A014664 -id:A014664 - cp's OEIS frontend

A139669 Number of isomorphism classes of finite groups of order 11*2^n.

1, 2, 4, 12, 42, 195, 1387, 19324, 1083472

Offset: 0

Anthony D. Elmendorf (aelmendo(AT)calumet.purdue.edu), Jun 12 2008

This appears to be the smallest possible number of groups of order q*2^n for an odd number q.
Apparently, a(n) is also the number of isomorphism classes of finite groups of order 19*2^n and, more generally, of order p*2^n for primes p such that p is congruent to 3 modulo 4 and p+1 is not a power of 2.
Comment from Miles Englezou, Sep 26 2024: (Start)
The comment above which starts "Apparently, ... power of 2." is not true. (For a proof see the Miles Englezou link). However, it is true that a(0) to a(8) are the smallest possible number of groups of order q*2^n for an odd number q, and this can be generalized in the way stated below. (For further details see the Miles Englezou link).
A correct generalization of the 9 terms:
 The number of groups of order q*2^n is the least possible for prime q such that q == 3 (mod 4) and where the least 2^m such that 2^m == 1 (mod q) is greater than 2^n. Or put another way, if A014664(A080148(n)) > n, then for q = A000040(A080148(n)) the number of groups of order q*2^n is the least possible. (End)

			a(2) is the number of groups of order 11*2^2=44, which is 4 and also the number of groups of order 19*2^2=76, 23*2^2=92, etc.

J. H. Conway et al., The Symmetries of Things, Peters, 2008, p. 206.

John H. Conway, Heiko Dietrich and E. A. O'Brien, Counting groups: gnus, moas and other exotica.
Miles Englezou, Proofs

Cf. A000001, A002145, A014664, A080148.

A139669 := n -> GroupTheory[NumGroups](11*2^n);

a(n) = A000001(11*2^n). - Max Alekseyev, Apr 26 2010

a(8) from Max Alekseyev, Dec 24 2014

A143584 Integers that are equal to the multiplicative order of 2 modulo some overpseudoprime to base 2.

Original entry on oeis.org

11, 23, 25, 28, 29, 35, 36, 37, 39, 41, 43, 44, 45, 47, 48, 50, 51, 52, 53, 55, 57, 58, 59, 60, 63, 64, 66, 67, 68, 70, 71, 72, 73, 74, 75, 76, 79, 81, 82, 83, 84, 87, 88, 91, 92, 94, 95, 96, 97, 99, 100, 101, 102, 103, 104, 105, 106, 108, 109, 110, 111, 112

Offset: 1

Vladimir Shevelev, Aug 25 2008

nonn

A064078(a(n)) is a composite number. The sequence has a positive density since it contains, in particular, numbers of the form 8n+20 for n >= 1 (C. Pomerance, private correspondence). Since, e.g., 38 is not in the sequence, there is not an overpseudoprime m such that ord_m(2)=38.
Phi_{a(n)}(2), the a(n)-th cyclotomic polynomial of x evaluated at x=2 has at least 2 distinct prime factors that are not prime factors of the Phi_k(2) for any positive integer k < a(n). For example, Phi_11(2) = 2^11 - 1 = 2047 = 23 * 89 and Phi_25(2) = 2^20 + 2^15 + 2^10 + 2^5 + 1 = 1082401 = 601 * 1801. Note that p = a(n) is prime if and only if Phi_p(2) = 2^p - 1 is composite. - David Terr, Sep 09 2018
It is easy to prove the statement above. We use the fact that Phi_j(n) and Phi_k(n) are coprime whenever j and k are coprime as well as the fact that an overpseudoprime has at least 2 distinct prime factors. - David Terr, Oct 10 2018
A number k is included iff either 2^k-1 has more than one primitive prime factor (cf. A086251, A161508) or the only primitive prime factor of 2^k-1 is a Wieferich prime (no examples known). - Jeppe Stig Nielsen, Sep 01 2020

Jeppe Stig Nielsen, Table of n, a(n) for n = 1..1000

Cf. A002326, A014664, A064078, A086251, A122929, A141232, A161508.

