A336280 Number of consecutive primes of the form k*prime(n) + 1, starting with the least such prime A035095(n), that divides the Wendt determinant A048954(prime(n)).

1, 1, 2, 1, 4, 3, 2, 3, 2, 3, 1, 5, 6, 2, 6, 3, 3, 3, 1, 6, 3, 5, 5, 7, 5, 5, 6, 7, 4, 7, 5, 10, 10, 4, 4, 6, 10, 3, 4, 12, 11, 5, 7, 8, 7, 8, 11, 4, 4, 4, 14, 7, 11, 7, 13, 11, 13, 7, 18, 18, 6, 7, 17, 12, 9, 7, 9, 14, 12, 9, 16, 14, 11, 13, 10

Offset

1,3

Comments

Michael B Rees has conjectured that:

1. for every prime p, the Wendt determinant Wendt(p) has all its prime factors that are greater than p of the form k*p + 1.

2. for every prime p = prime(n) and its corresponding Wendt determinant W(p) there exists a finite number of m consecutive primes (p_1,p_2,..,p_m) of the form k*p + 1 that will divide Wendt(p) where p_1 is always the least prime of the form k*p + 1.

This sequence gives the value m for each p = prime(n).

Links

Table of n, a(n) for n=1..75.

Gerard P. Michon, Factorization of Wendt's Determinant (table for n=1 to 114).

Eric Weisstein's World of Mathematics, Circulant matrix .

Wikipedia, Circulant matrix .

Example

a(6) = 3 gives p = prime(6) = 13 and W(13) = 3^6*53^2*79^2*131^2*521^2*8191. The sequence of primes of the form q = k*13 + 1, starting with the least such prime 53 that divide W(11) is (53, 79, 131). The sequence has 3 terms.

Mathematica

w[n_] := Module[{x}, Resultant[x^n-1, (1+x)^n-1, x]]; k[n_, m_] := Module[{p=Prime@n, q=0, lst={}}, Do[q++; While[! PrimeQ[p*q+1], q++]; AppendTo[lst, q], {m}]; lst];
lst1 = {}; Do[lst=k[n, 50]*Prime[n]+1; m = 1; Do[If[IntegerQ[w[Prime[n]]/lst[[m]]]&&m<=Length@lst, m++, Break[]], {Length@lst}]; AppendTo[lst1, m-1], {n, 1, 75}]; lst1

Crossrefs

Cf. A035095, A048954.

Sequence in context: A341707 A347820 A318569 * A362806 A277749 A227629

Adjacent sequences: A336277 A336278 A336279 * A336281 A336282 A336283

Keyword

nonn

Author

Frank M Jackson and Michael B Rees, Jul 15 2020

Status

approved