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%I #18 Jan 06 2024 21:39:18
%S 1,1,7,8,7,0,1,9,7,2,3,0,8,5,5,1,9,6,5,4,6,3,8,8,0,5,5,0,3,2,7,9,6,8,
%T 6,7,5,0,4,9,5,0,5,9,9,0,5,2,5,3,3,6,6,3,4,8,2,7,8,0,0,9,0,9,4,8,5,0,
%U 3,4,4,4,8,7,2,2,9,7,9,3,7,7,7,3,8,4
%N Digits occurring in A173616.
%C a(n) = A173616(n) - 10*A173616(n-1).
%C This is the 10-adic integer x such that x^9 == (10^n-9) mod 10^n for all n. It is the 10's complement of A225458. - _Aswini Vaidyanathan_, May 11 2013
%e 1111^1111=.........8711; 111^111=........711;
%e 10^(1-4)(8711-711)=8 ==> a(4)=8
%e Comment from _Aswini Vaidyanathan_, May 11 2013:
%e 1^9 == 1 (mod 10).
%e 11^9 == 91 (mod 100).
%e 711^9 == 991 (mod 1000).
%e 8711^9 == 9991 (mod 10000).
%e 78711^9 == 99991 (mod 100000).
%e 78711^9 == 999991 (mod 1000000).
%t repunit[n_] := Sum[10^i, {i,0,n-1}]; a[n_] := 10^(1-n)(PowerMod[repunit[n], repunit[n],10^n] - PowerMod[repunit[n-1], repunit[n-1], 10^(n-1)]); Table[a[n],{n,200}]
%o (PARI) n=0;for(i=1,100,m=(10^i-9);for(x=0,9,if(((n+(x*10^(i-1)))^9)%(10^i)==m,n=n+(x*10^(i-1));print1(x", ");break))) (From _Aswini Vaidyanathan_, May 11 2013)
%Y Cf. A173616, A225458.
%K nonn,base,easy
%O 1,3
%A _José María Grau Ribas_, Jul 23 2011
%E Edited by _N. J. A. Sloane_, May 12 2013
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