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A056968 10^(n-1) modulo n.

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%I A056968 #11 Nov 26 2024 02:35:47
%S A056968 0,0,1,0,0,4,1,0,1,0,1,4,1,10,10,0,1,10,1,0,16,10,1,16,0,10,19,20,1,
%T A056968 10,1,0,1,10,25,28,1,10,22,0,1,40,1,32,10,10,1,16,8,0,49,12,1,46,45,
%U A056968 24,43,10,1,40,1,10,37,0,55,10,1,48,31,20,1,64,1,10,25,12,67,4,1,0,73,10,1
%N A056968 10^(n-1) modulo n.
%H A056968 Robert Israel, <a href="/A056968/b056968.txt">Table of n, a(n) for n = 1..10000</a>
%F A056968 If n is of form 2^i*5^j then a(n)=0, otherwise a(n)=10^(n-1)+n-A053041(n)
%F A056968 From _Robert Israel_, Nov 25 2024: (Start)
%F A056968 If n is prime other than 2 or 5, then a(n) = 1.
%F A056968 If n = 2^i * 5^j * p where p is a prime > 10^(2^i * 5^j), then a(n) = 10^(2^i * 5^j).
%F A056968 If n = 2^i * 5^j * p where p is a prime and
%F A056968   2^(2^i * 5^j - 1 - i) * 5^(2^i * 5^j -1 - j) > p > 2^(2^i * 5^j-2 - u) * 5^(2^i * 5^j-1-j),
%F A056968 then a(n) =  10^(2^i * 5^j - 1) - 2^i * 5^j * p.
%F A056968 For example, with i = 0 and j = 1 we get a(5*p) = 10^4 - 5*p if p is a prime between 1000 and 2000.
%F A056968 (End)
%e A056968 a(6)=4 since 100000=6*16666+4
%p A056968 0, seq(10&^(n-1) mod n, n=2..100); # _Robert Israel_, Nov 25 2024
%t A056968 Table[PowerMod[10,n-1,n],{n,100}] (* _Harvey P. Dale_, Jul 17 2021 *)
%Y A056968 Cf. A003592, A053041, A056969.
%K A056968 nonn,look
%O A056968 1,6
%A A056968 _Henry Bottomley_, Jul 20 2000