A146326 -id:A146326 - cp's OEIS frontend

A146358 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 13: primes in A333640.

421, 757, 1021, 1097, 1117, 1301, 1553, 1973, 2069, 2237, 2273, 2789, 2861, 3373, 3461, 3517, 3877, 3917, 4133, 4397, 4481, 5521, 5573, 5717, 6221, 6317, 6637, 6997, 7253, 7517, 8741, 9049, 9173, 9437, 10181, 10949, 11597, 11789, 12497, 15473, 15797, 16141, 18353

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, A333640, A146348-A146360.

Select[Range[2*10^4], PrimeQ[#] && Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]] == 13 &] (* Amiram Eldar, Mar 30 2020 *)
Select[Prime[Range[2500]],Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]]==13&] (* Harvey P. Dale, Mar 05 2023 *)

Definition corrected, 3 terms added. - R. J. Mathar, Nov 08 2008

More terms from Amiram Eldar, Mar 30 2020

A146359 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 14: primes in A146337.

Original entry on oeis.org

179, 251, 307, 347, 467, 587, 683, 1987, 5099, 5683, 7883, 8059, 8707, 12227, 14867, 15083, 15227, 22283, 34883, 40627, 42787, 47819, 50147, 51683, 68147, 73547, 78467, 84523, 84979, 89051, 95219, 104947, 106451, 107699, 132707, 134291, 142811, 149939, 164051

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Amiram Eldar, Table of n, a(n) for n = 1..1000

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

A := proc(n) local c; try c := numtheory[cfrac](1/2+sqrt(n)/2,'periodic,quotients') ; RETURN(nops(c[2]) ); catch: RETURN(-1) end try ; end: isA146337 := proc(n) if A(n) = 14 then RETURN(true); else RETURN(false); fi; end: isA146359 := proc(n) RETURN(isprime(n) and isA146337(n)) ; end: for k from 1 do if isA146359(ithprime(k)) then printf("%d, ",ithprime(k)) ; fi; od: # R. J. Mathar, Nov 08 2008

Select[Range[2*10^4], PrimeQ[#] && Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]] == 14 &] (* Amiram Eldar, Mar 30 2020 *)

5813 and 6791 removed, extended beyond 8707 by R. J. Mathar, Nov 08 2008

More terms from Amiram Eldar, Mar 30 2020

A146361 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 16 : primes in A146339.

Original entry on oeis.org

191, 311, 431, 647, 1319, 1487, 2351, 5527, 9431, 19087, 21143, 24359, 27239, 29207, 32183, 34367, 36791, 38711, 41759, 42071, 43063, 43319, 49367, 58271, 58391, 59399, 62327, 65183, 66239, 77543, 82759, 84263, 87407, 90271, 93967, 94463, 97127, 100703, 101063

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Amiram Eldar, Table of n, a(n) for n = 1..1000

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

Select[Range[2*10^4], PrimeQ[#] && Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]] == 16 &] (* Amiram Eldar, Mar 30 2020 *)
Select[Prime[Range[10000]],Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]]==16&] (* Harvey P. Dale, Apr 12 2025 *)

3391 removed - R. J. Mathar, Sep 06 2009

More terms from Amiram Eldar, Mar 30 2020

A333640 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 13.

Original entry on oeis.org

421, 757, 1021, 1097, 1117, 1241, 1301, 1553, 1625, 1649, 1973, 2069, 2125, 2237, 2249, 2273, 2665, 2789, 2861, 3085, 3349, 3373, 3461, 3517, 3545, 3877, 3917, 4133, 4397, 4481, 4573, 4589, 4885, 5389, 5521, 5573, 5713, 5717, 6185, 6221, 6317, 6637, 6997, 7093

Offset: 1

Amiram Eldar, Mar 31 2020

nonn

For primes in this sequence see A146358.

			a(1) = 421 because the continued fraction of (1 + sqrt(421))/2 = 10, 1, 3, 6, 1, 1, 2, 2, 1, 1, 6, 3, 1, 19, 1, 3, 6, ... has a period (1, 3, 6, 1, 1, 2, 2, 1, 1, 6, 3, 1, 19) of length 13.

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

Cf. A000290, A078370, A146326-A146345, A146348-A146362.

Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 13 &]

A146329 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 4.

