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%I #15 May 06 2020 01:59:56
%S 1,1,4,1,4,4,3,2,36,4,25,4,8,3,8,4,8,36,9,4,6,25,11,4,20,8,108,3,14,8,
%T 1,8,50,8,12,36,432,9,8,8,80,6,28,25,72,11,23,8,21,20,8,8,208,108,50,
%U 3,18,14,29,8,30,1,6,16,8,50,44,8,22,12,5,36,81,432,40
%N a(n) = p(n, 2)*p(n, 5)/p(n, 10) where p(n, b) is the period of repeating digits of 1/n in base b.
%H Michel Marcus, <a href="/A334487/b334487.txt">Table of n, a(n) for n = 1..5000</a>
%F a(n) = A007733(n)*A007736(n)/A007732(n).
%t f[n_, p_] := MultiplicativeOrder[p, n/(p^IntegerExponent[n, p])]; a[n_] := f[n, 2] * f[n, 5] / MultiplicativeOrder[10, n / 2^IntegerExponent[n, 2] / 5^IntegerExponent[n, 5]]; Array[a, 100] (* _Amiram Eldar_, May 04 2020 *)
%o (PARI)
%o a2(n) = znorder(Mod(2,n/2^valuation(n,2))); \\ A007733
%o a5(n) = znorder(Mod(5,n/5^valuation(n,5))); \\ A007736
%o a10(n) = znorder(Mod(10,n/2^valuation(n,2)/5^valuation(n,5))); \\ A007732
%o a(n) = a2(n)*a5(n)/a10(n);
%Y Cf. A007733, A007736, A007732, A334488.
%K nonn,base
%O 1,3
%A _Michel Marcus_, May 03 2020
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