A146332 -id:A146332 - cp's OEIS frontend

A146352 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 7: primes in A146332.

89, 109, 113, 137, 373, 389, 509, 653, 797, 853, 997, 1009, 1493, 1997, 2309, 2621, 2677, 3797, 4973, 7817, 7873, 9829, 9833, 12197, 12269, 12821, 14009, 15773, 16661, 16673, 18253, 18269, 20389, 21557, 24197, 24533, 25037, 25741, 30677, 31973, 33941, 34253, 35977

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Amiram Eldar, Table of n, a(n) for n = 1..2000

Cf. A000290, A050950-A050969, A078370, A146326-A146345, A146348-A146360.

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146352 := proc(n) RETURN(isprime(n) and A146326(n) = 7) ; end: for n from 2 to 13000 do if isA146352(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Select[Range[2*10^4], PrimeQ[#] && Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]] == 7 &] (* Amiram Eldar, Mar 30 2020 *)

607 removed, 797 inserted by R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 30 2020

A146326 Length of the period of the continued fraction of (1+sqrt(n))/2.

Original entry on oeis.org

0, 2, 2, 0, 1, 4, 4, 4, 0, 2, 2, 2, 1, 4, 2, 0, 3, 6, 6, 4, 2, 6, 4, 4, 0, 2, 2, 4, 1, 2, 8, 4, 4, 4, 2, 0, 3, 6, 6, 8, 5, 4, 10, 6, 2, 8, 4, 4, 0, 2, 2, 4, 1, 6, 4, 2, 6, 6, 6, 4, 3, 4, 2, 0, 3, 6, 10, 6, 4, 6, 8, 4, 9, 6, 4, 8, 2, 4, 4, 4, 0, 2, 2, 2, 1, 6, 2, 8, 7, 2, 8, 8, 2, 12, 4, 8, 9, 4, 2, 0

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

First occurrence of n in this sequence see A146343.
Records see A146344.
Indices where records occurred see A146345.
a(n) =0 for n = k^2 (A000290).
a(n) =1 for n = 4 k^2 + 4 k + 5 (A078370). For primes see A005473.
a(n) =2 for n in A146327. For primes see A056899.
a(n) =3 for n in A146328. For primes see A146348.
a(n) =4 for n in A146329. For primes see A028871 - {2}.
a(n) =5 for n in A146330. For primes see A146350.
a(n) =6 for n in A146331. For primes see A146351.
a(n) =7 for n in A146332. For primes see A146352.
a(n) =8 for n in A146333. For primes see A146353.
a(n) =9 for n in A143577. For primes see A146354.
a(n)=10 for n in A146334. For primes see A146355.
a(n)=11 for n in A146335. For primes see A146356.
a(n)=12 for n in A146336. For primes see A146357.
a(n)=13 for n in A333640. For primes see A146358.
a(n)=14 for n in A146337. For primes see A146359.
a(n)=15 for n in A146338. For primes see A146360.
a(n)=16 for n in A146339. For primes see A146361.
a(n)=17 for n in A146340. For primes see A146362.

			a(2) = 2 because continued fraction of (1+sqrt(2))/2 = 1, 4, 1, 4, 1, 4, 1, ... has period (1,4) length 2.

R. J. Mathar, Table of n, a(n) for n = 1..20000.

Cf. A003285, A146343, A146344, A146345.

Cf. also A000290, A078370, A146327-A146342; A005473, A056899, A146348-A146362.

A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: seq(A146326(n),n=1..100) ; # R. J. Mathar, Sep 06 2009

Table[cf = ContinuedFraction[(1 + Sqrt[n])/2]; If[Head[cf[[-1]]] === List, Length[cf[[-1]]], 0], {n, 100}]
f[n_] := Length@ ContinuedFraction[(1 + Sqrt[n])/2][[-1]]; Array[f, 100] (* Robert G. Wilson v, Apr 11 2017 *)

a(n) = 0 iff n is a square (A000290). - Robert G. Wilson v, Apr 11 2017

a(39) and a(68) corrected by R. J. Mathar, Sep 06 2009

cp's OEIS Frontend

A146352 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 7: primes in A146332.

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