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%I #26 Mar 20 2024 05:13:32
%S 0,2,0,2,14,15,9,5,16,4,25,21,40,67,78,66,25,111,161,49,30,15,27,20,
%T 63,98,102,3,99,92,296,71,22,367,4,48,50,91,45,241,137,258,23,28,212,
%U 40,96,408,456,110,16,731,403,667,90,130,111,458,146,18,577,276,708
%N Smallest number k such that the reverse concatenation of natural numbers from k to 1 is divisible by prime(n), or 0 if no such number exists.
%C Conjecture: a(n) > 0 for all n > 3, since prime(1) = 2 and prime(3) = 5 are the only primes whose multiples cannot end in 1. - _Ryan Propper_, Jul 29 2005
%H Daniel Mondot, <a href="/A077183/b077183.txt">Table of n, a(n) for n = 1..1000</a>
%H Ralf Stephan, <a href="http://www.ark.in-berlin.de/sm.pdf">Factors and Primes in Two Smarandache Sequences</a>, Smar. Notions 9 (1998), pp. 4-10.
%e a(4) = 2 as 21 is divisible by prime(4) = 7.
%e The smallest reverse concatenation of natural numbers k..1 that is divisible by prime(5) = 11 is 1413121110987654321, so a(5) = k = 14.
%t Do[p = Prime[n]; k = 1; s = ToString[k]; While[Mod[ToExpression[s], p] > 0, k++; s = ToString[k] <> s]; Print[k], {n, 4, 50}] (* _Ryan Propper_, Jul 29 2005 *)
%Y Cf. A077180, A077181, A077182, A077185.
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, Nov 01 2002
%E Corrected and extended by _Ralf Stephan_, Mar 18 2003
%E Example clarified by _Harvey P. Dale_, Aug 22 2013