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%I #43 Jun 26 2019 08:37:47
%S 2,3,5,5,11,11,11,11,11,11,17,17,17,17,17,17,41,41,41,41,41,41,41,41,
%T 41,41,41,41,41,41,41,41,41,41,41,41,41,41,41,41
%N a(n) is the least k such that m^2 + m + k is prime for m = 0..n.
%C From Pieter Moree (moree(AT)mpim-bonn.mpg.de), Apr 16 2004: (Start)
%C The numbers 2, 3, 5, 11, 17, and 41 above are the only numbers B such that m^2+m+B is prime for m=0,...,B-2; this can be proved (see Mollin's paper) and is closely related to the celebrated Rabinowitsch criterion.
%C Since the value of m^2+m+B is B^2 for m=B-1, one cannot possibly do better than this.
%C An obvious question of course is whether for given n, a(n) exists at all. This is far from obvious. Assuming the generally believed k-tuplets conjecture, the answer is yes as was shown by Andrew Granville. For a proof (which is not very difficult) see the paper by Mollin.
%C It is also known, due to work of Lukes, Patterson and Williams, that any further elements in the above sequence, if they exist, exceed 10^18.
%C (End)
%C George Bright conjectured that a(n) exists for every n (private communication, 1974; see Dudley). - _Charles R Greathouse IV_, Sep 12 2013
%C Least prime in a succession of primes whose difference are n consecutive even numbers. - _Robert G. Wilson v_, Sep 30 2013
%C From _Altug Alkan_, Oct 06 2017: (Start)
%C Let b_i(n) be the least k such that i*(m^2 + m) + k is prime for m = 0..n and this sequence be the b_1(n) and b_3(n) be the A256302(n). Some initial values of b_i(n) for 2 <= i <= 7 are:
%C b_2(n): 2, 3, 7, 7, 7, 7, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19.
%C b_4(n): 2, 3, 5, 5, 23, 59, 59, 59, 59, 59, 59, 59, 59, 59, 653, 653, 653, 653.
%C b_5(n): 2, 3, 7, 7, 7, 7, 13, 13, 13, 13, 13, 13.
%C b_6(n): 2, 5, 5, 7, 7, 11, 11, 11, 11, 17, 17, 17, 17, 17, 17, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31, 31.
%C b_7(n): 2, 3, 5, 5, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17. (End)
%D Underwood Dudley, Mathematical Cranks, MAA: Washington, DC, 1992. See pp. 62f.
%D R. F. Lukes, C. D. Patterson, and H. C. Williams, Numerical sieving devices: their history and some applications, Nieuw Archief Wisk. 13 (1995), pp. 113-139.
%H R. A. Mollin, <a href="http://www.jstor.org/stable/2975080">Prime-producing quadratics</a>, Amer. Math. Monthly 104 (1997), 529-544.
%e a(1) = 3 because 0^2 + 0 + 3 = 3 is prime and 1^2 + 1 + 3 = 5 is prime and it is the smallest number with the required properties.
%e a(2) = 5 because 5, 7, and 11 are primes; a(3) = 5 because 5, 7, 11, and 17 are primes; a(4) = 11 because 11, 13, 17, 23, and 31 are prime. - _Robert G. Wilson v_, Sep 30 2013
%t allPrime[n_, k_] := And @@ PrimeQ[Table[m^2 + m + k, {m, 0, n}]]; Table[k = 0; While[! allPrime[n, k], k++]; k, {n, 0, 39}] (* _T. D. Noe_, Mar 05 2012 *)
%t f[n_] := Block[{p = FoldList[#1 + #2 &, 1, 2 Range@ n]}, While[ Union[ PrimeQ@ p][[1]] == False, p = p + 2]; p[[1]]]; f[0] = 2; Array[f, 40, 0] (* _Robert G. Wilson v_, Sep 30 2013 *)
%o (PARI) isok(k,n) = {for (m=0, n, if (!isprime(m^2 + m + k), return(0));); return (1);}
%o a(n) = {my(k = 0); while(!isok(k,n), k++); k;} \\ _Michel Marcus_, Oct 06 2017
%Y Cf. A014556.
%K nonn,more
%O 0,1
%A Gabriel Cunningham (gcasey(AT)mit.edu), Apr 12 2004
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