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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, 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
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
Sequence in context: A341707 A347820 A318569 * A362806 A277749 A227629
KEYWORD
nonn
AUTHOR
Frank M Jackson and Michael B Rees, Jul 15 2020
STATUS
approved