Cf. A131952 (for the corresponding maximal overpseudoprimes).

isok(k) = my(m=polcyclo(k,2)); m/=gcd(m,k); m!=1&&!isprime(m) \\ Jeppe Stig Nielsen, Sep 01 2020

Name edited by Michel Marcus, Oct 06 2018
More terms from Michel Marcus, Oct 11 2018
Data for terms >= 100 corrected by Jeppe Stig Nielsen, Sep 01 2020

A170820 Let p = n-th prime; a(n) = (p-1)/(order of (p+3)/2 mod p).

Original entry on oeis.org

2, 1, 1, 3, 1, 6, 2, 4, 1, 1, 1, 2, 2, 4, 1, 5, 2, 10, 2, 3, 1, 1, 12, 4, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 5, 2, 2, 4, 3, 42, 1, 1, 1, 1, 2, 8, 1, 1, 2, 4, 1, 1, 7, 2, 4, 6, 2, 2, 4, 30, 2, 1, 1, 1, 2, 1, 3, 2, 2, 2, 1, 25, 4, 11, 1, 10, 2, 3, 1, 1, 8, 10, 33, 1, 2, 3, 1, 6, 2, 4, 1, 2, 1, 2, 2, 1

Offset: 3

N. J. A. Sloane, Dec 24 2009

nonn

Alois P. Heinz, Table of n, a(n) for n = 3..65536
I. Anderson and D. A. Preece, Combinatorially fruitful properties of 3*2^(-1) and 3*2^(-2) modulo p, Discr. Math., 310 (2010), 312-324.

Cf. A014664, A001917, A170821, A170822.

with(numtheory); [seq((ithprime(n)-1)/order((ithprime(n)+3)/2,ithprime(n)),n=3..130)];

a[n_] := Module[{p=Prime[n]}, (p-1)/MultiplicativeOrder[(p+3)/2, p]]; Array[a, 100, 3] (* Amiram Eldar, Dec 03 2018 *)

a(n) = my(p=prime(n)); (p-1)/znorder(Mod((p+3)/2, p)); \\ Michel Marcus, Dec 03 2018

A240657 Least k such that 2^k == -1 (mod prime(n)), or 0 if no such k exists.

Original entry on oeis.org

0, 1, 2, 0, 5, 6, 4, 9, 0, 14, 0, 18, 10, 7, 0, 26, 29, 30, 33, 0, 0, 0, 41, 0, 24, 50, 0, 53, 18, 14, 0, 65, 34, 69, 74, 0, 26, 81, 0, 86, 89, 90, 0, 48, 98, 0, 105, 0, 113, 38, 0, 0, 12, 25, 8, 0, 134, 0, 46, 35, 47, 146, 51, 0, 78, 158, 15, 0, 173, 174, 44, 0

Offset: 1

T. D. Noe, Apr 14 2014

nonn

The least k, if it exists, such that prime(n) divides 2^k + 1.

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A014664 (order of 2 mod prime(n)), A072936 (zero terms).

Table[p = Prime[n]; s = Select[Range[p/2], PowerMod[2, #, p] == p - 1 &, 1]; If[s == {}, 0, s[[1]]], {n, 100}]

a(n) = A014664(n)/2 if A014664(n) is even, otherwise 0.

A270600 a(n) is the smallest k > 1 not already in the sequence such that p = prime(n) is a factor of 2^k-1.

Original entry on oeis.org

2, 4, 3, 10, 12, 8, 18, 11, 28, 5, 36, 20, 14, 23, 52, 58, 60, 66, 35, 9, 39, 82, 22, 48, 100, 51, 106, 72, 56, 7, 130, 68, 138, 148, 15, 104, 162, 83, 172, 178, 180, 95, 96, 196, 99, 210, 37, 226, 76, 29

Offset: 2

Felix Fröhlich, Mar 20 2016

nonn

First differs from A014664 at n = 24.
a(n) is the smallest r such that p = prime(n) appears in row r of A060443 and r has not been the smallest such r for any previous p.
6 is not a term of the sequence, since 2^6-1 = 3^2 * 7, but 3 and 7 are already factors of 2^4-1 and 2^3-1 respectively.
Apart from 6, are there any other k > 1 that are not terms of the sequence?