Original entry on oeis.org

6, 7, 8, 14, 20, 23, 24, 28, 32, 33, 34, 42, 47, 48, 52, 55, 60, 62, 69, 72, 75, 78, 79, 80, 95, 98, 110, 119, 120, 126, 133, 135, 136, 138, 140, 141, 142, 156, 167, 168, 174, 180, 189, 194, 205, 210, 213, 215, 219, 220, 222, 223, 224, 248, 252, 254, 272, 287, 288

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

For primes in this sequence see A028871 - {2}.

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

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

Cf. A000290, A078370, A146326-A146345, A028871, A146348-A146360.

isA146329 := proc(n) RETURN(A146326(n) = 4) ; end:
for n from 2 to 400 do if isA146329(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

cf4Q[n_]:=Module[{s=(1+Sqrt[n])/2},If[IntegerQ[s],1,Length[ ContinuedFraction[ s][[2]]]]==4]; Select[Range[300],cf4Q] (* Harvey P. Dale, Dec 14 2017 *)

39, 68, 150, 155, etc. removed by R. J. Mathar, Sep 06 2009

A146334 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 10.

Original entry on oeis.org

43, 67, 116, 129, 134, 161, 162, 184, 218, 242, 243, 246, 270, 274, 297, 301, 314, 338, 339, 345, 354, 356, 407, 411, 451, 452, 459, 465, 475, 498, 515, 517, 532, 534, 561, 563, 590, 591, 595, 597, 603, 611, 638, 648, 657, 665, 669, 671, 690, 705, 715

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

For primes in this sequence see A146355.

			a(1) = 43 because continued fraction of (1+Sqrt[43])/2 = 3, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, 5, 1, 3, 1, 1, 12, 1, 1, 3, 1, ... has period (1, 3, 1, 1, 12, 1, 1, 3, 1, 5) length 10.

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: isA146334 := proc(n) RETURN(A146326(n) = 10) ; end: for n from 2 to 715 do if isA146334(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009

cf10Q[n_]:=Module[{s=(1+Sqrt[n])/2,x},x=If[IntegerQ[s],1,Length[ ContinuedFraction[ s][[2]]]];x==10]; Select[Range[750],cf10Q] (* Harvey P. Dale, Sep 22 2015 *)

284 removed by R. J. Mathar, Sep 06 2009

A107356 Period of continued fraction for (1 + square root of n-th squarefree integer)/2.

Original entry on oeis.org

2, 2, 1, 4, 4, 2, 2, 1, 4, 2, 3, 6, 2, 6, 4, 2, 1, 2, 8, 4, 4, 2, 3, 6, 6, 5, 4, 10, 8, 4, 2, 1, 4, 6, 6, 6, 3, 4, 3, 6, 10, 4, 6, 8, 9, 6, 2, 4, 4, 2, 2, 1, 6, 2, 7, 8, 2, 12, 4, 9, 3, 6, 12, 6, 18, 6, 7, 4, 6, 7, 6, 6, 14, 4, 2, 2, 12, 10, 6, 6, 4, 10, 7, 4, 18, 4, 4, 2, 3, 6, 5, 20, 14, 8, 5, 12, 6, 10

Offset: 1

Steven Finch, May 24 2005

nonn

			a(7) = 2 because 11 is the 7th smallest squarefree integer and (1 + sqrt 11)/2 = [2,6,3,6,3,6,3,... ] thus has an eventual period of 2. We omit 1 from the list of squarefree integers.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Ron Knott, An Introduction to Continued Fractions
K. S. Williams and N. Buck, Comparison of the lengths of the continued fractions of sqrt(D) and (1/2)*(1+sqrt(D)), Proc. Amer. Math. Soc. 120 (1994) 995-1002.

Cf. A005117, A035015, A146326.

(* first do *) Needs["NumberTheory`NumberTheoryFunctions`"] (* then *) s = Drop[ Select[ Range[162], SquareFreeQ[ # ] &], 1]; Length[ ContinuedFraction[ # ][[2]]] & /@ ((1 + Sqrt[s])/2) (* Robert G. Wilson v, May 27 2005 *)
Length[ContinuedFraction[(Sqrt[#]+1)/2][[2]]]&/@Select[Range[ 2,200], SquareFreeQ] (* Harvey P. Dale, Aug 16 2021 *)

a(n) = A146326(A005117(n+1)). - R. J. Mathar, Sep 24 2009

More terms from Robert G. Wilson v, May 27 2005

A146364 a(n) = smallest primes whose continued fraction have different period.

