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.

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A146331 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 6.

Original entry on oeis.org

18, 19, 22, 38, 39, 44, 54, 57, 58, 59, 66, 68, 70, 74, 86, 102, 105, 107, 111, 112, 114, 115, 130, 131, 146, 147, 148, 150, 159, 164, 175, 178, 183, 186, 187, 198, 203, 253, 258, 260, 264, 267, 273, 275, 278, 294, 303, 308, 309, 326, 327, 330, 333, 341, 346
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

For primes in this sequence see A146351.

Examples

			a(2) = 19 because continued fraction of (1+sqrt(19))/2 = 2, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1, 3, 1, 2, 8, 2, 1 ... has period (1, 2, 8, 2, 1, 3) length 6.

Links

Crossrefs

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

Programs

  • Maple
    A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146331 := proc(n) RETURN(A146326(n) = 6) ; end: for n from 2 to 380 do if isA146331(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009
  • Mathematica
    cf6Q[n_]:=Module[{c=(1+Sqrt[n])/2},!IntegerQ[c]&&Length[ContinuedFraction[ c][[2]]]==6]; Select[Range[400],cf6Q] (* Harvey P. Dale, May 30 2012 *)

Extensions

39, 68, 150, 203, etc. added by R. J. Mathar, Sep 06 2009

A146332 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 7.

Original entry on oeis.org

89, 109, 113, 137, 373, 389, 509, 653, 685, 797, 853, 925, 949, 997, 1009, 1105, 1145, 1165, 1261, 1493, 1997, 2309, 2621, 2677, 2885, 2941, 3133, 3277, 3445, 3653, 3797, 4325, 4505, 4745, 4825, 4973, 5353, 5429, 5765, 6305, 6437, 6845, 7085, 7373, 7817, 7873
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

For primes in this sequence see A146352.

Examples

			a(4) = 137 because continued fraction of (1+sqrt(137))/2 = 6, 2, 1, 5, 5, 1, 2, 11, 2, 1, 5, 5, 1, 2, 11, 2, 1, 5, 5, 1, 2, 11 ... has period (2, 1, 5, 5, 1, 2, 11) length 7.

Links

Crossrefs

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

Programs

  • Maple
    A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146332 := proc(n) RETURN(A146326(n) = 7) ; end: for n from 2 to 1100 do if isA146332(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009
  • Mathematica
    Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 7 &]  (* Amiram Eldar, Mar 31 2020 *)

Extensions

997 added by R. J. Mathar, Sep 06 2009
More terms from Amiram Eldar, Mar 31 2020

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

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

For primes in this sequence see A146353.

Examples

			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.

Links

Crossrefs

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

Programs

  • Maple
    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
  • Mathematica
    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 *)

Extensions

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

A146336 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 12.

Original entry on oeis.org

94, 103, 124, 127, 128, 158, 160, 177, 190, 204, 208, 209, 216, 236, 239, 247, 263, 295, 296, 302, 316, 332, 351, 364, 376, 384, 385, 415, 423, 426, 432, 460, 464, 479, 492, 535, 544, 585, 606, 608, 609, 636, 639, 666, 668, 684, 696, 706, 734, 736, 744, 750
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

For primes in this sequence see A146357.

Examples

			a(2) = 103 because continued fraction of (1+sqrt(103))/2 = 15, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9, 1, 1, 2, 1, 6, 20, 6, 1, 2 ... has period (1, 1, 2, 1, 6, 20, 6, 1, 2, 1, 1, 9) length 12.

Links

Crossrefs

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

Programs

  • Mathematica
    cf12Q[n_]:=!OddQ[Sqrt[n]]&&Length[ContinuedFraction[(1+Sqrt[n])/2][[2]]]==12; Select[Range[800],cf12Q] (* Harvey P. Dale, Oct 15 2011 *)

A146337 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 14.

Original entry on oeis.org

118, 154, 179, 201, 212, 244, 251, 262, 286, 292, 307, 340, 347, 388, 403, 418, 422, 430, 467, 471, 474, 494, 497, 500, 519, 543, 548, 566, 587, 594, 598, 670, 683, 687, 692, 698, 699, 703, 713, 722, 742, 745, 754, 831, 833, 847, 873, 879, 932, 939, 945
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

For primes in this sequence see A146359.

Examples

			a(1) = 421 because continued fraction of (1+sqrt(421))/2 = 17, 5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13, 5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13, 5, 3, 1, 1, 1, 2, 26... has period (5, 3, 1, 1, 1, 2, 26, 2, 1, 1, 1, 3, 5, 13) length 14.

Links

Crossrefs

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

Programs

  • Maple
    A := proc(n) option remember ; 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: for k from 1 do if isA146337(k) then printf("%d, ",k) ; fi; od: # R. J. Mathar, Nov 08 2008
  • Mathematica
    cf14Q[n_]:=Module[{s=(1+Sqrt[n])/2},!IntegerQ[s]&&Length[ ContinuedFraction[ s][[2]]] == 14]; Select[Range[1000],cf14Q] (* Harvey P. Dale, Oct 15 2015 *)

Extensions

More terms from R. J. Mathar, Nov 08 2008

A146338 Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 15.

