A256302 -id:A256302 - cp's OEIS frontend

A092749 a(n) is the least k such that m^2 + m + k is prime for m = 0..n.

2, 3, 5, 5, 11, 11, 11, 11, 11, 11, 17, 17, 17, 17, 17, 17, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41, 41

Offset: 0

Gabriel Cunningham (gcasey(AT)mit.edu), Apr 12 2004

From Pieter Moree (moree(AT)mpim-bonn.mpg.de), Apr 16 2004: (Start)
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.
Since the value of m^2+m+B is B^2 for m=B-1, one cannot possibly do better than this.
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.
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.
(End)
George Bright conjectured that a(n) exists for every n (private communication, 1974; see Dudley). - Charles R Greathouse IV, Sep 12 2013
Least prime in a succession of primes whose difference are n consecutive even numbers. - Robert G. Wilson v, Sep 30 2013
From Altug Alkan, Oct 06 2017: (Start)
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:
b_2(n): 2, 3, 7, 7, 7, 7, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19.
b_4(n): 2, 3, 5, 5, 23, 59, 59, 59, 59, 59, 59, 59, 59, 59, 653, 653, 653, 653.
b_5(n): 2, 3, 7, 7, 7, 7, 13, 13, 13, 13, 13, 13.
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.
b_7(n): 2, 3, 5, 5, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17. (End)

			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.
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

Underwood Dudley, Mathematical Cranks, MAA: Washington, DC, 1992. See pp. 62f.
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.

R. A. Mollin, Prime-producing quadratics, Amer. Math. Monthly 104 (1997), 529-544.

Cf. A014556.

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 *)
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 *)

isok(k,n) = {for (m=0, n, if (!isprime(m^2 + m + k), return(0));); return (1);}
a(n) = {my(k = 0); while(!isok(k,n), k++); k;} \\ Michel Marcus, Oct 06 2017

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A092749 a(n) is the least k such that m^2 + m + k is prime for m = 0..n.

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