 A381009 * A381016 A381017 A381018

Keyword

nonn,base,easy

Author

Marco Ripà, Feb 11 2025

Status

approved