Original entry on oeis.org

2, 5, 7, 17, 19, 31, 41, 43, 73, 89, 103, 139, 151, 179, 191, 193, 211, 241, 271, 331, 337, 379, 409, 421, 433, 463, 487, 491, 521, 541, 571, 601, 619, 631, 673, 739, 751, 769, 823, 919, 929, 937, 1033, 1039, 1051, 1201, 1249, 1291, 1321, 1399, 1439, 1471, 1531, 1579, 1609, 1699, 1747, 1753, 1759

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

This sequence is sorted A146363.

Robert Israel, Table of n, a(n) for n = 1..1500

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

g:= proc(n) local c;
      c:= NumberTheory:-ContinuedFraction((1+sqrt(n))/2);
      nops(Term(c,periodic)[2]);
end proc:
R:= NULL: S:= {}: count:= 0:
p:= 1:
while count < 100 do
  p:= nextprime(p);
  v:= g(p);
  if not member(v,S) then
    R:= R,p; count:= count+1; S:= S union {v};
    if count mod 20 = 0 then printf("%d %d\n",count,p) fi
  fi
od:
R; # Robert Israel, Jun 14 2024

$MaxExtraPrecision = 300; s = 10; aa = {}; Do[k = ContinuedFraction[(1 + Sqrt[n])/2, 1000]; If[Length[k] < 190, AppendTo[aa, 0], m = 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; s = s + 1; While[k[[s ]] != k[[s + m]] || k[[s + m]] != k[[s + 2 m]] || k[[s + 2 m]] != k[[s + 3 m]] || k[[s + 3 m]] != k[[s + 4 m]], m++ ]; AppendTo[aa, m]], {n, 1, 1200}]; Print[aa]; bb = {}; Do[k = 1; yes = 0&&PeimeQ[k]; Do[If[aa[[k]] == n && yes == 0, AppendTo[bb, k]; yes = 1], {k, 1, Length[aa]}], {n, 1, 22}]; Sort[bb] (*Artur Jasinski*)

More terms from Robert Israel, Jun 14 2024

A146478 a(n) = length of period of continued fraction (1 + sqrt(prime(n)))/2.

Original entry on oeis.org

2, 2, 1, 4, 2, 1, 3, 6, 4, 1, 8, 3, 5, 10, 4, 1, 6, 3, 10, 8, 9, 4, 2, 7, 9, 3, 12, 6, 7, 7, 12, 6, 7, 18, 5, 20, 5, 18, 4, 1, 14, 5, 16, 15, 3, 20, 26, 4, 2, 1, 9, 12, 19, 14, 3, 12, 5, 24, 9, 15, 18, 1, 14, 16, 19, 3, 34, 21, 14, 9, 9, 4, 20, 7, 30, 8, 7, 5, 3, 27, 18, 13, 16, 23, 4, 2, 19, 23, 3

Offset: 1

Artur Jasinski, Oct 30 2008

nonn

Subsequence of A146326 (length of period continued fraction of (1 + sqrt(n))/2).

Zak Seidov, Table of n, a(n) for n = 1..1000

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

A146478 := proc(p) local c; c := numtheory[cfrac](1/2+sqrt(p)/2,'periodic','quotients') ; nops(c[2]) ; end: for n from 1 to 100 do printf("%d,",A146478(ithprime(n))) ; od: # R. J. Mathar, Nov 05 2008

Table[Length[ContinuedFraction[(1+Sqrt[Prime[n]])/2][[2]]],{n,100}] (* Zak Seidov, Mar 22 2011 *)

a(59) changed from 7 to 9 by R. J. Mathar, Nov 05 2008

cp's OEIS Frontend

A146358 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 13: primes in A333640.

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A146359 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 14: primes in A146337.

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A146361 Primes p such that continued fraction of (1 + sqrt(p))/2 has period 16 : primes in A146339.

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A333640 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 13.

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A146329 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 4.

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A146334 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 10.

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A107356 Period of continued fraction for (1 + square root of n-th squarefree integer)/2.

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A146364 a(n) = smallest primes whose continued fraction have different period.

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