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A088145 Let p = prime(n); then a(n) = (Sum(primitive roots of p) - moebius(p-1))/p. 3

%I #15 Feb 12 2013 17:18:33

%S 0,1,1,1,2,2,4,3,6,6,4,6,8,6,13,12,15,8,10,15,12,14,21,20,16,20,18,27,

%T 18,24,19,27,32,24,36,22,24,28,46,42,46,24,42,32,42,35,27,34,58,36,56,

%U 53,32,52,64,71,66,39,44,48,48,72,48,66,48,78,44,48,88,56,80

%N Let p = prime(n); then a(n) = (Sum(primitive roots of p) - moebius(p-1))/p.

%C Gauss proved that the sum of the primitive roots of p is congruent to moebius(p-1) modulo p, for all primes p. - _Jonathan Sondow_, Feb 09 2013

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Primitive_root_modulo_n#Arithmetic_facts">Primitive root</a>

%e The primitive roots of prime(4) = 7 are 3 and 5, and moebius(7-1) = A008683(6) = 1, so a(4) = (3+5-1)/7 = 7/7 = 1. - _Jonathan Sondow_, Feb 10 2013

%t PrimitiveRootQ[ a_Integer, p_Integer ] := Block[ {fac, res}, fac = FactorInteger[ p - 1 ]; res = Table[ PowerMod[ a, ( p - 1)/fac[ [ i, 1 ] ], p ], {i, Length[ fac ]} ]; ! MemberQ[ res, 1 ] ] PrimitiveRoots[ p_Integer ] := Select[ Range[ p - 1 ], PrimitiveRootQ[ #, p ] & ] Table[ (Total[ PrimitiveRoots[ Prime[ n ] ] ] - MoebiusMu[ Prime[ n ] - 1 ])/Prime[ n ], {n, 1, 100} ]

%t a[n_] := With[{p = Prime[n]}, Select[Range[p - 1], MultiplicativeOrder[#, p] == p - 1 &]]; Table[(Sum[a[n][[i]], {i, Length[a[n]]}] - MoebiusMu[Prime[n] - 1])/Prime[n], {n, 1,10}] (* _Jonathan Sondow_, Feb 09 2013 *)

%Y Cf. A008683, A060749, A123475, A222009.

%K nonn

%O 1,5

%A _Ed Pegg Jr_, Nov 03 2003

%E Definition corrected by _Jonathan Sondow_, Feb 09 2013

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Last modified April 23 20:33 EDT 2024. Contains 371916 sequences. (Running on oeis4.)