Original entry on oeis.org

193, 281, 481, 1417, 1861, 1933, 2089, 2141, 2197, 2437, 2741, 2837, 3037, 3065, 3121, 3413, 3589, 3625, 3785, 3925, 3977, 4001, 4637, 4777, 4877, 5317, 5821, 5941, 6025, 6641, 6653, 6749, 7673, 8117, 8177, 8345, 10069, 10273, 10457, 11197, 11281, 11549, 11821
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

For primes in this sequence see A146360.

Examples

			a(1) = 193 because continued fraction of (1+sqrt(193))/2 = 7, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13, 2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13, ... has period (2, 4, 6, 1, 2, 1, 1, 1, 1, 2, 1, 6, 4, 2, 13) length 15.

Links

Crossrefs

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

Programs

  • Maple
    A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end:
isA146338 := proc(n) RETURN(A146326(n) = 15) ; end:
for n from 2 to 4000 do if isA146338(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Sep 06 2009
  • Mathematica
    Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 15 &]  (* Amiram Eldar, Mar 31 2020 *)

Extensions

Extended beyond 3 terms by R. J. Mathar, Sep 06 2009
More terms from Amiram Eldar, Mar 31 2020

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

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

For primes in this sequence see A146361.

Examples

			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.

Links

Crossrefs

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

Programs

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

Extensions

311 inserted, sequence extended by R. J. Mathar, Sep 06 2009
More terms from Amiram Eldar, Mar 31 2020

A146340 Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 17.

Original entry on oeis.org

521, 617, 709, 1433, 1597, 2549, 2909, 2965, 3161, 3581, 3821, 4013, 4285, 4649, 5501, 5585, 5693, 5813, 6197, 6409, 7825, 7853, 8093, 8125, 8573, 8917, 9281, 9665, 9677, 9925, 10265, 10597, 10973, 11273, 12085, 12805, 13061, 13109, 13613, 13957, 14677
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

For primes in this sequence see A146362.

Examples

			a(1) = 521 because continued fraction of (1+sqrt(521))/2 = 11, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21, 1, 10, 2, 5, ... has period (1, 10, 2, 5, 4, 2, 1, 1, 1, 1, 2, 4, 5, 2, 10, 1, 21) length 17.

Links

Crossrefs

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

Programs

  • Maple
    A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146340 := proc(n) RETURN(A146326(n) = 17) ; end: for n from 2 do if isA146340(n) then printf("%d,\n",n) ; fi; od: # R. J. Mathar, Sep 06 2009
  • Mathematica
    cf17Q[n_]:=Module[{s=(1+Sqrt[n])/2},If[IntegerQ[s], 1,Length[ ContinuedFraction[ s][[2]]]]==17]; Select[Range[5000],cf17Q] (* Harvey P. Dale, Dec 20 2017 *)

Extensions

998 and 1006 removed, sequence extended by R. J. Mathar, Sep 06 2009
More terms from Harvey P. Dale, Dec 20 2017

A146343 a(n) = smallest number k such that the continued fraction of (1 + sqrt(k))/2 has period n.

Original entry on oeis.org

5, 2, 17, 6, 41, 18, 89, 31, 73, 43, 265, 94, 421, 118, 193, 172, 521, 106, 241, 151, 337, 489, 433, 268, 929, 211, 409, 334, 673, 379, 937, 463, 601, 331, 769, 721, 2297, 619, 1033, 718, 1777, 394, 1753, 604, 1993, 634, 1249, 526, 3649, 694
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Links

Crossrefs

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

Programs

  • Maple
    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: A146343 := proc(n) for k from 1 do if A(k) = n then RETURN(k); fi; od: end: for n from 1 to 30 do printf("%d,",A146343(n)) ; od: # R. J. Mathar, Nov 08 2008
  • Mathematica
    nn = 50; t = Table[0, {nn}]; cnt = 0; k = 1; While[cnt < nn, k++; cf = ContinuedFraction[(1 + Sqrt[k])/2]; If[Head[cf[[-1]]] === List, len = Length[cf[[-1]]]; If[len <= nn && t[[len]] == 0, t[[len]] = k; cnt++]]]; t

Extensions

a(6) changed to 18, a(25) to 929, a(28) to 334 by R. J. Mathar, Nov 08 2008
Extended by T. D. Noe, Mar 22 2011

A146350 Primes p such that continued fraction of (1+sqrt(p))/2 has period 5 : primes in A146330.

Original entry on oeis.org

41, 149, 157, 181, 269, 397, 761, 941, 1013, 2081, 2153, 2477, 2693, 3181, 3221, 3533, 4253, 4409, 5273, 5297, 5741, 6949, 8069, 8501, 8597, 9293, 10301, 10357, 10957, 11321, 12281, 12589, 13313, 17477, 19477, 19949, 20369, 21433, 22397, 23957, 26309
Offset: 1

Views

Author

Artur Jasinski, Oct 30 2008

Keywords

Comments

A050954 is subset of this sequence.

Links

Crossrefs

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

Programs

  • Mathematica
    Select[Prime[Range[2000]],Length[ContinuedFraction[(1+Sqrt[#])/2][[2]]] == 5&] (* Harvey P. Dale, Aug 13 2016 *)

Extensions

More terms from Harvey P. Dale, Aug 13 2016
Previous Showing 11-20 of 36 results. Next

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