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%I #34 Sep 03 2021 03:48:53
%S 9,89,289,5289,45289,745289,2745289,92745289,392745289,7392745289,
%T 97392745289,597392745289,7597392745289,87597392745289,
%U 8087597392745289,48087597392745289,748087597392745289,10748087597392745289,610748087597392745289,5610748087597392745289
%N Values of n such that 9^n ends in n, or expomorphic numbers relative to "base" 9.
%C Definition: For positive integers b (as base) and n, the positive integer (allowing initial zeros) k(n) is expomorphic relative to base b (here 9) if k(n) has exactly n decimal digits and if b^k(n) == k(n) (mod 10^n) or, equivalently, b^k(n) ends in k(n). [See Crux Mathematicorum link.]
%C For sequences in the OEIS, no term is allowed to begin with a digit 0 (except for the 1-digit number 0 itself). However, in the problem as defined in the Crux Mathematicorum article, leading 0 digits are allowed; under that definition a(n) = k(n) until the first k(n) which begins with digit 0. When k(n) begins with 0, then, a(n) is the next term of the sequence k(n) which doesn't begin with digit 0.
%C Conjecture: if k(n) is expomorphic relative to "base" b, then, the next one in the sequence, k(n+1), consists of the last n+1 digits of b^k(n).
%C a(n) is the backward concatenation of A133619(0) through A133619(n-1). So, a(1) = 9, a(2) = 89 and so on, with recognition of the former comments about the OEIS and terms beginning with 0. - _Davis Smith_, Mar 07 2019
%H Davis Smith, <a href="/A306686/b306686.txt">Table of n, a(n) for n = 1..944</a>
%H Charles W. Trigg, <a href="https://cms.math.ca/crux/backfile/Crux_v7n06_Jun.pdf">Problem 559</a>, Crux Mathematicorum, page 192, Vol. 7, Jun. 81.
%H Emil Vaughan, <a href="https://m500.org.uk/wp-content/uploads/2016/04/M226WEB.pdf">Problem 226.8 - 999 nines</a>, M500 Magazine of the Open University, number 226, February 2009, page 21; and Tony Forbes, <a href="https://m500.org.uk/wp-content/uploads/2016/04/M195WEB.pdf">Solution 226.8 - 999 nines</a>, M500 Magazine of the Open University, number 232, February 2010, pages 8-9, calculating a(9) = 392745289.
%e 9^9 = 387420489 ends in 9, so 9 is a term; 9^89 = .....289 ends in 89, so 89 is another term.
%o (PARI) tetrmod(b, n, m)=my(t=b); for(i=1, n, if(i>1, t=lift(Mod(b,m)^t), t)); t
%o for(n=1, 21,if(tetrmod(9,n,10^n)!=tetrmod(9,n-1,10^(n-1)),print1(tetrmod(9,n,10^(n-1)),", "))) \\ _Davis Smith_, Mar 09 2019
%Y Cf. A064541 (base 2), A183613 (base 3), A288845 (base 4), A306570 (base 5), A290788 (base 6), A321970 (base 7), A289138 (smallest expomorphic number in base n).
%Y Cf. A003226 (automorphic numbers), A033819 (trimorphic numbers).
%Y Cf. A133619 (leading digits).
%K nonn,base
%O 1,1
%A _Bernard Schott_, Mar 05 2019
%E a(8)-a(20) from _Davis Smith_, Mar 07 2019
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