A349473 -id:A349473 - cp's OEIS frontend

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

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

Amiram Eldar, Nov 25 2021

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.

			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,
...

Amiram Eldar, Table of n, a(n) for n = 1..10489 (rows 1..2000)

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

row[n_] := ContinuedFraction[DivisorSigma[1, n]/n]; Table[row[k], {k, 1, 32}] // Flatten

row(n) = contfrac(sigma(n)/n); \\ Michel Marcus, Nov 25 2021

A349474 a(n) is the length of the continued fraction of the harmonic mean of the divisors of n.

Original entry on oeis.org

1, 2, 2, 4, 3, 1, 3, 3, 2, 3, 3, 4, 3, 2, 2, 7, 3, 4, 3, 3, 5, 3, 3, 2, 6, 3, 4, 1, 3, 2, 3, 2, 3, 4, 3, 8, 3, 4, 5, 4, 3, 2, 3, 2, 3, 4, 3, 5, 6, 4, 3, 4, 3, 4, 2, 5, 5, 7, 3, 3, 3, 5, 7, 7, 3, 3, 3, 3, 3, 3, 3, 4, 3, 5, 7, 4, 4, 4, 3, 4, 6, 6, 3, 2, 4, 6, 3

Offset: 1

Amiram Eldar, Nov 19 2021

nonn

a(n) = 1 if and only if n is a harmonic number (A001599).

a(n) <= 2 if and only if n is in A348865.

			a(1) = 1 since the harmonic mean of the divisors of 1 is 1 and its continued fraction has 1 element, {1}.
a(2) = 2 since the harmonic mean of the divisors of 2 is 4/3 = 1 + 1/3 and its continued fraction has 2 elements, {1, 3}.
a(4) = 4 since the harmonic mean of the divisors of 4 is 12/7 = 1 + 1/(1 + 1/(2 + 1/2)) and its continued fraction has 4 elements, {1, 1, 2, 2}.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Row length of A349473.

Cf. A001599, A071862, A099377, A099378, A348865.

a[n_] := Length @ ContinuedFraction[DivisorSigma[0, n] / DivisorSigma[-1, n]]; Array[a, 100]

A349476 Numbers k such that the continued fraction of the harmonic mean of the divisors of k contains a single distinct element.

Original entry on oeis.org

1, 6, 15, 28, 30, 140, 270, 496, 545, 672, 792, 1365, 1638, 2970, 3515, 6200, 8128, 8190, 18600, 18620, 27846, 30240, 32760, 44950, 46359, 55860, 59670, 105664, 117800, 167400, 173600, 237510, 242060, 253539, 332640, 360360, 539400, 681156, 691782, 695520, 726180, 753480, 950976

Offset: 1

Amiram Eldar, Nov 19 2021

nonn

All the harmonic numbers (A001599) are terms of this sequence.
The least term with m elements in the continued fraction of the harmonic mean of its divisors for m = 1, 2, 3, and 4 is 1, 15, 792 and 545, respectively.
Are there terms with more than 4 elements? There are no such terms below 2*10^9.

			15 is a term since the harmonic mean of its divisors is 5/2 = 2 + 1/2.
545 is a term since the harmonic mean of its divisors is 109/33 = 3 + 1/(3 + 1/(3 + 1/3)).
792 is a term since the harmonic mean of its divisors is 528/65 = 8 + 1/(8 + 1/8).

Amiram Eldar, Table of n, a(n) for n = 1..150

Cf. A099377, A099378, A349473.

c[n_] := ContinuedFraction[DivisorSigma[0, n] / DivisorSigma[-1, n]]; q[n_] := Length[Union[c[n]]] == 1; Select[Range[10^6], q]

A349499 Numbers k such that A349474(k) = A349474(k+1).

Original entry on oeis.org

2, 7, 10, 14, 19, 22, 56, 59, 60, 63, 65, 66, 67, 68, 69, 70, 76, 77, 81, 104, 109, 113, 114, 123, 125, 137, 138, 154, 155, 164, 171, 184, 185, 187, 190, 195, 199, 210, 217, 221, 230, 232, 236, 248, 251, 255, 257, 274, 276, 280, 281, 282, 290, 293, 295, 301, 306

Offset: 1

Amiram Eldar, Nov 20 2021

nonn

			2 is a term since A349474(2) = A349474(3) = 2.
7 is a term since A349474(7) = A349474(8) = 3.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A349473, A349474.

c[n_] := Length @ ContinuedFraction[DivisorSigma[0, n]/DivisorSigma[-1, n]]; Select[Range[350], c[#] == c[# + 1] &]

A349477 Numbers k such that the sequence of elements of the continued fraction of the harmonic mean of the divisors of k is palindromic.

