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.

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A335368 Harmonic numbers k with a record number of primes p not dividing k such that k*p is also a harmonic number.

Original entry on oeis.org

1, 28, 1638, 30240, 2178540, 2457000, 32997888, 142990848, 1307124000, 71271827200, 547929930240, 2198278051200, 2567400675840, 54409216942080
Offset: 1

Views

Author

Amiram Eldar, Jun 03 2020

Keywords

Comments

The corresponding record values are 0, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 17.
If k is a harmonic number (A001599) and p is a prime that does not divide k, then k*p is a harmonic number if and only if (p+1)/2 is a divisor of the harmonic mean of the divisors of k, k*tau(k)/sigma(k) = k*A000005(k)/A000203(k).

Examples

			1 is the first harmonic number, and it has 0 primes p such that 1*p = p is a harmonic number, since a prime number cannot be harmonic. The next harmonic number k with at least one prime p such that k*p is also a harmonic number is 28, since 28 * 5 = 140 is a harmonic number.
		

Crossrefs

Programs

  • Mathematica
    harmNums = Cases[Import["https://oeis.org/A001599/b001599.txt", "Table"], {, }][[;; , 2]]; harMean[n_] := n*DivisorSigma[0, n]/DivisorSigma[1, n]; primeCount[n_] := Module[{d = Divisors[harMean[n]]}, Length @ Select[2*d - 1, PrimeQ[#] && ! Divisible[n, #] &]]; primeCountMax = -1; seq = {}; Do[If[(pc = primeCount[harmNums[[k]]]) > primeCountMax, primeCountMax = pc; AppendTo[seq, harmNums[[k]]]], {k, 1, Length[harmNums]}]; seq

A335370 Harmonic numbers m with a record number k of distinct prime numbers p_i (i = 1..k) that do not divide m such that m*p_1, m*p_1*p_2, ... , m*p_1*...*p_k are all harmonic numbers.

Original entry on oeis.org

1, 28, 1638, 6200, 2457000, 4713984, 1381161600, 10200236032
Offset: 1

Views

Author

Amiram Eldar, Jun 03 2020

Keywords

Comments

If m is a harmonic number (A001599), then it is possible to generate a new harmonic number m*p if p is a prime number that does not divide m and (p+1)/2 is a divisor of the harmonic mean of the divisors of m, h(m) = m * tau(m)/sigma(m) = m * A000005(m)/A000203(m).
The terms of this sequence begin a chain of harmonic numbers of a record length. In each chain, each member, except the first, is generated from its predecessor by multiplying it by a prime that does not divide it.
The corresponding record values of k are 0, 1, 2, 3, 4, 6, 7, 8, ...
The list of primes or their order may not be unique.

Examples

			28 is the least harmonic number with one prime, p = 5, such that 28*p = 140 is also a harmonic number.
1638 is the least harmonic number with 2 primes, 5 and 29, such that 1638*5 = 8190 and 1638*5*29 = 237510 are also harmonic numbers.
.
n  a(n)         k   primes p_i, i = 1..k                 number of permutations
-------------------------------------------------------------------------------
1  1            0   -                                         -
2  28           1   5                                         1
3  1638         2   5, 29                                     1
4  6200         3   19, 37, 73                                1
5  2457000      4   11, 19, 37, 73                            4
6  4713984      6   5, 7, 13, 19, 37, 73                      15
                    5, 7, 19, 37, 73, 1021                    5
7  1381161600   7   11, 19, 37, 43, 73, 6277, 12553           10
                    11, 19, 37, 43, 3181, 6361, 12721         6
8  10200236032  8   3, 5, 79, 157, 313, 1877, 7507, 15013     5
		

Crossrefs

Programs

  • Mathematica
    harmNums = Cases[Import["https://oeis.org/A001599/b001599.txt", "Table"], {, }][[;; , 2]]; harMean[n_] := n * DivisorSigma[0, n]/DivisorSigma[1, n]; harmGen[n_] := Module[{d = Divisors[harMean[n]]}, n * Select[2*d - 1, PrimeQ[#] && ! Divisible[n, #] &]]; harmGens[s_] := Union @ Flatten[harmGen /@ s]; lenmax = -1; seq = {}; Do[len = -3 + Length @ FixedPointList[harmGens, {harmNums[[k]]}]; If[len > lenmax, lenmax = len; AppendTo[seq, harmNums[[k]]]], {k, 1, Length[harmNums]}]; seq

A335371 Harmonic numbers with a record number of harmonic numbers that can be generated from them using an iterative process of multiplying by primes (see Comments).

