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%I #29 Apr 29 2023 23:02:24
%S 4,2,6,0,9,8,2,1,2,8,1,9,9,5,2,6,5,2,2,9,3,7,7,9,9,1,6,6,0,1,4,0,0,9,
%T 0,1,6,9,8,0,3,2,3,2,4,3,2,4,7,5,5,0,0,0,1,1,8,3,6,8,0,8,5,9,0,5,6,6,
%U 1,2,6,0,0,9,8,9,0,5,8,3,9,2,0,8,9,6,1,8,0,1,9,1,3,7,0,0,3,5,9,3
%N 10-adic integer x=.....06619977392256259918212890624 satisfying x^3 = x.
%C Let a,b be integers defined in A018247, A018248 satisfying a^2=a, b^2=b, obviously a^3=a, b^3=b; let c,d,e,f be integers defined in A091661, A063006, A091663, A091664; then c^3=c, d^3=d, e^3=e, f^3=f, c+d=1, a+e=1, b+f=1, b+c=a, d+f=e, a+f=c, a=f+1, b=e+1, cd=-1, af=-1, gh=-1 where -1=.....999999999.
%H Seiichi Manyama, <a href="/A091664/b091664.txt">Table of n, a(n) for n = 0..9999</a> (terms 0..999 from Paul D. Hanna)
%F x = r^2 where r = ...1441224165530407839804103263499879186432 (A120817). x = 10-adic lim_{n->oo} 4^(5^n). - _Paul D. Hanna_, Jul 06 2006
%F For n > 0, a(n) = 9 - A018248(n) = A018247(n). - _Seiichi Manyama_, Jul 28 2017
%e x equals the limit of the (n+1) trailing digits of 4^(5^n):
%e 4^(5^0) = (4), 4^(5^1) = 10(24), 4^(5^2) = 1125899906842(624), ...
%e x = ...0557423423230896109004106619977392256259918212890624.
%t To calculate c, d, e, f use Mathematica algorithms for a, b and equations: c=a-b, d=1-c, e=b-1, f=a-1.
%o (PARI) {a(n)=local(b=4,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) (A091664_vec(n)=Vecrev(digits(lift(chinese(Mod(0,2^n),Mod(-1,5^n))))))(99) \\ _M. F. Hasler_, Jan 26 2020
%Y Cf. A018247, A018248, A120817, A120818, A091661, A063006, A091663.
%K base,nonn
%O 0,1
%A Edoardo Gueglio (egueglio(AT)yahoo.it), Jan 28 2004
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