A146348 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 3.

17, 37, 61, 101, 197, 257, 317, 401, 461, 557, 577, 677, 773, 1129, 1297, 1429, 1601, 1877, 1901, 2917, 3137, 4357, 4597, 5417, 5477, 6053, 7057, 8101, 8761, 8837, 10733, 11621, 12101, 13457, 13877, 14401, 15277, 15377, 15877, 16333, 16901, 17737, 17957, 18329, 21317, 22501, 23593, 24337, 25601, 28901, 30137, 30977, 32401, 33857, 41453, 41617, 42437, 44101

Offset: 1

Author

Artur Jasinski, Oct 30 2008

Keywords

• nonn

Comments

Primes in A146328. Finite A050952 is subset of this sequence.

From Michel Lagneau, Sep 03 2014: (Start)

The primes of the form p = n^2+1 for n>2 are in the sequence, and the continued fraction of (1+sqrt(p))/2 is [n/2; 1, 1, n-1, 1, 1, n-1, 1, 1, ...] with the period (1, 1, n-1).

We observe that the other primes {61, 317, 461, 557, 773, 1129, 1429, ...} are prime divisors of composite numbers of the form k^2+1 where k = 11, 114, 48, 118, 317, 168, 620, ... .

(End)

Another possibly infinite subset of the sequence is primes of the form 100*k^2-44*k+5, where the continued fraction is [5*k-1; 2, 2, 10*k-3, ...] with period [2, 2, 10*k-3]. This includes {61, 317, 773, 1429, 4597, 6053, ...}. - Robert Israel, Sep 03 2014

Links

• Amiram Eldar, Table of n, a(n) for n = 1..2500 (terms 1..200 from Zak Seidov)

Crossrefs

Cf. A000290, A050952, A078370, A146326-A146345, A146348-A146360, A146363.

Programs

Maple

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146348 := proc(n) RETURN(isprime(n) and A146326(n) = 3) ; end: for n from 2 to 4000 do if isA146348(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Mathematica

okQ[n_] := Length[ContinuedFraction[(1 + Sqrt[n])/2][[2]]] == 3; Select[Prime[Range[100]], okQ]

Extensions

1019 removed; more terms added by R. J. Mathar, Sep 06 2009

More terms from Zak Seidov, Mar 09 2011