Original entry on oeis.org

1, 28, 1638, 6200, 950976, 2178540, 2457000, 4713984, 45532800, 142990848, 459818240, 1381161600, 10200236032, 57575890944, 109585986048, 513480135168, 1553357978368, 10881843388416, 43947421401888
Offset: 1

Views

Author

Amiram Eldar, Jun 03 2020

Keywords

Comments

If m is a harmonic number (A001599), then it is possible to generate a new harmonic number m*p if p is a prime number that does not divide m and (p+1)/2 is a divisor of the harmonic mean of the divisors of m, h(m) = m * tau(m)/sigma(m) = m * A000005(m)/A000203(m).
Given a harmonic number m, in the first iteration a finite set of new harmonic numbers, {m*p_1, m*p_2, ...} is being generated. In the second iteration, a new set of harmonic number is being generated from each of the harmonic numbers from the previous iteration, a union of these sets is be calculated (removing duplicates). The process is terminated when no more harmonic numbers can be generated. The total number of harmonic numbers from all the iterations is being counted. The terms of this sequence have a record count of new harmonic numbers.
The corresponding record values of k are 0, 1, 3, 5, 8, 12, 17, 36, 38, 40, 44, 62, 70, 82, 156, 226, 281, 335, 358, ...

Examples

			1638 is a term since a record number of 3 new harmonic numbers can be generated from it. In the first iteration 2 new harmonic numbers can be generated: 1638 * 5 = 8190, and 1638 * 17 = 27846. In the second iteration, a new harmonic number can be generated from 8190: 8190 * 29 = 237510.
		

Crossrefs

Programs

  • Mathematica
    harmNums = Cases[Import["https://oeis.org/A001599/b001599.txt", "Table"], {, }][[;; , 2]]; harMean[n_] := n * DivisorSigma[0, n]/DivisorSigma[1, n]; harmGen[n_] := Module[{d = Divisors[harMean[n]]}, n * Select[2*d - 1, PrimeQ[#] && ! Divisible[n, #] &]]; harmGens[s_] := Union@Flatten[harmGen /@ s]; lenmax = -1; seq = {}; Do[len = Length @ Union @ Flatten @ FixedPointList[harmGens, {harmNums[[k]]}]; If[len > lenmax, lenmax = len; AppendTo[seq, harmNums[[k]]]], {k, 1, Length[harmNums]}]; seq

A348414 Numbers with record values of the denominator of the harmonic mean of their divisors.

Original entry on oeis.org

1, 2, 4, 8, 16, 36, 64, 100, 128, 144, 256, 324, 400, 512, 576, 900, 1296, 1600, 1936, 2304, 3600, 5184, 6400, 8100, 9216, 10404, 11664, 14400, 17424, 20736, 22500, 30276, 32400, 41616, 46656, 57600, 76176, 90000, 108900, 129600, 166464, 202500, 230400, 260100
Offset: 1

Views

Author

Amiram Eldar, Oct 17 2021

Keywords

Comments

The corresponding record values are 1, 3, 7, 15, 31, 91, 127, 217, 255, 403, ... (see the link for more values).

Examples

			The first 4 terms of A099378 are 1, 3, 2 and 7. The record values, 1, 3 and 7, occur at 1, 2 and 4, the first 3 terms of this sequence.
		

Crossrefs

Programs

  • Mathematica
    d[n_] := Denominator[DivisorSigma[0, n]/DivisorSigma[-1, n]]; dm = 0; s = {}; Do[dn = d[n]; If[dn > dm, dm = dn; AppendTo[s, n]], {n, 1, 10^6}]; s

A348825 Numbers k such that the denominator of the harmonic mean of the divisors of k is larger than k.

Original entry on oeis.org

2, 4, 8, 9, 16, 25, 36, 64, 80, 81, 100, 104, 121, 128, 144, 208, 225, 256, 272, 289, 320, 324, 400, 484, 512, 529, 576, 625, 729, 841, 900, 1024, 1088, 1089, 1156, 1250, 1296, 1300, 1332, 1575, 1600, 1664, 1681, 1856, 1936, 2025, 2116, 2196, 2209, 2304, 2368
Offset: 1

Views

Author

Amiram Eldar, Nov 01 2021

Keywords

Comments

Numbers k such that A099378(k) > k.
This sequence is infinite. For example, if p is a prime of the form 6*k-1, then p^2 is a term.
2 is the only prime term, since the denominator of the harmonic mean of the divisors of an odd prime p is (p+1)/2 < p.

