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
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.
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
KEYWORD
nonn,base,easy,new
AUTHOR
Marco Ripà, Feb 11 2025
STATUS
approved