A050975 -id:A050975 - cp's OEIS frontend

A053446 Multiplicative order of 3 mod m, where gcd(m, 3) = 1.

1, 1, 2, 4, 6, 2, 4, 5, 3, 6, 4, 16, 18, 4, 5, 11, 20, 3, 6, 28, 30, 8, 16, 12, 18, 18, 4, 8, 42, 10, 11, 23, 42, 20, 6, 52, 20, 6, 28, 29, 10, 30, 16, 12, 22, 16, 12, 35, 12, 18, 18, 30, 78, 4, 8, 41, 16, 42, 10, 88, 6, 22, 23, 36, 48, 42, 20, 100, 34, 6, 52, 53, 27, 20, 12, 112, 44

Offset: 1

David W. Wilson

nonn

Essentially the same as A050975. - R. J. Mathar, Oct 13 2008

Let k, m be any positive numbers not divisible by 3. Let k <+> m denote that of the two numbers k + m, k + 2*m which is divisible by 3. Finally, for a number t divisible by 3, let t* = t/3^s where s is the 3-adic order of t. Let u = u(n) be the n-th number which is not divisible by 3. Consider the following algorithm of the calculating a(n), similar to the algorithm in A002326: Compute successively r_1 = (1 <+> u)*, r_2 = (r_1 <+> u)*, ..., r_h = (r_(h-1) <+> u)* and finish as soon as r_h = 1. Then a(n) = s(1 <+> u) + s(r_1 <+> u) + ... + s(r_(h-1) <+> u). Note that by a similar algorithm one can compute an arbitrary multiplicative order of a mod b, where gcd(a, b) = 1. - Vladimir Shevelev, Oct 06 2017

			From _Vladimir Shevelev_, Oct 06 2017: (Start)
7 is the fifth number not divisible by 3. According to the algorithm in the comment we have in the form of a "finite continued fraction"
    1 + 14
    ------ + 7
       3
    ---------- + 14
          3
    ----------------- + 7
            3^2
    ---------------------- = 1
               3^2
Summing the exponents of 3 in the denominators, we obtain a(5) = 1 + 1 + 2 + 2 = 6. (End)

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A001651, A002326, A050975.

List(Filtered([1..130],n->Gcd(n,3)=1),n->OrderMod(3,n)); # Muniru A Asiru, Feb 26 2019

MultiplicativeOrder[3, #] & /@ Select[ Range@ 115, GCD[3, #] == 1 &] (* Robert G. Wilson v, Apr 05 2011 *)

lista(nn) = {for (n=1, nn, if (gcd(n,3) == 1, print1(znorder(Mod(3, n)), ", ")););} \\ Michel Marcus, Feb 06 2015

[Mod(3,n).multiplicative_order() for n in (1..115) if gcd(n,3) == 1] # Peter Luschny, Oct 07 2017

a(n) = multiplicative order of 3 modulo floor((3*n-1)/2) = A001651(n), for n >= 1. - Wolfdieter Lang, Sep 28 2020

A050977 Haupt-exponents of 5 modulo integers relatively prime to 5.

Original entry on oeis.org

1, 2, 1, 2, 6, 2, 6, 5, 2, 4, 6, 4, 16, 6, 9, 6, 5, 22, 2, 4, 18, 6, 14, 3, 8, 10, 16, 6, 36, 9, 4, 20, 6, 42, 5, 22, 46, 4, 42, 16, 4, 52, 18, 6, 18, 14, 29, 30, 3, 6, 16, 10, 22, 16, 22, 5, 6, 72, 36, 9, 30, 4, 39, 54, 20, 82, 6, 42, 14, 10, 44, 12, 22, 6, 46, 8, 96, 42, 30, 25, 16

Offset: 1

Eric W. Weisstein

nonn

R. J. Mathar, Table of n, a(n) for n = 1..7999
Eric Weisstein's World of Mathematics, Multiplicative Order

Cf. A002326 (base 2), A050975 (base 3), A002329.

n := 1 :
for i from 2 to 10000 do
    if igcd(i,5) = 1 then
        printf("%d %d\n",n,numtheory[order](5,i)) ;
        n := n+1 ;
    end if;
end do: # R. J. Mathar, Oct 14 2014

A297363 Numbers k such that (3^ord(3, k) - 1)/k is prime, where ord(3, k) is the multiplicative order of 3 (mod k).

Original entry on oeis.org

1, 4, 13, 16, 22, 40, 46, 56, 94, 104, 121, 160, 364, 526, 862, 968, 1093, 1312, 1514, 3146, 3194, 3280, 3742, 4376, 5368, 7280, 7702, 8744, 9841, 28418, 29524, 40880, 69022, 75920, 88573, 106288, 157394

Offset: 1

Amiram Eldar, Dec 29 2017

The corresponding primes are 2, 2, 2, 5, 11, 2, 3851, 13, 1001523179, 7, 2, 41, 2, 605199588591144003100881306574406851660288427740394885828171, ...

			46 is in the sequence since ord(3, 46) = 11 and (3^11 - 1)/46 = 3851 is prime.

Cf. A050975, A053446.

aQ[n_] := PrimeQ[(3^MultiplicativeOrder[3, n] - 1)/n]; Select[ Range[10000], aQ ]

isok(n) = (gcd(n,3) == 1) && isprime((3^znorder(Mod(3, n)) - 1)/n); \\ Michel Marcus, Dec 30 2017

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A053446 Multiplicative order of 3 mod m, where gcd(m, 3) = 1.

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A050977 Haupt-exponents of 5 modulo integers relatively prime to 5.

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A297363 Numbers k such that (3^ord(3, k) - 1)/k is prime, where ord(3, k) is the multiplicative order of 3 (mod k).

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