Felix Fröhlich, Table of n, a(n) for n = 2..10000

Cf. A014664, A060443.

nextk(p, k) = my(kk=k+1); while(Mod(2, p)^kk!=1, kk++); kk
isinvec(v, k) = for(x=1, #v, if(v[x]==k, return(1))); return(0)
addtovec(v, k) = my(w=vector(1)); w[1]=k; v=concat(v, w); v
addnextknotinvec(v, p) = my(kk=2); while(isinvec(v, kk), kk=nextk(p, kk)); addtovec(v, kk)
my(i=0, v=[], p=3); while(i < 50, v=addnextknotinvec(v, p); p=nextprime(p+1); i++); v

A278967 a(n) = least multiple of the n-th prime that has the minimum Hamming weight (=A278966(n)).

Original entry on oeis.org

2, 3, 5, 7, 33, 65, 17, 513, 69, 16385, 31, 262145, 1025, 129, 517, 67108865, 536870913, 1073741825, 8589934593, 8449, 73, 1027, 2199023255553, 89, 16777217, 1125899906842625, 515, 9007199254740993, 262145, 16385, 127, 36893488147419103233, 17179869185, 590295810358705651713, 18889465931478580854785

Offset: 1

Charles R Greathouse IV, Dec 02 2016

nonn

Apart from the first term, all terms are odd.

			2 = 2^1 has Hamming weight 1 and so a(1) = 2.
3 = 2^1 + 2^0 has Hamming weight 2, and any multiple of 3 has a Hamming weight at least as high, so a(2) = 3.
5 = 2^2 + 2^0 has Hamming weight 2 and so similarly a(3) = 5.
7 = 2^2 + 2^1 + 2^0 has Hamming weight 3, and all powers of 2 are 1, 2, or 4 mod 7, and so all multiples of 7 have Hamming weight at least 3, so a(4) = 7.
11 = 2^3 + 2^1 + 2^0 has Hamming weight 3 but 33 = 2^5 + 2^0 has Hamming weight 2 so a(5) = 33.

Contains A143027 as subsequence.

Cf. A278966, A278968, A086342.

min1s(p)=my(o=znorder(Mod(2,p)), v1=Set(powers(Mod(2,p),o)), v=v1, s=1); while(!setsearch(v,Mod(0,p)), v=setbinop((x,y)->x+y,v,v1); s++); s
a(n,p=prime(n))=my(m=min1s(p),t=p,k=2*p); while(hammingweight(t)>m, t+=k); t

a(n) = 2^(A014664(n)/2) + 1 whenever A014664(n) is even. Also, a(n) = prime(n) whenever prime(n) is in A143027. - Max Alekseyev, May 22 2025

a(23)-a(25) from Charles R Greathouse IV, Dec 09 2016

Name clarified and terms a(26) onward added by Max Alekseyev, May 22 2025

A333245 Primes p such that the order of 2 mod p is less than the square root of p.

Original entry on oeis.org

31, 127, 257, 683, 1103, 1801, 2089, 2113, 2351, 2731, 3191, 4051, 4513, 5419, 6361, 8191, 9719, 11119, 11447, 13367, 14449, 14951, 20231, 20857, 23279, 23311, 26317, 29191, 30269, 32377, 37171, 38737, 39551, 43441, 43691, 49477, 54001, 55633, 55871, 59393

Offset: 1

Charles R Greathouse IV, Mar 12 2020

nonn

			The order of 2 mod 31 is 5, and sqrt(31) = 5.56776436283..., which is more than 5, so 31 is in the sequence.
The order of 2 mod 37 is 36, and sqrt(37) = 6.08276253..., which is significantly less than 36, so 37 is not in the sequence.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A014664, A002326.

