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%I #26 Apr 13 2024 22:59:09
%S 2,3,29,5,23,269,272879,149,61463,929,7426253,2609,233,59
%N a(n) is the least prime p such that p + 4*k*(k+1) is prime for 0 <= k <= n-1 but not for k=n.
%C a(15) > 3277860277, a(16) > 3103623446, a(17) > 2853255995,
%C a(18) = 653, a(19) > 2480173428, a(20) > 2058783580, a(21) > 1894529774, a(22) > 1896261075, a(23) > 1836831342, a(24), ..., a(100) > 15000000.
%C Other than a(1)-a(14) and a(18), no terms < 24870000007. - _Michael S. Branicky_, Apr 12 2024
%C From _David A. Corneth_, Apr 12 2024: (Start)
%C Using remainders mod q we can restrict the search. For example for a(15) a term can only be 2, 3 or 5 (mod 7). Or maybe 7 itself. If a(15) = p == 1 (mod 7) then for k = 3 we have q + 4*3*(3+1) == 0 mod 7. Similarily number 0, 4 and 6 (mod 7) produce a multiple of 7 where they should not.
%C Doing so for various primes mod q we can reduce the number of remainders and with that the search space by combining the possible remainders using the Chinese Remainder Theorem (CRT).
%C So the possible remainders mod 2 are 1. The possible remainders mod 3 are 2. Using the CRT, a number of the form 1 (mod 2) and 2 (mod 3) simultaneously is of the form 5 (mod 6).
%C a(15) > 2.3*10^13 if it exists. (End)
%p f:= proc(p) local k;
%p for k from 1 while isprime(p+k*(k+1)*4) do od:
%p k
%p end proc:
%p A:= Vector(12): count:= 0:
%p for i from 1 while count < 12 do
%p v:= f(ithprime(i));
%p if A[v] = 0 then count:= count+1; A[v]:= ithprime(i) fi
%p od:
%p convert(A,list);
%t Table[p=1;m=4;Monitor[Parallelize[While[True,If[And[MemberQ[PrimeQ[Table[p+m*k*(k+1),{k,0,n-1}]],False]==False,PrimeQ[p+m*n*(n+1)]==False],Break[]];p++];p],p],{n,1,10}]
%o (PARI) isok(p, n) = for (k=0, n-1, if (! isprime(p + 4*k*(k+1)), return(0))); return (!isprime(p + 4*n*(n+1)));
%o a(n) = my(p=2); while (!isok(p, n), p=nextprime(p+1)); p; \\ _Michel Marcus_, Mar 12 2024
%o (Python)
%o from sympy import isprime, nextprime
%o from itertools import count, islice
%o def f(p):
%o k = 1
%o while isprime(p+4*k*(k+1)): k += 1
%o return k
%o def agen(verbose=False): # generator of terms
%o adict, n, p = dict(), 1, 1
%o while True:
%o p = nextprime(p)
%o v = f(p)
%o if v not in adict:
%o adict[v] = p
%o if verbose: print("FOUND", v, p)
%o while n in adict:
%o yield adict[n]; n += 1
%o print(list(islice(agen(), 14))) # _Michael S. Branicky_, Apr 12 2024
%Y Cf. A164926, A370387.
%K nonn,more
%O 1,1
%A _J.W.L. (Jan) Eerland_, Mar 08 2024
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