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%I #72 Feb 26 2024 02:17:45
%S 6,7,3,9,0,1,7,8,7,1,8,0,0,4,7,3,4,7,7,0,6,2,2,0,0,8,3,3,9,8,5,9,9,0,
%T 9,8,3,0,1,9,6,7,6,7,5,6,7,5,2,4,4,9,9,9,8,8,1,6,3,1,9,1,4,0,9,4,3,3,
%U 8,7,3,9,9,0,1,0,9,4,1,6,0,7,9,1,0,3,8,1,9,8,0,8,6,2,9,9,6,4,0,6,9,0,6,3,7,5,3,2
%N The 10-adic integer x = ...1787109376 satisfies x^2 = x.
%C The 10-adic numbers a and b defined in A018247 and this sequence satisfy a^2=a, b^2=b, a+b=1, ab=0. - _Michael Somos_
%D W. W. R. Ball, Mathematical Recreations & Essays, N.Y. Macmillan Co, 1947.
%D R. Cuculière, Jeux Mathématiques, in Pour la Science, No. 6 (1986), 10-15.
%D V. deGuerre and R. A. Fairbairn, Automorphic numbers, J. Rec. Math., 1 (No. 3, 1968), 173-179.
%D M. Kraitchik, Sphinx, 1935, p. 1.
%D A. M. Robert, A Course in p-adic Analysis, Springer, 2000; see pp. 63, 419.
%H Seiichi Manyama, <a href="/A018248/b018248.txt">Table of n, a(n) for n = 0..9999</a> (terms 0..999 from Paul D. Hanna).
%H Anonymous, <a href="http://freespace.virgin.net/anthony.edey/automorph.htm">Automorphic numbers (2)</a>
%H Peter Bala, <a href="/A018248/a018248.pdf">A note on A018248</a>
%H V. deGuerre and R. A. Fairbairn, <a href="/A003226/a003226.pdf">Automorphic numbers</a>, Jnl. Rec. Math., 1 (No. 3, 1968), 173-179.
%H MathOverflow, <a href="https://mathoverflow.net/questions/156301/distribution-of-digits-of-pq-adic-idempotents-aka-automorphic-numbers">Distribution of digits of pq-adic idempotents (a.k.a. "automorphic numbers")</a>, 2014.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/AutomorphicNumber.html">Automorphic numbers (1)</a>
%H <a href="/index/Ar#automorphic">Index entries for sequences related to automorphic numbers</a>
%F x = r^4 where r=...1441224165530407839804103263499879186432 (A120817). x = 10-adic limit_{n->oo} 6^(5^n). - _Paul D. Hanna_, Jul 06 2006
%F For n >= 2, the final n+1 digits of either 2^(10^n), 4^(10^n) or 6^(10^n), when read from right to left, give the first n+1 entries in the sequence. - _Peter Bala_, Nov 05 2022
%e x equals the limit of the (n+1) trailing digits of 6^(5^n):
%e 6^(5^0)=(6), 6^(5^1)=77(76), 6^(5^2)=28430288029929701(376), ...
%e x = ...9442576576769103890995893380022607743740081787109376.
%e From _Peter Bala_, Nov 05 2022: (Start)
%e Trailing digits of 2^(10^n), 4^(10^n) and 6^(10^n) for n = 5:
%e 2^(10^5) = ...9883(109376);
%e 4^(10^5) = ...7979(109376);
%e 6^(10^5) = ...4155(109376). (End)
%p a := proc (n) option remember; if n = 1 then 2 else irem(a(n-1)^10, 10^n) end if; end proc:
%p # display the digits of a(100) from right to left
%p S := convert(a(100), string):
%p with(ListTools):
%p the_List := [seq(parse(S[i]), i = 1..length(S))]:
%p Reverse(the_List); # _Peter Bala_, Nov 04 2022
%t b = {6}; g[n_] := Block[{k = 0, c}, While[c = FromDigits[Prepend[b, k]]; Mod[c^2, 10^n] != c, k++ ]; b = Prepend[b, k]]; Do[ g[n], {n, 2, 105}]; Reverse[b]
%t With[{n = 150}, Reverse[IntegerDigits[PowerMod[16, 5^n, 10^n]]]] (* _IWABUCHI Yu(u)ki_, Feb 16 2024 *)
%o (PARI) {a(n)=local(b=6,v=[]);for(k=1,n+1,b=b^5%10^k;v=concat(v,(10*b\10^k)));v[n+1]} \\ _Paul D. Hanna_, Jul 06 2006
%o (PARI) Vecrev(digits(lift(chinese(Mod(0, 2^100), Mod(1, 5^100))))) \\ _Seiichi Manyama_, Aug 07 2019
%Y A016090 gives associated automorphic numbers.
%Y Cf. A018247, A033819.
%Y The difference between this sequence & A018247 is A075693 and their product is A075693.
%Y Cf. A120817, A120818, A091664.
%Y The six examples given by deGuerre and Fairbairn are A055620, A054869, A018247, A018248, A259468, A259469.
%K nonn,base
%O 0,1
%A Yoshihide Tamori (yo(AT)salk.edu)
%E More terms from _David W. Wilson_
%E Edited by _David W. Wilson_, Sep 26 2002
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