A146327
Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 2.
Original entry on oeis.org
2, 3, 10, 11, 12, 15, 21, 26, 27, 30, 35, 45, 50, 51, 56, 63, 77, 82, 83, 84, 87, 90, 93, 99, 117, 122, 123, 132, 143, 165, 170, 171, 182, 195, 221, 226, 227, 228, 230, 231, 235, 237, 240, 245, 255, 285, 290, 291, 306, 323, 357, 362, 363, 380, 399, 437, 442, 443
Offset: 1
a(1) = 2 because continued fraction of (1 + sqrt(2))/2 = 1, 4, 1, 4, 1, 4, 1, ... has repeating part (1,4), period 2.
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A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146327 := proc(n) RETURN(A146326(n) = 2) ; end: for n from 2 to 450 do if isA146327(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009
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Select[Range[1000], 2 == Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] &]
A146328
Numbers k such that the continued fraction of (1 + sqrt(k))/2 has period 3.
Original entry on oeis.org
17, 37, 61, 65, 101, 145, 185, 197, 257, 317, 325, 401, 461, 485, 557, 577, 677, 773, 785, 901, 985, 1025, 1129, 1157, 1297, 1429, 1445, 1601, 1765, 1877, 1901, 1937, 2117, 2285, 2305, 2501, 2705, 2873, 2917, 3077, 3137, 3281, 3293, 3341, 3365, 3601, 3845, 4045, 4097, 4357, 4597, 4625, 4901
Offset: 1
a(1) = 3 because continued fraction of (1+sqrt(17))/2 = 2, 1, 1, 3, 1, 1, 3, ... has period (1,1,3) length 3.
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A146326 := proc(n) if not issqr(n) then numtheory[cfrac]( (1+sqrt(n))/2, 'periodic','quotients') ; nops(%[2]) ; else 0 ; fi; end: isA146328 := proc(n) RETURN(A146326(n) = 3) ; end: for n from 2 to 1801 do if isA146328(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009
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okQ[n_] := Module[{cf = ContinuedFraction[(1 + Sqrt[n])/2]}, Length[cf] > 1 && Length[cf[[2]]] == 3]; Select[Range[5000], okQ]
A146330
Numbers k such that continued fraction of (1 + sqrt(k))/2 has period 5.
Original entry on oeis.org
41, 149, 157, 181, 269, 397, 425, 493, 565, 697, 761, 941, 1013, 1037, 1325, 1565, 1781, 1825, 2081, 2153, 2165, 2173, 2465, 2477, 2693, 2725, 3181, 3221, 3533, 3869, 4253, 4409, 5165, 5213, 5273, 5297, 5741, 5837, 6485, 6757, 6949, 7045, 7325, 7465, 8021, 8069
Offset: 1
a(1) = 41 because continued fraction of (1+sqrt(41))/2 = 3, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, 2, 1, 5, 1, 2, ... has period (1,2,2,1,5) length 5.
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isA146330 := proc(n) RETURN(A146326(n) = 5) ; end:
for n from 2 to 2000 do if isA146330(n) then printf("%d,",n) ; fi; od: # R. J. Mathar, Sep 06 2009
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Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 5 &] (* Amiram Eldar, Mar 31 2020 *)
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
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.
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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
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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 *)
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
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.
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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
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Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 7 &] (* 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
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.
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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
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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
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
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.
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
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.
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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
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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 *)
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
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.
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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
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Select[Range[10^4], !IntegerQ @ Sqrt[#] && Length[ContinuedFraction[(1 + Sqrt[#])/2][[2]]] == 15 &] (* 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
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.
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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
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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
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