A336280 - OEIS

%I #11 Jul 25 2020 10:41:01

%S 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,

%T 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,

%U 17,12,9,7,9,14,12,9,16,14,11,13,10

%N 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)).

%C Michael B Rees has conjectured that:

%C 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.

%C 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.

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

%H Gerard P. Michon, <a href="http://www.numericana.com/data/wendt.htm">Factorization of Wendt's Determinant</a> (table for n=1 to 114).

%H Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CirculantMatrix.html">Circulant matrix </a>.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Circulant_matrix">Circulant matrix </a>.

%e 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.

%t 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];

%t 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

%Y Cf. A035095, A048954.

%K nonn

%O 1,3

%A _Frank M Jackson_ and Michael B Rees, Jul 15 2020