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 1-4 of 4 results.

A206578 The least number with exactly n ones in the continued fraction of its square root.

Original entry on oeis.org

2, 3, 14, 7, 13, 91, 43, 115, 94, 819, 133, 1075, 211, 1219, 309, 871, 421, 1147, 244, 3427, 478, 2575, 991, 8791, 604, 3799, 886, 5539, 1381, 8851, 1279, 7303, 1561, 19519, 1759, 10339, 1831, 12871, 2038, 13771, 1999, 8611, 1516, 15871, 2731, 20875, 1726
Offset: 0

Views

Author

T. D. Noe, Feb 24 2012

Keywords

Comments

It appears that only the odd-numbered terms 3 and 7 are prime; all other primes occur at even-numbered terms 0, 4, 6, 12, 16, 22, 28, 30, 34, ... In terms 0 to 1000, there are 268 primes and 632 semiprimes.

Links

Crossrefs

Cf. A013647-A013650 (0-3), A020440-A020446 (4-10), A031779-A031868 (11-100).
Cf. A206582 (n twos), A206583 (n threes), A206584 (n fours), A206585 (n fives).

Programs

  • Mathematica
    nn = 50; zeros = nn; t = Table[0, {nn}]; k = 2; While[zeros > 0, If[! IntegerQ[Sqrt[k]], cnt = Count[ContinuedFraction[Sqrt[k]][[2]], 1]; If[cnt <= nn && t[[cnt]] == 0, t[[cnt]] = k; zeros--]]; k++]; Join[{2}, t]
  • Python
    from sympy import continued_fraction_periodic
def A206578(n):
    m = 1
    while True:
        s = continued_fraction_periodic(0,1,m)[-1]
        if isinstance(s,list) and s.count(1) == n:
            return m
        m += 1 # Chai Wah Wu, Jun 12 2017

A206585 The least number s > 1 having exactly n fives in the periodic part of the continued fraction of sqrt(s).

Original entry on oeis.org

2, 27, 67, 664, 331, 6487, 1237, 6019, 1999, 6331, 3964, 23983, 4204, 22075, 9739, 64639, 10684, 26419, 17971, 80719, 22969, 140971, 28414, 310759, 34189, 290779, 39181, 228691, 46099, 261691, 56884, 416707, 61429, 136579, 76651, 535375, 75916, 296839, 87151
Offset: 0

Views

Author

T. D. Noe, Mar 19 2012

Keywords

Links

Crossrefs

Cf. A206578 (n ones), A206582 (n twos), A206583 (n threes), A206584 (n fours).

Programs

  • Mathematica
    nn = 50; zeros = nn; t = Table[0, {nn}]; k = 2; While[zeros > 0, If[! IntegerQ[Sqrt[k]], cnt = Count[ContinuedFraction[Sqrt[k]][[2]], 5]; If[cnt <= nn && t[[cnt]] == 0, t[[cnt]] = k; zeros--]]; k++]; Join[{2}, t]
  • Python
    from sympy import continued_fraction_periodic
def A206585(n):
    i = 2
    while True:
        s = continued_fraction_periodic(0,1,i)[-1]
        if isinstance(s, list) and s.count(5) == n:
            return i
        i += 1 # Chai Wah Wu, Jun 10 2017

Extensions

Definition clarified by Chai Wah Wu, Jun 10 2017

A206583 The least number s having exactly n threes in the continued fraction of sqrt(s).

Original entry on oeis.org

2, 11, 28, 335, 69, 1507, 268, 1963, 589, 7675, 1318, 2899, 1549, 11575, 1789, 17371, 3331, 22459, 3319, 23791, 6211, 18019, 6379, 63379, 6694, 17131, 9199, 75331, 10189, 49471, 10429, 75091, 16069, 105991, 17509, 114559, 14221, 152707, 25141, 159031, 20959
Offset: 0

Views

Author

T. D. Noe, Mar 19 2012

Keywords

Crossrefs

Cf. A206578 (n ones), A206582 (n twos), A206584 (n fours), A206585 (n fives).

Programs

  • Mathematica
    nn = 50; zeros = nn; t = Table[0, {nn}]; k = 2; While[zeros > 0, If[! IntegerQ[Sqrt[k]], cnt = Count[ContinuedFraction[Sqrt[k]][[2]], 3]; If[cnt <= nn && t[[cnt]] == 0, t[[cnt]] = k; zeros--]]; k++]; Join[{2}, t]

A206584 The least number s having exactly n fours in the continued fraction of sqrt(s).

Original entry on oeis.org

2, 5, 52, 329, 379, 639, 631, 3937, 1471, 2461, 2524, 6931, 3259, 11809, 5659, 16756, 8779, 14749, 10399, 24721, 14452, 25429, 16669, 31021, 16879, 45301, 21061, 45109, 31054, 72721, 28414, 81061, 40189, 86041, 34654, 109909, 44371, 74881, 42046, 79501, 53551
Offset: 0

Views

Author

T. D. Noe, Mar 19 2012

Keywords

Crossrefs

Cf. A206578 (n ones), A206582 (n twos), A206583 (n threes), A206585 (n fives).

Programs

  • Mathematica
    nn = 50; zeros = nn; t = Table[0, {nn}]; k = 2; While[zeros > 0, If[! IntegerQ[Sqrt[k]], cnt = Count[ContinuedFraction[Sqrt[k]][[2]], 4]; If[cnt <= nn && t[[cnt]] == 0, t[[cnt]] = k; zeros--]]; k++]; Join[{2}, t]
With[{tbl=Table[{n,Count[If[IntegerQ[Sqrt[n]],{1},ContinuedFraction[Sqrt[n]][[2]]],4]},{n,2,110000}]},Table[SelectFirst[tbl,#[[2]]==k&],{k,0,40}]][[All,1]] (* Harvey P. Dale, Oct 05 2022 *)
Showing 1-4 of 4 results.

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