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.

A349685 Irregular triangle read by rows: the n-th row contains the elements in the continued fraction of the abundancy index of n.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 3, 1, 5, 2, 1, 7, 1, 1, 7, 1, 2, 4, 1, 1, 4, 1, 11, 2, 3, 1, 13, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 15, 1, 17, 2, 6, 1, 19, 2, 10, 1, 1, 1, 10, 1, 1, 1, 1, 3, 1, 23, 2, 2, 1, 4, 6, 1, 1, 1, 1, 1, 2, 1, 2, 13, 2, 1, 29, 2, 2, 2, 1, 31, 1, 1, 31
Offset: 1

Views

Author

Amiram Eldar, Nov 25 2021

Keywords

Comments

The abundancy index of n is sigma(n)/n = A000203(n)/n = A017665(n)/A017666(n).
For a prime p, the p-th row has a length 2 with a(p, 1) = 1 and a(p, 2) = p.
For multiply-perfect numbers m (A007691), the m-th row has a length 1, since their abundancy index is an integer. In particular, for a perfect number m (A000396), the m-th row has a length 1 with a(m, 1) = 2.

Examples

			The first ten rows of the triangle are:
1,
1, 2,
1, 3,
1, 1, 3,
1, 5,
2,
1, 7,
1, 1, 7,
1, 2, 4,
1, 1, 4,
...

Links

Crossrefs

Cf. A000203, A000396, A007691, A017665, A017666, A071862 (row lengths), A071865, A349473.

Programs

  • Mathematica
    row[n_] := ContinuedFraction[DivisorSigma[1, n]/n]; Table[row[k], {k, 1, 32}] // Flatten
  • PARI
    row(n) = contfrac(sigma(n)/n); \\ Michel Marcus, Nov 25 2021

A349686 Numbers k such that the continued fraction of the abundancy index of k contains a single distinct element.

Original entry on oeis.org

1, 6, 24, 28, 30, 120, 140, 348, 496, 672, 1080, 2480, 6048, 6200, 6552, 6786, 8128, 30240, 32760, 40640, 143880, 238080, 435708, 514080, 523776, 524160, 556920, 805728, 1997868, 2178540, 4713984, 23569920, 33550336, 37035180, 38958426, 45532800, 91963648, 142990848
Offset: 1

Views

Author

Amiram Eldar, Nov 25 2021

Keywords

Comments

All the multiply-perfect numbers (A007691) are terms of this sequence, since the continued fraction of their abundancy index contains a single element.
Up to 4*10^10 the continued fractions of the abundancy indices of the terms have lengths 1, 2, 3, 5 or 11. The least terms that are corresponding to these lengths are 1, 24, 30, 348 and 1997868, respectively. Are there terms with other lengths?

Examples

			24 is a term since the continued fraction of its abundancy index sigma(24)/24 = 5/2 = 2 + 1/2 has the elements {2, 2}.
30 is a term since the continued fraction of its abundancy index sigma(30)/30 = 12/5 = 2 + 1/(2 + 1/2) has the elements {2, 2, 2}.
143880 is a term since the continued fraction of its abundancy index sigma(143880)/143880 = 360/109 = 3 + 1/(3 + 1/(3 + 1/(3 + 1/3))) has the elements {3, 3, 3}.

Crossrefs

Cf. A000203, A007691, A071862, A071865, A017665, A017666, A349685.

Programs

  • Mathematica
    c[n_] := ContinuedFraction[DivisorSigma[1, n] / n]; q[n_] := Length[Union[c[n]]] == 1; Select[Range[10^6], q]
  • PARI
    isok(k) = #Set(contfrac(sigma(k)/k)) == 1; \\ Michel Marcus, Nov 25 2021

A342867 a(n) is the least number k such that the continued fraction for phi(k)/k contains exactly n elements.

Original entry on oeis.org

1, 2, 3, 15, 35, 33, 65, 215, 221, 551, 455, 2001, 3417, 3621, 11523, 16705, 16617, 69845, 107545, 157285, 324569, 358883, 1404949, 1569295, 3783970, 3106285, 7536065, 12216295, 10589487, 24038979, 57759065, 51961945, 177005465, 131462695, 741703701, 1467144445
Offset: 1

Views

Author

Amiram Eldar, Mar 27 2021

Keywords

Comments

a(n) is the least number k such that A342866(k) = n.
All the terms above 3 are composite numbers.

Crossrefs

Cf. A000010, A007694, A076512, A109395, A342866.
Cf. A071865 (similar, with sigma(k)/k).

Programs

  • Mathematica
    f[n_] := Length @ ContinuedFraction[EulerPhi[n]/n]; seq[max_] := Module[{s = Table[0, {max}], c = 0, n  = 1, i}, While[c < max, i = f[n]; If[i <= max && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[20]
  • PARI
    a(n) = my(k=1); while (#contfrac(eulerphi(k)/k) != n, k++); k; \\ Michel Marcus, Mar 30 2021

Formula

a(2) = 2 since 2 is the least number k such that A342866(k) = 2.

A349475 a(n) is the least number k such that A349474(k) = n, or -1 if no such k exists.

Original entry on oeis.org

1, 2, 5, 4, 21, 25, 16, 36, 106, 712, 1588, 3775, 900, 4356, 18496, 14400, 45700, 87003, 135445, 229543, 554216, 937019, 1764724, 3431952, 3431088, 10217808, 21357233, 36972202, 42436276, 79056144, 235027304, 261540000, 530582544, 705929608, 1371526825, 1127941321
Offset: 1

Views

Author

Amiram Eldar, Nov 19 2021

Keywords

Examples

			a(3) = 5 since 5 is the least number k such that A349474(k) = 3.

Crossrefs

Cf. A071865, A349473, A349474.

Programs

  • Mathematica
    cflen[n_] := Length @ ContinuedFraction[DivisorSigma[0, n] / DivisorSigma[-1, n]]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = cflen[n]; If[i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; TakeWhile[s, # > 0 &]]; seq[20, 10^7]
Showing 1-4 of 4 results.

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