A284043 Starts of a run of at least n consecutive numbers k for which k^2 - k + 41 is composite.

41, 41, 122, 162, 299, 326, 326, 1064, 1064, 1064, 1064, 1064, 5664, 5664, 5664, 5664, 9265, 9265, 9265, 22818, 22818, 37784, 37784, 47494, 100202, 100202, 100202, 167628, 167628, 167628, 167628, 167628, 167628, 167628, 167628, 176956, 176956, 176956, 1081297

Offset: 1

Author

Amiram Eldar, Jun 14 2017

Keywords

• nonn

Comments

This sequence is inspired by the problem proposed by Sidney Kravitz in 1963: "It is known that f(n)=n^2-n+41 yields prime numbers for n=1, 2, ..., 40. Find a sequence of 40 consecutive values of n for which f(n) is composite." Lawrence A. Ringenberg and others suggested the solution that starts at f(1)*f(2)*...*f(40)+1 (about 4.890... * 10^101). B. A. Hausmann suggested the smaller solution that starts at f(1)*f(2)*...*f(20)-19 (about 3.213... * 10^42). The smallest solution is a(40) = 1081297.

Examples

			The values of f(n)=n^2-n+41 at 122, 123 and 124 are: 14803 = 113*131, 15047 = 41*367 and 15293 = 41*373. This is the first case of 3 consecutive composite values, thus a(3) = 122.
		

References

• Thomas Koshy, Elementary Number Theory with Applications, Academic Press, 2nd edition, 2007, Chapter 2, p. 147, exercise 50.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..130 (terms below 10^10)
• Sidney Kravitz, Problem 527, Mathematics Magazine, Vol. 36, No. 4 (1963), p. 264.
• Lawrence A. Ringenberg et al., A Prime Generator, Solutions to Problem 527, Mathematics Magazine, Vol. 37, No. 2 (1964), pp. 122-123.

Crossrefs

Cf. A002837, A005846, A007634, A056561, A145292, A202018.

Programs

Mathematica

f[n_] := n^2 - n + 41; a = PrimeQ[f[Range[1, 10^7]]]; b = Split[a]; c = Length /@ b; d = Accumulate[c]; nc = Length[c]; e = {}; For[len = 0, len < 100, len++; k = 2;  While[k <= nc && c[[k]] < len, k += 2]; If[k <= nc && c[[k]] >= len, ind = d[[k - 1]] + 1; e = AppendTo[e, ind]]]; e