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

%I A381015 #21 Mar 02 2025 23:33:35
%S A381015 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,
%T A381015 27,28,29,31,31,32,33,34,35,36,37,38,39,41,41,42,43,44,45,46,47,48,49,
%U A381015 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
%N A381015 a(n) = n + (number of trailing 0's of n).
%C A381015 Constant congruence speed of (10^n + 1)^n, i.e., a(n) = A373387((10^n + 1)^n).
%C A381015 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.
%C A381015 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.
%H A381015 Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/5031005/non-existence-of-perfect-powers-of-the-form-10n1-or-2-cdot-10n1">Non-existence of perfect powers of the form 10^n+1 or 2*10^n+1</a>.
%H A381015 Marco Ripà, <a href="https://doi.org/10.59400/jam1771">On the relation between perfect powers and tetration frozen digits</a>, Journal of AppliedMath, 2024, 2(5), 1771, see Theorem 2.
%H A381015 Wikipedia, <a href="https://en.wikipedia.org/wiki/Catalan%27s_conjecture">Catalan's_conjecture</a>.
%F A381015 a(n) = n + A122840(n).
%F A381015 a(n) = A373387(A121520(n)).
%e A381015 a(10) = 11 since A373387((10^10 + 1)^10) = 11.
%t A381015 a[n_]:=n+IntegerExponent[n,10]; Array[a,77] (* _Stefano Spezia_, Feb 13 2025 *)
%o A381015 (PARI) a(n) = n + valuation(n, 10); \\ _Michel Marcus_, Feb 13 2025
%Y A381015 Cf. A121520, A122840, A317905, A372490, A373387, A379243.
%K A381015 nonn,base,easy
%O A381015 1,2
%A A381015 _Marco Ripà_, Feb 11 2025