A133870 -id:A133870 - cp's OEIS frontend

A133869 Numbers k such that 32*k + 1 is prime.

3, 6, 8, 11, 14, 18, 20, 21, 24, 29, 36, 38, 39, 44, 50, 53, 59, 63, 65, 66, 71, 81, 83, 84, 86, 95, 98, 99, 104, 105, 108, 113, 125, 129, 134, 140, 141, 146, 150, 156, 161, 165, 170, 174, 183, 186, 191, 198, 204, 209, 218, 228, 231, 233, 234, 239, 240, 245, 246

Offset: 1

Zak Seidov, Sep 27 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A133870 (corresponding primes).

[n: n in [0..1500] | IsPrime(32*n+1)]; // Vincenzo Librandi, Dec 16 2010

Select[Range[800], PrimeQ[32 # + 1] &] (* Vincenzo Librandi, May 23 2014 *)

is(n)=isprime(32*n+1) \\ Charles R Greathouse IV, Jun 06 2017

A163592 Numbers n such that n^6 + 545 is prime.

Original entry on oeis.org

84, 1302, 10584, 11382, 12012, 12558, 13356, 15498, 19362, 20286, 20496, 22092, 23142, 23772, 25452, 26418, 26502, 26544, 28644, 29274, 29778, 31374, 31962, 35406, 36876, 37338, 39522, 40152, 40488, 41286, 42924, 43428, 45108, 46116, 47754, 47796, 48678

Offset: 1

Arkadiusz Wesolowski, Aug 01 2009

nonn

Arkadiusz Wesolowski, Table of n, a(n) for n = 1..1000
G. L. Honaker, Jr. and Chris Caldwell, Prime Curios! 545

Cf. A133870, A161998, A066386, A126893, A119276.

[n : n in [0..48678 by 2] | IsPrime(n^6+545)]; // Arkadiusz Wesolowski, Mar 05 2011

Select[Range[0, 48678, 2], PrimeQ[#^6 + 545] &] (* Arkadiusz Wesolowski, Mar 05 2011 *)

forstep(n=0, 48678, 2, if(isprime(n^6+545), print1(n, ", "))); \\ Arkadiusz Wesolowski, Mar 05 2011

A263770 Smallest prime q such that (prime(n)^2 + q*prime(n))/(prime(n) + 1) is an integer.

Original entry on oeis.org

7, 5, 7, 17, 13, 29, 19, 41, 73, 31, 97, 191, 43, 89, 97, 109, 61, 311, 137, 73, 149, 241, 337, 181, 197, 103, 313, 109, 331, 229, 257, 397, 139, 281, 151, 457, 317, 821, 337, 349, 181, 547, 193, 389, 199, 401, 1061, 449, 229, 461, 937, 241, 727, 757, 1033, 1321, 271, 1361, 557

Offset: 1

Juri-Stepan Gerasimov, Oct 25 2015

nonn

Least prime q such that q == 1 (mod prime(n) + 1).

Consists of the initial terms of sequences A002476, A002144, A002476, A007519, A068228, A140444, A061237, A141881, A107008, A132230, A133870, A141868, A124826, A142292, A142398, A141948, A088955, A142003, ...

Cf. A263729, A263730, A263769.

Table[q = 2; While[! IntegerQ[(Prime[n]^2 + q Prime@ n)/(Prime@ n + 1)], q = NextPrime@ q]; q, {n, 59}] (* Michael De Vlieger, Oct 26 2015 *)

a(n) = {p = prime(n); q = 2; while ((p^2 + p*q) % (p + 1), q = nextprime(q+1)); q;} \\ Michel Marcus, Oct 26 2015

5 is in this sequence because (prime(2)^2 + 5*prime(2))/(prime(2) + 1) = 6 and 5 is prime.

A376349 Number of isomorphism classes k of groups G of order p*2^n when G contains a unique Sylow p subgroup and the maximal 2^m dividing p-1 is such that 2^m >= 2^n.

