|
%I #33 Sep 03 2022 21:45:06
%S 0,2,3,4,5,7,9,10,11,12,14,16,17,18,19,21,23,24,25,26,28,30,31,32,33,
%T 35,37,38,39,40,42,44,45,46,47,49,51,52,53,54,56,58,59,60,61,63,65,66,
%U 67,68,70,72,73,74,75,77,79
%N Numbers that are congruent to {0, 2, 3, 4, 5} mod 7.
%C Conjecture: Apart from 0, and the further exclusions noted below, the sequence gives the values of c/6 such that an infinite number of primes, p, result in both p^3+c and p^3-c being positive primes. Taking the complement we say: the excluded c/6 values are {1,6} mod 7. See A005097 for a conjecture on the modulo patterns of excluded c/6 values for the general case of p^q + c and p^q - c both prime, for any q > 0, and see A047222 for q=2. Note that polynomial factorization also excludes a few c/6 values. This occurs here when c is an even cube (A016743), which requires a further exclusion of certain c/6 values in this sequence when (6c)^3/6 == 0 (mod 7), or c/6 = {0, 12348, 98784, ...}. - _Richard R. Forberg_, Jun 28 2016
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).
%F G.f.: x^2*(2*x^2 + 3*x + 2)*(x^2 - x + 1) / ( (x^4 + x^3 + x^2 + x + 1)*(x-1)^2 ). - _R. J. Mathar_, Dec 04 2011
%F a(n) = a(n-1) + a(n-5) - a(n-6). - _Wesley Ivan Hurt_, Sep 03 2022
%t Table[7 n + {0, 2, 3, 4, 5}, {n, 0, 12}] // Flatten (* or *)
%t Select[Range[0, 79], ! MemberQ[{1, 6}, Mod[#, 7]] &] (* or *)
%t Rest@ CoefficientList[Series[x^2 (2 x^2 + 3 x + 2) (x^2 - x + 1)/((x^4 + x^3 + x^2 + x + 1) (x - 1)^2), {x, 0, 57}], x] (* _Michael De Vlieger_, Jul 25 2016 *)
%Y Cf. A047222, A005097.
%K nonn
%O 1,2
%A _N. J. A. Sloane_
|
