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%I #29 Oct 15 2023 05:14:04
%S 41,41,122,162,299,326,326,1064,1064,1064,1064,1064,5664,5664,5664,
%T 5664,9265,9265,9265,22818,22818,37784,37784,47494,100202,100202,
%U 100202,167628,167628,167628,167628,167628,167628,167628,167628,176956,176956,176956,1081297
%N Starts of a run of at least n consecutive numbers k for which k^2 - k + 41 is composite.
%C 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.
%D Thomas Koshy, Elementary Number Theory with Applications, Academic Press, 2nd edition, 2007, Chapter 2, p. 147, exercise 50.
%H Amiram Eldar, <a href="/A284043/b284043.txt">Table of n, a(n) for n = 1..130</a> (terms below 10^10)
%H Sidney Kravitz, <a href="http://www.jstor.org/stable/2687926">Problem 527</a>, Mathematics Magazine, Vol. 36, No. 4 (1963), p. 264.
%H Lawrence A. Ringenberg et al., <a href="http://www.jstor.org/stable/2688531">A Prime Generator</a>, Solutions to Problem 527, Mathematics Magazine, Vol. 37, No. 2 (1964), pp. 122-123.
%e 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.
%t 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
%Y Cf. A002837, A005846, A007634, A056561, A145292, A202018.
%K nonn
%O 1,1
%A _Amiram Eldar_, Jun 14 2017
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