A071862 -id:A071862 - 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]

A071865 Smallest k such that the simple continued fraction for Sum(d|k, 1/d) contains exactly n elements.

Original entry on oeis.org

1, 2, 4, 14, 22, 26, 75, 195, 330, 324, 935, 1598, 3422, 3663, 10191, 14066, 12099, 53661, 121555, 182169, 235509, 307615, 633945, 2097595, 2072198, 2643298, 6544282, 8675343, 13670722, 17573794, 85112326, 77295778, 235873898, 362150458, 544042486, 1457255474

Offset: 1

Benoit Cloitre, Jun 09 2002

			sum(d|195, 1/d) = 112/65 and 112/65 continued fraction is [1, 1, 2, 1, 1, 1, 1, 3] which contains 8 elements. There is no smaller number than 195 with this property hence a(8)=195.

Cf. A071862.

a = Table[0, {50}]; Do[b = Length[ ContinuedFraction[ Apply[ Plus, 1/Divisors[n]]]]; If[ a[[b]] == 0, a[[b]] = n], {n, 1, 10^7}]

for(n=1,21,s=1; while(length(contfrac(sumdiv(s,d,1/d)))

More terms from Robert G. Wilson v, Jun 11 2002

a(29)-a(36) from Michel Marcus, Sep 17 2012

A342866 The number of elements in the continued fraction for phi(n)/n, where phi is the Euler totient function (A000010).

Original entry on oeis.org

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

Offset: 1

Amiram Eldar, Mar 27 2021

			a(2) = 2 since the continued fraction of phi(2)/2 = 1/2 = 0 + 1/2 has 2 elements: {0, 2}.
a(3) = 3 since the continued fraction of phi(3)/3 = 2/3 = 0 + 1/(1 + 1/2) has 3 elements: {0, 1, 2}.
a(15) = 4 since the continued fraction of phi(15)/15 = 8/15 = 0 + 1/(1 + 1/(1 + 1/7)) has 4 elements: {0, 1, 1, 7}.

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

Cf. A000010, A007694, A076512, A109395, A342867.

Cf. A071862 (similar, with sigma(n)/n).

a[n_] := Length @ ContinuedFraction[EulerPhi[n]/n]; Array[a, 100]

a(n) = #contfrac(eulerphi(n)/n); \\ Michel Marcus, Mar 30 2021

a(n) = 2 if and only if n is in A007694.

a(p) = 3 for an odd prime p.

A071864 Nonprime n such that the number of elements in the continued fraction for Sum_{d|n} 1/d equals tau(n), the number of divisors of n.

Original entry on oeis.org

1, 4, 9, 14, 15, 21, 25, 49, 51, 55, 57, 63, 95, 98, 99, 115, 116, 121, 147, 161, 169, 172, 175, 188, 195, 203, 236, 244, 245, 247, 265, 284, 287, 289, 297, 299, 322, 328, 329, 351, 356, 361, 363, 370, 371, 374, 387, 406, 412, 413, 418, 423, 425, 437, 465, 488

Offset: 1

Benoit Cloitre, Jun 09 2002

If p is prime p^2 is in the sequence since the continued fraction for Sum_{d|n} 1/d is [1, p-1, p+1] and there are 3 divisors for p^2.

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

Cf. A000005, A017665, A017666, A071862.

aQ[n_] := ! PrimeQ[n] && Length@ContinuedFraction[DivisorSigma[1, n]/n] ==  DivisorSigma[0, n]; Select[Range[488], aQ] (* Amiram Eldar, Aug 30 2019 *)

for(n=1,1000,if(length(contfrac(sumdiv(n,d,1/d)))==numdiv(n)*(1-isprime(n)),print1(n,",")))

A215012 Composite numbers n such that sigma(n)/n leaves a remainder which divides n.

Original entry on oeis.org

12, 18, 20, 24, 40, 56, 88, 104, 180, 196, 224, 234, 240, 360, 368, 420, 464, 540, 600, 650, 780, 992, 1080, 1344, 1504, 1872, 1888, 1890, 1952, 2016, 2184, 2352, 2376, 2688, 3192, 3276, 3724, 3744, 4284, 4320, 4680

Offset: 1

J. M. Bergot, Jul 31 2012

nonn

The numbers and the program were provided by Charles R Greathouse IV.

If n belongs to the sequence, then sigma(n) = d*n + rem, so sigma(n)/n = d + rem/n. Since rem is a divisor of n, n = rem*r, thus rem/n = 1/r. Then sigma(n)/n = d + 1/r and contfrac(sigma(n)/n) = [d, r], and length(contfrac(sigma(n)/n)) = 2. That is, A071862(n) = 2. - Michel Marcus, Aug 29 2012

			24 has the divisors 1,2,3,4,6,12,24, which sum to be 60. Divide 60 by 24 and the remainder is 12, which is a divisor of 24.

Donovan Johnson, Table of n, a(n) for n = 1..1000

Cf. A000203, A071862.

a={}; For[n=1, n<=5000, n++, If[!PrimeQ[n], {s=DivisorSigma[1, n]; If[Mod[n, Mod[s,n]] == 0, AppendTo[a,n]]; }]; ]; a  (* John W. Layman, Jul 31 2012 *)
Select[Range[5000],CompositeQ[#]&&Mod[#,Mod[DivisorSigma[1,#],#]]==0&] // Quiet (* Harvey P. Dale, May 24 2019 *)

is(n)=my(t=sigma(n)%n);t && n%t==0 && !isprime(n)

Terms a(24)-a(41) from John W. Layman, Jul 31 2012

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

Amiram Eldar, Nov 25 2021

nonn

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?

			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}.

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

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

isok(k) = #Set(contfrac(sigma(k)/k)) == 1; \\ Michel Marcus, Nov 25 2021

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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A071865 Smallest k such that the simple continued fraction for Sum(d|k, 1/d) contains exactly n elements.

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A342866 The number of elements in the continued fraction for phi(n)/n, where phi is the Euler totient function (A000010).

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A071864 Nonprime n such that the number of elements in the continued fraction for Sum_{d|n} 1/d equals tau(n), the number of divisors of n.

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A215012 Composite numbers n such that sigma(n)/n leaves a remainder which divides n.

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

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