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Let s(m) = the set of k > 0 such that k^m ends with k. Then a(n) = least m such that s(m) = s(n).
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%I #21 Feb 18 2024 09:05:52

%S 1,2,3,2,5,6,3,2,9,2,11,2,5,2,3,6,17,2,3,2,21,2,3,2,9,26,3,2,5,2,11,2,

%T 33,2,3,6,5,2,3,2,41,2,3,2,5,6,3,2,17,2,51,2,5,2,3,6,9,2,3,2,21,2,3,2,

%U 65,6,3,2,5,2,11,2,9,2,3,26,5,2,3,2,81,2

%N Let s(m) = the set of k > 0 such that k^m ends with k. Then a(n) = least m such that s(m) = s(n).

%C See A227070 for more details and for the numbers n such that n = a(n).

%C The entries in the b-file have been tentatively obtained by comparing the terms < 10^30 in the sets s(n). - _Giovanni Resta_, Jul 30 2013

%H Giovanni Resta, <a href="/A227071/b227071.txt">Table of n, a(n) for n = 1..10000</a> (warning: contains tentative terms)

%F Conjecture: a(n+1) = A132741(n) + 1. - _Eric M. Schmidt_, Jul 30 2013

%t ts = {{}}; t2 = {1}; te = {1}; Do[s = Select[Range[0, 10^7], PowerMod[#, n, 10^IntegerLength[#]] == # &]; If[MemberQ[ts, s], AppendTo[t2, te[[Position[ts, s, 1, 1][[1, 1]]]]], AppendTo[ts, s]; AppendTo[te, n]; AppendTo[t2, n]], {n, 2, 82}]; t2

%Y Cf. A003226 (n=2), A033819 (n=3), A068407 (n=5), A068408 (n=6).

%Y Cf. A072496 (n=11), A072495 (n=21), A076650 (n=26).

%Y Cf. A227070 (n such that n = a(n)).

%K nonn,base

%O 1,2

%A _T. D. Noe_, Jul 29 2013

%E Mathematica program and some entries corrected by _Giovanni Resta_, Jul 30 2013