login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Number of distinct primes among the cubes mod n.
1

%I #13 Jun 28 2018 02:44:35

%S 0,0,1,1,2,3,0,3,0,4,4,4,1,2,6,5,6,1,2,7,2,8,8,8,8,2,2,2,9,10,3,10,11,

%T 11,3,2,4,5,3,11,12,4,3,13,3,14,14,14,4,14,15,4,15,4,16,5,5,16,16,16,

%U 6,6,0,17,5,18,5,18,19,5

%N Number of distinct primes among the cubes mod n.

%C This is to cubes A000578 as A132213 is to squares A000290.

%C It seems that the size of a(n) as compared to its surrounding elements is dependent on whether or not n is in A088232. If n is in A088232 the sequence assumes "big" values, otherwise the values will be "small". - _Stefan Steinerberger_, Nov 24 2007

%C If n is in A088232, a(n) = A000720(n-1) - A056170(n). - _Robert Israel_, Jun 28 2018

%H Robert Israel, <a href="/A132385/b132385.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Card{p = k^3 mod n, for primes p and for all integers k}.

%e a(10) = 4 because the cubes mod 10 repeat 0, 1, 8, 7, 4, 5, 6, 3, 2, 9, 0, 1, 8, 7, 4, 5, ... of which the 4 distinct primes are {2, 3, 5, 7}.

%p f:= proc(n)

%p if numtheory:-phi(n) mod 3 = 0 then nops(select(isprime, {seq(i^3 mod n, i=0..n-1)}))

%p else numtheory:-pi(n-1) - nops(select(t -> t[2]>1, ifactors(n)[2]))

%p fi

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Jun 28 2018

%t Table[Length[Select[Union[Table[Mod[i^3, n], {i, 0, n}], Table[Mod[i^3, n], {i, 0, n}]], PrimeQ[ # ] &]], {n, 1, 70}] (* _Stefan Steinerberger_, Nov 12 2007 *)

%Y Cf. A000040, A000578, A000720, A056170, A132213.

%K easy,nonn,look

%O 1,5

%A _Jonathan Vos Post_, Nov 07 2007

%E More terms from _Stefan Steinerberger_, Nov 12 2007

%E Spelling/notation corrections by _Charles R Greathouse IV_, Mar 18 2010