Original entry on oeis.org

1, 6, 8, 10, 15, 16, 21, 28, 30, 39, 48, 56, 57, 64, 93, 111, 129, 140, 183, 184, 192, 195, 200, 201, 210, 219, 220, 237, 270, 291, 309, 327, 345, 381, 417, 453, 471, 489, 496, 543, 545, 574, 579, 597, 600, 633, 645, 669, 672, 687, 723, 765, 792, 795, 798, 813

Offset: 1

Amiram Eldar, Nov 19 2021

nonn

All the harmonic numbers (A001599) are terms of this sequence.

			8 is a term since the sequence of elements of the continued fraction of the harmonic mean of the divisors of 8, 32/15 = 2 + 1/(7 + 1/2), is {2, 7, 2}, which is palindromic.

Amiram Eldar, Table of n, a(n) for n = 1..10000

A001599 and A349476 are subsequences.

Cf. A099377, A099378, A349473.

q[n_] := PalindromeQ[ContinuedFraction[DivisorSigma[0, n] / DivisorSigma[-1, n]]]; Select[Range[1000], q]

A349502 Numbers k such that the continued fraction of the harmonic mean of the divisors of k contains distinct elements.

Original entry on oeis.org

1, 2, 3, 6, 9, 14, 20, 24, 28, 32, 33, 35, 42, 44, 45, 51, 52, 55, 60, 65, 66, 68, 69, 70, 72, 84, 87, 88, 91, 95, 99, 104, 110, 114, 115, 117, 120, 123, 125, 126, 128, 135, 136, 138, 140, 141, 145, 152, 153, 156, 159, 170, 174, 177, 180, 182, 185, 186, 187, 188

Offset: 1

Amiram Eldar, Nov 20 2021

nonn

All the harmonic numbers (A001599) are terms of this sequence.

			2 is a term since the harmonic mean of the divisors of 2 is 4/3 = 1 + 1/3 and the elements of the continued fraction, {1, 3}, are different.
4 is not a term since the harmonic mean of the divisors of 4 is 12/7 = 1 + 1/(1 + 1/(2 + 1/2)) and the elements of the continued fraction, {1, 1, 2, 2}, are not distinct.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001599, A349473, A349476.

c[n_] := ContinuedFraction[DivisorSigma[0, n]/DivisorSigma[-1, n]]; q[n_] := Length[(cn = c[n])] == Length[DeleteDuplicates[cn]]; Select[Range[200], q]

A349497 a(n) is the smallest element in the continued fraction of the harmonic mean of the divisors of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1

Offset: 1

Amiram Eldar, Nov 20 2021

nonn

			a(2) = 1 since the continued fraction of the harmonic mean of the divisors of 2, 4/3 = 1 + 1/3, has 2 elements, {1, 3}, and the smallest of them is 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001248, A001599, A001600, A099377, A099378, A129521, A349473.

a[n_] := Min[ContinuedFraction[DivisorSigma[0, n] / DivisorSigma[-1, n]]]; Array[a, 100]

a(p) = 1 for a prime p.
a(p^2) = 1 for a prime p != 3.
a(A129521(n)) = 1 for n > 3.
For a harmonic number m = A001599(k), a(m) = A099377(m) = A001600(k).

A353345 Numbers k such that the elements of the continued fractions of the harmonic means of the divisors of k and k+1 are anagrams of each other.

Original entry on oeis.org

688126, 29900656, 35217656, 71624168, 154979487, 527560886, 871173148, 1370592266, 2461226804, 3232529461, 3232684430, 3431178214, 3471121856, 3486231973, 3527029430, 5732671200, 6258062402, 8784477355, 9334188311, 12670993089, 12707869077, 15120804392, 16317131894

Offset: 1

Amiram Eldar, Apr 15 2022

nonn

			688126 is a term since sequences of elements of the continued fractions of the harmonic means of the divisors of 688126 and 688127, 688126/70281 and 688127/77304, are {9, 1, 3, 1, 3, 1, 2, 9, 1, 1, 6, 8} and {8, 1, 9, 6, 3, 1, 2, 1, 3, 1, 1, 9} respectively, and they are anagrams of each other.

Cf. A069567, A099377, A099378, A349473, A353346, A353347.

Subsequence of A349499.

h[n_] := Sort[ContinuedFraction[DivisorSigma[0, n]/DivisorSigma[-1, n]]]; seq[max_] := Module[{s = {}, n = 2, c = 0, h1 = h[1], h2}, While[n < max, h2 = h[n]; If[h1 == h2, AppendTo[s, n - 1]]; h1 = h2; n++]; s]; seq[4*10^7]

A349478 a(n) is the least number k such that the sequence of elements of the continued fraction of the harmonic mean of the divisors of k is palindromic with length n, or -1 if no such k exists.

