A146353 -id:A146353 - cp's OEIS frontend

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

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

A146333 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 8.

Original entry on oeis.org

31, 40, 46, 71, 76, 88, 91, 92, 96, 104, 108, 152, 153, 155, 176, 188, 192, 200, 206, 207, 234, 238, 261, 266, 276, 279, 280, 282, 320, 328, 335, 336, 348, 366, 378, 383, 386, 392, 408, 414, 450, 476, 477, 480, 488, 501, 503, 504, 505, 540, 542, 555, 558, 581

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

For primes in this sequence see A146353.

			a(1) = 31 because continued fraction of (1+sqrt(31))/2 = 3, 3, 1, 1, 10, 1, 1, 3, 5, 3, 1, 1, 10, 1, 1, 3, 5, 3, 1, 1, 10, 1, ... has period (3, 1, 1, 10, 1, 1, 3, 5) length 8.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)

Cf. A000290, 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: isA146333 := proc(n) RETURN(A146326(n) = 8) ; end: for n from 2 to 700 do if isA146333(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

cf8Q[n_]:=Module[{sqrt=Sqrt[n]},!IntegerQ[sqrt]&&Length[ ContinuedFraction[ (1+sqrt)/2][[2]]]==8]; Select[Range[600],cf8Q] (* Harvey P. Dale, Sep 06 2012 *)

155 and 279 etc. added, 311 etc. removed by R. J. Mathar, Sep 06 2009

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions