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

Author

Marco Ripà, Feb 11 2025

Keywords

• nonn
• base
• easy

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.

Examples

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

Links

• 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.

Crossrefs

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

Programs

Mathematica

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

PARI

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

Formula

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

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