A146361 -id:A146361 - 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

A146339 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 16.

Original entry on oeis.org

172, 191, 217, 232, 249, 310, 311, 329, 343, 344, 355, 369, 391, 393, 416, 428, 431, 446, 496, 513, 520, 524, 536, 537, 550, 559, 589, 647, 655, 679, 682, 686, 700, 704, 748, 760, 768, 775, 802, 816, 848, 851, 872, 927, 995, 996, 1036, 1058, 1079, 1080, 1120, 1136

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

For primes in this sequence see A146361.

			a(1) = 191 because continued fraction of (1+sqrt(191))/2 = 7, 2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13, 2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13, 2, 2, 3, 1, 1, 4, 1, 26... has period (2, 2, 3, 1, 1, 4, 1, 26, 1, 4, 1, 1, 3, 2, 2, 13) length 16.

Amiram Eldar, Table of n, a(n) for n = 1..10000

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:
isA146339 := proc(n) RETURN(A146326(n) = 16) ; end:
for n from 2 to 1000 do if isA146339(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

Select[Range[1000], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 16 &]  (* Amiram Eldar, Mar 31 2020 *)

311 inserted, sequence extended by R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 31 2020

cp's OEIS Frontend

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

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