Original entry on oeis.org

1, 2, 5, 15, 54, 247, 1684, 21820, 1118964

Offset: 0

Miles Englezou, Sep 19 2024

A Sylow p subgroup is a subgroup of order p^r that necessarily exists when r is a maximal power of p. It is not necessarily unique, but when it is unique it is normal in G.
The condition that G of order p*2^n contains a unique Sylow p subgroup places an upper bound on the number of isomorphism classes of G; it is equivalent to stating that the minimal 2^r such that 2^r == 1 (mod p) be such that 2^r > 2^n. The condition that the maximal 2^m dividing p-1, i.e. for p == 1 (mod 2^m), is such that 2^m >= 2^n ensures a lower bound which is equal to the upper bound. See the Miles Englezou link for a proof.
If we relax the two conditions and just consider an arbitrary odd prime p and the number of isomorphism classes for |G| = p*2^n, it is likely that the set of such numbers is unique to p. Since every odd prime has a minimal 2^r such that 2^r == 1 (mod p) (a consequence of Fermat's little theorem), when 2^r = 2^n for |G| = p*2^n, the number of isomorphism classes will differ from a(n) due to the existence of groups where the Sylow p subgroup is not unique.

			a(2) = 5 since D_(p*2^2), C_(p*2^2), C_(p*2^1) x C_2, and two semidirect products C_p : C_4 are all the groups of order p*2^2 for p satisfying the two conditions.
Table showing minimal 2^r and maximal 2^m (as defined in the Comments) for some primes:
---------------------------------------------------------------------------
p |      Minimal 2^r == 1 (mod p)       |   Maximal 2^m, p == 1 (mod 2^m)  |
---------------------------------------------------------------------------
2 |             2^0  = 1                |              2^0 = 1             |
3 |             2^2  = 4                |              2^1 = 2             |
5 |             2^4  = 16               |              2^2 = 4             |
7 |             2^3  = 8                |              2^1 = 2             |
11|             2^10 = 1024             |              2^1 = 2             |
13|             2^12 = 4096             |              2^2 = 4             |
17|             2^8  = 256              |              2^4 = 16            |
19|             2^18 = 262144           |              2^1 = 2             |
23|             2^11 = 2048             |              2^1 = 2             |
29|             2^28 = 268435456        |              2^2 = 4             |
31|             2^5  = 32               |              2^1 = 2             |
37|             2^36 = 68719476736      |              2^2 = 4             |
---------------------------------------------------------------------------
Table of primes satisfying 2^r > 2^n, and 2^m >= 2^n:
-------------------------------------------------------------------------------
   2^n   |                          primes                           |   a(n)  |
-------------------------------------------------------------------------------
2^0 = 1  |  all primes                                    = A000040  | 1       |
2^1 = 2  |  all primes > 2                                = A065091  | 2       |
2^2 = 4  |  5, 13, 17, 29, 37, 41, 53, ...                = A002144  | 5       |
2^3 = 8  |  17, 41, 73, 89, 97, 113, 137, ...             = A007519  | 15      |
2^4 = 16 |  17, 97, 113, 193, 241, 257, 337 ...           = A094407  | 54      |
2^5 = 32 |  97, 193, 257, 353, 449, 577, 641, ...         = A133870  | 247     |
2^6 = 64 |  193, 257, 449, 577, 641, 769, 1153, ...       = A142925  | 1684    |
2^7 = 128|  257, 641, 769, 1153, 1409, 2689, 3329, ...    = A208177  | 21820   |
2^8 = 256|  257, 769, 3329, 7937, 9473, 14081, 14593 ...  = A105131  | 1118964 |
-------------------------------------------------------------------------------

Miles Englezou, Proof

Cf. A000001, A000040, A065091, A002144, A007519, A094407, A133870, A142925, A208177, A105131, A139669, A014664, A023506, A058384.

S:=[];
for i in [0..8] do
    n:=7681*2^i; # 7681 is an appropriate prime for reproducing up to a(8)
    S:=Concatenation(S,[NrSmallGroups(n)]);
od;
Print(S);

a(n) = A000001(p*2^(n)) for every p satisfying the two conditions mentioned in Comments.

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A133869 Numbers k such that 32*k + 1 is prime.

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A163592 Numbers n such that n^6 + 545 is prime.

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A263770 Smallest prime q such that (prime(n)^2 + q*prime(n))/(prime(n) + 1) is an integer.

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A376349 Number of isomorphism classes k of groups G of order p*2^n when G contains a unique Sylow p subgroup and the maximal 2^m dividing p-1 is such that 2^m >= 2^n.

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