cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A146354 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 9: primes in A143577.

Original entry on oeis.org

73, 97, 233, 277, 349, 353, 613, 821, 877, 1181, 1277, 1613, 1637, 1693, 2357, 2777, 3557, 3989, 4157, 4517, 4889, 4933, 5261, 6113, 7213, 9133, 9181, 9749, 10313, 10909, 11057, 11213, 11257, 12161, 12301, 13033, 16217, 16741, 17989, 19469
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Links

Crossrefs

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

Programs

  • Mathematica
    Select[Prime[Range[2300]],Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]] == 9&] (* Harvey P. Dale, Aug 22 2011 *)

Extensions

A-number in definition corrected. 1613 and 4933 inserted, 9421 deleted, extended beyond 9749 by R. J. Mathar, Nov 09 2008

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

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

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.

Examples

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

Links

Crossrefs

Cf. A003285, A146343, A146344, A146345.
Cf. also A000290, A078370, A146327-A146342; A005473, A056899, A146348-A146362.

Programs

  • Maple
    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
  • Mathematica
    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 *)

Formula

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

Extensions

a(39) and a(68) corrected by R. J. Mathar, Sep 06 2009
Showing 1-2 of 2 results.

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