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Numbers k such that k is a substring of k^k.
9

%I #45 Sep 18 2023 18:05:01

%S 1,5,6,9,10,11,16,17,19,21,24,25,28,31,32,33,35,36,37,39,41,42,43,45,

%T 46,48,49,50,51,52,53,54,56,57,59,60,61,63,64,66,71,73,74,75,76,77,78,

%U 79,81,82,83,84,86,91,92,93,94,95,96,97,98,99,100,101,103,108,111,115

%N Numbers k such that k is a substring of k^k.

%C For typical large k, the string corresponding to k^k has length on the order of k log_10(k); heuristically, each substring of length d = log_10(k) has probability 10^(-d) ~ 1/k of matching k, and the probability that none of these matches is about exp(-log_10(k)) = k^(-log_10(e)) ~ k^(-0.434). Thus we should expect that most large k are in the sequence, but infinitely many are not. - _Robert Israel_, Jul 14 2015

%H Charles R Greathouse IV, <a href="/A049329/b049329.txt">Table of n, a(n) for n = 1..10000</a>

%p filter:= proc(n) local L, Ln;

%p L:= convert(n,string);

%p Ln:= convert(n^n,string);

%p StringTools:-Search(L,Ln) <> 0

%p end proc:

%p select(filter, [$1..1000]); # _Robert Israel_, Jul 13 2015

%t ssQ[n_] := Module[{idn = IntegerDigits[n]}, MemberQ[Partition[ IntegerDigits[ n^n], Length[idn], 1], idn]]; Select[Range[120], ssQ] (* _Harvey P. Dale_, Apr 01 2011 *)

%t Select[Range[120],SequenceCount[IntegerDigits[#^#],IntegerDigits[#]]>0&] (* _Harvey P. Dale_, Sep 18 2023 *)

%Y Cf. A000312.

%K base,nonn,easy,nice

%O 1,2

%A _David W. Wilson_