using Nemo
function isA333245(n)
    ! isprime(n) && return false
    s, m, N = 0, 1, n
    r = isqrt(n)
    while true
        k = N + m
        v = valuation(k, 2)
        s += v
        s > r && return false
        m = k >> v
        m == 1 && break
    end
    return true
end
print([n for n in 3:2:60000 if isA333245(n)]) # Peter Luschny, Mar 16 2020

q:= p-> is(numtheory[order](2, p)^2Alois P. Heinz, Mar 16 2020

Select[Prime[Range[6000]], MultiplicativeOrder[2, #] < Sqrt[#] &] (* Amiram Eldar, Mar 16 2020 *)

list(lim)=my(v=List(),t,p,o); forfactored(P=30,lim\1, if(vecsum(P[2][,2])==1, t=znorder(Mod(2,p=P[1]),o); if(t^2

A367318 Lesser of twin primes p such that p and p+2 are both in A115591.

Original entry on oeis.org

191, 311, 1487, 1871, 2711, 2999, 3167, 3767, 4967, 5519, 7559, 8087, 10271, 11351, 11831, 13679, 15647, 18311, 18911, 21647, 22271, 22367, 23687, 25799, 26711, 27239, 27527, 27791, 29399, 29879, 31727, 31847, 33287, 34367, 35591, 38447, 38567, 40127, 40847, 42071

Offset: 1

Amiram Eldar, Nov 14 2023

nonn

Primes p such that p+2 is also a prime and (p-1)/ord(2, p) = (p+1)/ord(2, p+2) = 2, where ord(2,k) is the multiplicative order of 2 modulo k.
Equivalently, lesser of twin primes p such that ord(2, p+2) = ord(2, p) + 1,
Equal consecutive values in A001917 that correspond to twin primes (p, p+2) are either 1 if p is in A319248, or 2 if p is in this sequence.
Terms are congruent to 23 modulo 24. - Jianing Song, Nov 01 2024

Amiram Eldar, Table of n, a(n) for n = 1..10000

Subsequence of A001359 and A115591.

Cf. A001122, A001917, A002326, A014664, A319248, A333743.

Select[Prime[Range[2, 4400]], PrimeQ[# + 2] && MultiplicativeOrder[2, # + 2] == MultiplicativeOrder[2, #] + 1 &]

is(n) = isprime(n) && isprime(n+2) && znorder(Mod(2, n + 2)) == znorder(Mod(2, n)) + 1;

A367319 Base-2 Fermat pseudoprimes k such that (k-1)/ord(2, k) > (m-1)/ord(2, m) for all base-2 Fermat pseudoprimes m < k, where ord(2, k) is the multiplicative order of 2 modulo k.

Original entry on oeis.org

341, 1105, 1387, 2047, 4369, 4681, 5461, 13981, 15709, 35333, 42799, 60787, 126217, 158369, 215265, 256999, 266305, 486737, 617093, 1082401, 1398101, 2113665, 2304167, 4025905, 4188889, 4670029, 6236473, 6242685, 8388607, 13757653, 16843009, 17895697, 22369621

Offset: 1

Amiram Eldar, Nov 14 2023

nonn

Amiram Eldar, Table of n, a(n) for n = 1..182 (terms below 2^64)
Amiram Eldar, Table of n, a(n), (a(n)-1)/ord(2, a(n)) for n = 1..182

Subsequence of A001567.

