A309498 - OEIS

%I #32 May 23 2024 03:55:48

%S 1,1,1,1,3,1,5,15,4,2,2,1,13,5,1,6,2,13,1,2,1,3,9,5,10,5,1,5,2,9,6,8,

%T 4,2,7,3,1,1,10,11,2,7,2,1,4,5,13,4,4,3,1,3,7,2,4,6,3,7,5,2,20,6,4,2,

%U 6,1,2,1,4,3,4,3,5,5,5,16,2,14,3,3,2,2,5,5

%N Least number k > 0 such that 4*p^2*k^2 + 1 is prime, where p = prime(n) is the n-th prime.

%C Gagola calculated the first 669 terms of this sequence (for all the primes p < 5000) using an HP 9830 in 1981. She found that the largest value of k was only 45 and that 84% of the values of k were less than or equal to 10.

%C The Generalized Dickson Conjecture implies that the sequence contains each positive integer infinitely many times. - _Robert Israel_, Aug 05 2019

%H Jason Yuen, <a href="/A309498/b309498.txt">Table of n, a(n) for n = 1..10000</a>

%H Gloria Gagola, <a href="https://www.jstor.org/stable/2689383">Progress on primes</a>, News and Letters, Mathematics Magazine, Vol. 54, No. 1 (1981), p. 43.

%e a(1) = 1 since 4*1^2*prime(1)^2 + 1 = 4*1*2^2 + 1 = 17 is prime.

%p f:= proc(n) local q,k;

%p   q:= 4*ithprime(n)^2;

%p   for k from 1 do

%p      if isprime(q*k^2+1) then return k fi

%p   od

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Aug 05 2019

%t a[n_] := Module[{k = 1, p = Prime[n]}, While[!PrimeQ[4 * k^2 * p^2 + 1], k++]; k]; Array[a, 100]

%o (PARI) a(n) = my(k=1, p=prime(n)); while (!isprime(4*p^2*k^2 + 1), k++); k; \\ _Michel Marcus_, Aug 05 2019

%Y Cf. A002496, A005574, A035092, A052291.

%K nonn,easy

%O 1,5

%A _Amiram Eldar_, Aug 05 2019