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 1-6 of 6 results.

A353346 Numbers k such that the elements of the continued fraction of the abundancy index of k and k+1 are anagrams of each other.

Original entry on oeis.org

2084564, 11784194, 13667268, 52820326, 68397891, 101183694, 128247668, 135641787, 137681487, 170542955, 266319572, 284966486, 384109196, 386860419, 482419526, 483785771, 546800667, 579468939, 606809224, 622241109, 703636544, 737703005, 829965829, 830993564, 834224684, 973986250
Offset: 1

Views

Author

Amiram Eldar, Apr 15 2022

Keywords

Examples

			2084564 is a term since the sequences of elements of the continued fractions of sigma(2084564)/2084564 = 941472/521141 and sigma(2084565)/2084565 = 1270656/694855, {1, 1, 4, 5, 1, 8, 1, 5, 2, 4, 1, 7, 3} and {1, 1, 4, 1, 5, 8, 1, 2, 4, 7, 3, 1, 5} respectively, are anagrams of each other.

Links

Crossrefs

Cf. A000203 (sigma), A069567, A017665, A017666, A349685, A353345, A353347.

Programs

  • Mathematica
    ab[n_] := Sort[ContinuedFraction[DivisorSigma[-1, n]]]; seq[max_] := Module[{s = {}, n = 2, c = 0, ab1 = ab[1], ab2}, While[n < max, ab2 = ab[n]; If[ab1 == ab2, AppendTo[s, n - 1]]; ab1 = ab2; n++]; s]; seq[1.4*10^7]

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

A349688 Numbers k such that the sequence of elements of the continued fraction of the abundancy index of k is palindromic.

Original entry on oeis.org

1, 6, 24, 28, 30, 42, 54, 66, 70, 78, 84, 90, 96, 102, 114, 120, 138, 140, 174, 186, 220, 222, 246, 258, 264, 270, 282, 308, 318, 330, 342, 348, 354, 364, 366, 402, 426, 438, 474, 476, 496, 498, 532, 534, 582, 606, 618, 642, 644, 654, 660, 672, 678, 744, 760, 762
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.

Examples

			24 is a term since the sequence of elements of the abundancy index of 24, sigma(24)/24 = 5/2 = 2 + 1/2, is {2, 2}, which is palindromic.
42 is a term since the sequence of elements of the abundancy index of 42, sigma(42)/42 = 16/7 = 2 + 1/(3 + 1/2), is {2, 3, 2}, which is palindromic.

Links

Crossrefs

Cf. A349685.
A007691 and A349686 are subsequences.
Similar sequence: A349477.

Programs

  • Mathematica
    q[n_] := PalindromeQ[ContinuedFraction[DivisorSigma[1, n]/n]]; Select[Range[1000], q]
  • PARI
    isok(k) = my(v=contfrac(sigma(k)/k)); v == Vecrev(v); \\ Michel Marcus, Nov 25 2021

A349690 Numbers k such that the continued fraction of the abundancy index of k contains distinct elements.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 11, 12, 13, 17, 18, 19, 20, 23, 25, 27, 28, 29, 31, 33, 37, 40, 41, 43, 47, 49, 53, 56, 59, 60, 61, 67, 71, 73, 77, 79, 80, 81, 83, 88, 89, 91, 97, 101, 103, 104, 107, 109, 113, 120, 121, 125, 127, 131, 137, 139, 145, 149, 151, 155, 157, 163
Offset: 1

Views

Author

Amiram Eldar, Nov 25 2021

Keywords

Comments

All the primes (A000040) are terms of this sequence, since the continued fraction of the abundancy index of a prime p is {1, p}.
All the multiply-perfect numbers (A007691) are terms of this sequence, since the continued fraction of their abundancy index contains a single element.

Examples

			2 is a term since the abundancy index of 2 is 3/2 = 1 + 1/2 and the elements of the continued fraction, {1, 2}, are different.
4 is not a term since the abundancy index of 4 is 7/4 = 1 + 1/(1 + 1/3) and the elements of the continued fraction, {1, 1, 3}, are not distinct.

Links

Crossrefs

Cf. A007691, A349502, A349685, A349686.

Programs

  • Mathematica
    c[n_] := ContinuedFraction[DivisorSigma[1, n]/n]; q[n_] := Length[(cn = c[n])] == Length[DeleteDuplicates[cn]]; Select[Range[200], q]
  • PARI
    isok(k) = my(v=contfrac(sigma(k)/k)); #v == #Set(v); \\ Michel Marcus, Nov 25 2021

A349689 a(n) is the least number k such that the sequence of elements of the abundancy index of k is palindromic with length n, or -1 if no such k exists.