Original entry on oeis.org

1, 15, 8, 545, 21, 1131, 16, 98124, 28676, 1109305, 28672, 16837500, 1231932, 477021580, 6129711, 734420331, 441972042, 4343866215, 42741916965, 96692841558, 2193739177

Offset: 1

Amiram Eldar, Nov 19 2021

a(23) = 60755428490.

No more terms below 10^11.

			The elements of the continued fractions of the harmonic mean of the divisors of the terms are:
   n         a(n)   elements
  --  -----------   -------------------------------------------
   1            1   1
   2           15   2,2
   3            8   2,7,2
   4          545   3,3,3,3
   5           21   2,1,1,1,2
   6         1131   5,2,1,1,2,5
   7           16   2,1,1,2,1,1,2
   8        98124   17,1,1,3,3,1,1,17
   9        28676   6,1,2,3,1,3,2,1,6
  10      1109305   6,1,1,1,1,1,1,1,1,6
  11        28672   11,2,1,1,1,10,1,1,1,2,11
  12     16837500   24,1,1,1,2,1,1,2,1,1,1,24
  13      1231932   18,1,1,1,1,1,8,1,1,1,1,1,18
  14    477021580   38,2,3,1,1,1,1,1,1,1,1,3,2,38
  15      6129711   14,2,2,1,1,1,1,9,1,1,1,1,2,2,14
  16    734420331   20,2,1,1,1,1,1,1,1,1,1,1,1,1,2,20
  17    441972042   15,1,3,2,2,1,1,2,15,2,1,1,2,2,3,1,15
  18   4343866215   18,1,1,7,1,8,2,1,1,1,1,2,8,1,7,1,1,18
  19  42741916965   94,1,1,7,4,1,1,1,1,3,1,1,1,1,4,7,1,1,94
  20  96692841558   28,2,4,1,1,4,1,1,1,6,6,1,1,1,4,1,1,4,2,28
  21   2193739177   19,1,1,1,3,1,1,1,1,1,9,1,1,1,1,1,3,1,1,1,19

Cf. A099377, A099378, A349473, A349477.

cfhm[n_] := ContinuedFraction[DivisorSigma[0, n]/DivisorSigma[-1, n]]; seq[len_, nmax_] := Module[{s = Table[0, {len}], c = 0, n = 1, i, cf}, While[c < len && n < nmax, cf = cfhm[n]; If[PalindromeQ[cf] && (i = Length[cf]) <= len && s[[i]] == 0, c++; s[[i]] = n]; n++]; TakeWhile[s, # > 0 &]]; seq[11, 10^7]

A349503 a(n) is the least number k such that the continued fraction of the harmonic mean of the divisors of k contains n elements that are all distinct.

Original entry on oeis.org

1, 2, 20, 52, 156, 768, 8244, 25808, 406764, 3610688, 41395016, 453695175, 3325792768

Offset: 1

Amiram Eldar, Nov 20 2021

			The elements of the continued fractions of the harmonic mean of the divisors of the first 13 terms:
   n        a(n)  elements
  --  ----------  -----------------------------
   1           1  1
   2           2  1,3
   3          20  2,1,6
   4          52  3,5,2,4
   5         156  4,1,3,2,5
   6         768  6,1,3,4,2,13
   7        8244  7,11,8,3,1,13,2
   8       25808  5,6,3,13,1,2,4,7
   9      406764  7,8,3,6,9,2,1,4,12
  10     3610688  7,18,5,2,3,6,1,4,13,11
  11    41395016  7,1,12,8,4,2,3,5,19,6,10
  12   453695175  16,5,8,1,10,48,7,13,2,3,6,4
  13  3325792768  19,1,21,7,6,3,12,13,5,9,2,8,4

Cf. A349473, A349502.

cflen[n_] := Module[{cf = ContinuedFraction[DivisorSigma[0, n]/DivisorSigma[-1, 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[10, 10^7]

cp's OEIS Frontend

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

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A349474 a(n) is the length of the continued fraction of the harmonic mean of the divisors of n.

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A349476 Numbers k such that the continued fraction of the harmonic mean of the divisors of k contains a single distinct element.

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A349499 Numbers k such that A349474(k) = A349474(k+1).

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A349477 Numbers k such that the sequence of elements of the continued fraction of the harmonic mean of the divisors of k is palindromic.

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A349502 Numbers k such that the continued fraction of the harmonic mean of the divisors of k contains distinct elements.

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A349497 a(n) is the smallest element in the continued fraction of the harmonic mean of the divisors of n.

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A353345 Numbers k such that the elements of the continued fractions of the harmonic means of the divisors of k and k+1 are anagrams of each other.

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A349478 a(n) is the least number k such that the sequence of elements of the continued fraction of the harmonic mean of the divisors of k is palindromic with length n, or -1 if no such k exists.