OFFSET
1,1
COMMENTS
Since all consecutive primes, 2 < p < q, are odd, therefore gcd(p+1, q+1) must be even.
LINKS
Zak Seidov, Robert G. Wilson v, and Charles R Greathouse IV, Table of n, a(n) for n = 1..200 (1..100 terms from Seidov, 101..140 from Wilson, 141..200 from Greathouse)
FORMULA
Conjecture: a(n) = least k such that A001223(k) = 2n and A000040(k) == -1 (mod 2n). - Zak Seidov, Aug 16 2015
EXAMPLE
a(1) = 2, the first entry in A066940,
a(2) = 4, the first entry in A066941,
a(3) = 9, the first entry in A066942,
a(4) = 72, the first entry in A066943,
a(5) = 34, the first entry in A066944.
That is to say that the first k-th prime that has gcd(prime(k+1)+1, prime(k)+1)) of 2, 4, 6, 8, 10, ..., are k = 2, 4, 9, 72, 34, ..., and the prime_k = 3, 7, 23, 359, 139, 467, 293, ... (A067604).
If the floor of GCD is used, then a(0) equals 1.
MATHEMATICA
t = 0*Range@ 70; p = 3; q = 5; While[p < 15*10^6, d = GCD[p + 1, q + 1]/2; If[ t[[d]] == 0, t[[d]] = PrimePi@ p]; p = q; q = NextPrime@ q]; t
PROG
(PARI) a(n) = p=2; q=3; k=1; while(gcd(p+1, q+1) != 2*n, k++; p=q; q = nextprime(p+1); ); k; \\ Michel Marcus, Aug 16 2015
(PARI) a(n)=my(p=2, k=2*n, t); forprime(q=3, , t++; if((q-p)%k==0 && (p+1)%k==0, return(t)); p=q) \\ Charles R Greathouse IV, Aug 17 2015
(PARI) a(n)=my(k=2*n); forstep(p=k-1, oo, k, if(isprime(p) && (nextprime(p+1)-p)%k==0, return(primepi(p)))) \\ Charles R Greathouse IV, Aug 17 2015
(MATLAB)
P = primes(10^8);
G = gcd(P(1:end-1)+1, P(2:end)+1);
A = zeros(1, 66);
for n = 1:66
A(n) = find(G == 2*n, 1, 'first');
end
A % Robert Israel, Aug 16 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Jan 31 2002
EXTENSIONS
Edited by Robert G. Wilson v, Aug 17 2015 at the direction of Zak Seidov
STATUS
approved