Original entry on oeis.org

1, 24, 30, 90, 96, 342, 744, 812160, 330, 147258, 32784, 3314062080, 25896, 565632, 116412, 210317184, 145176, 6182491392, 963108
Offset: 1

Views

Author

Amiram Eldar, Nov 25 2021

Keywords

Comments

a(21) = 7094832, a(23) = 24167070, a(25) = 858983598, a(27) = 1137635260, a(29) = 1402857468, a(31) = 45230309244, and there are no more terms below 1.6*10^11.

Examples

			The elements of the continued fractions of the abundancy index of the terms are:
   n        a(n)  elements
  --    --------  -------------------------------------
   1           1  1
   2          24  2,2
   3          30  2,2,2
   4          90  2,1,1,2
   5          96  2,1,1,1,2
   6         342  2,3,1,1,3,2
   7         744  2,1,1,2,1,1,2
   8      812160  3,1,1,1,1,1,1,3
   9         330  2,1,1,1,1,1,1,1,2
  10      147258  2,3,1,2,5,5,2,1,3,2
  11       32784  2,1,1,2,2,1,2,2,1,1,2
  12  3314062080  4,2,1,1,2,1,1,2,1,1,2,4
  13       25896  2,1,2,1,1,1,2,1,1,1,2,1,2
  14      565632  2,1,7,1,1,2,1,1,2,1,1,7,1,2
  15      116412  2,2,1,1,1,1,1,8,1,1,1,1,1,2,2
  16   210317184  3,1,1,2,3,3,2,1,1,2,3,3,2,1,1,3
  17      145176  2,1,1,1,1,1,1,1,4,1,1,1,1,1,1,1,2
  18  6182491392  3,1,1,2,7,3,2,2,1,1,2,2,3,7,2,1,1,3
  19      963108  2,1,1,1,1,2,1,1,1,3,1,1,1,2,1,1,1,1,2

Crossrefs

Cf. A000203, A017665, A017666, A349685, A349688.
Similar sequence: A349478.

Programs

  • Mathematica
    cfai[n_] := ContinuedFraction[DivisorSigma[1, n]/n]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i, cf}, While[c < len && n < nmax, cf = cfai[n]; If[PalindromeQ[cf] && (i = Length[cf]) <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; TakeWhile[s, # > 0 &]]; seq[11, 10^6]
  • PARI
    isok(k, n) = my(v=contfrac(sigma(k)/k)); (#v == n) && (v == Vecrev(v));
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Nov 25 2021

A349691 a(n) is the least number k such that the continued fraction of the abundancy index of k contains n elements that are all distinct, or -1 if no such k exists.

Original entry on oeis.org

1, 2, 9, 176, 155, 2450, 21500, 118993, 767700, 12409639, 56024339, 857777653, 8648737607
Offset: 1

Views

Author

Amiram Eldar, Nov 25 2021

Keywords

Comments

a(14) > 4*10^10, if it exists.

Examples

			The elements of the continued fractions of the abundancy index of the first 13 terms are:
   n        a(n)  elements
  --  ----------  -----------------------------
   1           1  1
   2           2  1,2
   3           9  1,2,4
   4         176  2,8,1,4
   5         155  1,4,5,3,2
   6        2450  2,6,9,8,1,4
   7       21500  2,4,3,1,6,9,5
   8      118993  1,6,5,2,13,3,10,4
   9      767700  3,7,4,6,12,10,5,1,2
  10    12409639  1,10,12,6,3,2,4,14,5,7
  11    56024339  1,6,12,4,8,5,9,3,7,10,2
  12   857777653  1,14,3,5,12,4,6,2,7,9,10,8
  13  8648737607  1,12,6,13,2,4,10,7,11,3,9,8,5

Crossrefs

Cf. A349503, A349685, A349690.

Programs

  • Mathematica
    cflen[n_] := Module[{cf = ContinuedFraction[DivisorSigma[1, n]/n], len}, If[(len = Length[cf]) == Length[DeleteDuplicates[cf]], len, 0]]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i}, While[c < len && n < nmax, i = cflen[n]; If[i > 0 && i <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; TakeWhile[s, # > 0 &]]; seq[9, 10^6]
  • PARI
    isok(k, n) = my(v=contfrac(sigma(k)/k)); (#v == n) && (#Set(v) == n);
a(n) = my(k=1); while (!isok(k, n), k++); k; \\ Michel Marcus, Nov 25 2021
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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