Cf. A002326, A014664, A226216, A300101, A367320.

pspQ[n_] := CompositeQ[n] && PowerMod[2, n - 1, n] == 1; seq[kmax_] := Module[{s = {}, r, rm = 0}, Do[If[pspQ[k], r = (k - 1)/MultiplicativeOrder[2, k]; If[r > rm, rm = r; AppendTo[s, k]]], {k, 1, kmax}]; s]; seq[10^6]

ispsp(n) = n > 1 && n % 2 && Mod(2, n)^(n-1) == 1 && !isprime(n);
lista(kmax) = {my(r, rm = 0); for(k = 1, kmax, if(ispsp(k), r = (k-1)/znorder(Mod(2, k)); if(r > rm, rm = r; print1(k, ", "))));}

A250208 Ratio of the primitive part of 2^n-1 to the product of primitive prime factors of 2^n-1.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Eric Chen, Mar 02 2015

nonn

As with A178764, it can be shown that all terms are either 1 or prime.
a(2*3^n)   =  3 (n>=1).
a(4*5^n)   =  5 (n>=1).
a(3*7^n)   =  7 (n>=1).
a(10*11^n) = 11 (n>=1).
a(12*13^n) = 13 (n>=1).
a(8*17^n)  = 17 (n>=1).
a(18*19^n) = 19 (n>=1).
...
a(A014664(k)*prime(k)^n) = prime(k).
For other n (while Phi_n(2) is squarefree), a(n) = 1.
a(n) != 1 for n = {6, 18, 20, 21, 54, 100, 110, 136, 147, 155, 156, 162, ...}.
At least, a(A049093(n)) = 1. (In fact, since Phi_n(2) is not completely factored for n = 991, 1207, 1213, 1217, 1219, 1229, 1231, 1237, 1243, 1249, ..., so it is unknown whether they are squarefree or not, but it is likely that Phi_n(2) is squarefree for all n except 364 and 1755 (because it is likely 1093 and 3511 are the only two Wieferich primes), so a(991), a(1207), a(1213), ..., are likely to be 1.)

			a(11) = 1 since Phi_11(2) = (2^11-1)/(2-1) = 2047, and the primitive prime factors of 2^11-1 are 23 and 89, so a(11) = 2047/(23*89) = 1.
a(18) = 3 since Phi_18(2) = 2^6 - 2^3 + 1 = 57, and the only primitive prime factor of 2^18-1 is 19, so a(18) = 57/19 = 3.

Eric Chen, Table of n, a(n) for n = 1..990
Eric Chen, Table of n, a(n) for n = 1..1280 status (with unknown terms but all are conjectured to be 1)

Cf. A178764.

Cf. A001220, A237043, A093106, A049094, A049093, A014491, A072226, A161508.

a250208[n_] = If[n == 364, 1093, If[n == 1755, 3511, GCD[Cyclotomic[n, 2], n]]]; Table[a250208[n], {n, 0, 200}]

a(n) = if (n==364, 1093, if (n==1755, 3511, gcd(polcyclo(n, 2), n)));

isprimitive(p, n) = {for (r=1, n-1, if (((2^r-1) % p) == 0, return (0)); ); return (1); }
ppf(n) = {my(pf = factor(2^n-1)[,1]); prod(k=1,#pf, if (isprimitive(pf[k], n), pf[k], 1));}
a(n) = if (issquarefree(m=polcyclo(n,2)), gcd(m, n), m/ppf(n)); \\ Michel Marcus, Mar 06 2015

a(n) = A019320(n) / A064078(n) while Phi_n(2) is squarefree.
a(n) = GCD(Phi_n(2), n) while Phi_n(2) is squarefree.
Notice: a(364) = 1093, a(1755) = 3511. (See A001220.)

cp's OEIS Frontend

A139669 Number of isomorphism classes of finite groups of order 11*2^n.

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A143584 Integers that are equal to the multiplicative order of 2 modulo some overpseudoprime to base 2.

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A170820 Let p = n-th prime; a(n) = (p-1)/(order of (p+3)/2 mod p).

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A240657 Least k such that 2^k == -1 (mod prime(n)), or 0 if no such k exists.

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A270600 a(n) is the smallest k > 1 not already in the sequence such that p = prime(n) is a factor of 2^k-1.

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A278967 a(n) = least multiple of the n-th prime that has the minimum Hamming weight (=A278966(n)).

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A333245 Primes p such that the order of 2 mod p is less than the square root of p.

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A367318 Lesser of twin primes p such that p and p+2 are both in A115591.

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