Examples

			2 is a term since the harmonic mean of the divisors of 2 is 4/3 and 3 > 2.
4 is a term since the harmonic mean of the divisors of 4 is 12/7 and 7 > 4.
		

Crossrefs

Programs

  • Mathematica
    q[n_] := Denominator[DivisorSigma[0, n]/DivisorSigma[-1, n]] > n; Select[Range[2500], q]
  • PARI
    isok(k) = my(d=divisors(k)); (denominator(#d/sum(i=1, #d, 1/d[i])) > k); \\ Michel Marcus, Nov 01 2021

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

Views

Author

Amiram Eldar, Nov 19 2021

Keywords

Comments

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

Examples

			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.
		

Crossrefs

A001599 and A349476 are subsequences.

Programs

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

A330606 Numbers k such that k*d(k) and sigma(k) are relatively prime, where d(k) is the number of divisors of k (A000005) and sigma(k) is their sum (A000203).

Original entry on oeis.org

1, 2, 4, 8, 9, 16, 25, 36, 64, 81, 100, 121, 128, 144, 225, 256, 289, 324, 400, 484, 512, 529, 576, 625, 729, 841, 900, 1024, 1089, 1156, 1250, 1296, 1600, 1681, 1936, 2025, 2116, 2209, 2304, 2401, 2500, 2601, 2809, 3025, 3364, 3481, 3600, 4096, 4356, 4624, 4761
Offset: 1

Views

Author

Amiram Eldar, Dec 20 2019

Keywords

Comments

If p is prime and p == 2 (mod 3) then p^2 is in the sequence.
Let E(x) = #{n | a(n) <= x} be the number of terms of this sequence up to x. Kanold proved that there are two constants 0 < c1 < c2 and a positive number x_0 such that c1 < E(x)/sqrt(x/log(x)) < c2 for x > x_0. De Koninck and Kátai proved that there is a positive constant c such that E(x) = c * (1 + o(1)) * sqrt(x/log(x)).
Apparently most of the terms are squares or powers of 2. Terms that are not included 1250, 4802, 31250, 57122, ...
Numbers k such that A099377(k) = A038040(k) and A099378(k) = A000203(k). - Amiram Eldar, Nov 02 2021

References

  • József Sándor, Dragoslav S. Mitrinovic, Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, page 75.

Crossrefs

Programs

  • Magma
    [k:k in [1..5000]| Gcd(k*NumberOfDivisors(k),DivisorSigma(1,k)) eq 1]; // Marius A. Burtea, Dec 20 2019
  • Mathematica
    Select[Range[10^4], CoprimeQ[# * DivisorSigma[0, #], DivisorSigma[1, #]] &]

A348411 Numbers whose divisors have a harmonic mean with a denominator 2.

Original entry on oeis.org

3, 15, 42, 84, 135, 308, 420, 546, 1428, 1488, 1890, 2295, 2660, 3780, 6210, 7440, 9424, 12180, 13392, 18018, 20832, 24384, 24570, 43152, 43400, 64260, 66960, 77490, 90090, 98420, 121920, 127710, 155610, 200340, 204600, 227664, 316992, 348688, 353400, 461776, 483210
Offset: 1

Views

Author

Amiram Eldar, Oct 17 2021

Keywords

Comments

Numbers k such that A099378(k) = 2.
The odd terms seem to be relatively rare: 3, 15, 135, 2295, 544635, 9258795, 22330035, 39118408875, ...
If k is in this sequence, then 2*k is in A348412.

Examples

			3 is a term since the harmonic mean of its divisors, {1, 3}, is 3/2.
15 is a term since the harmonic mean of its divisors, {1, 3, 5, 15}, is 5/2.
		

Crossrefs

Similar sequences: A159907, A330598.

Programs

  • Maple
    filter:= proc(n) local L,h;
      L:= map(t->1/t,numtheory:-divisors(n));
      denom(nops(L)/convert(L,`+`))=2;
    end proc:
    select(filter, [$1..10^6]); # Robert Israel, Oct 17 2021
  • Mathematica
    Select[Range[10^5], Denominator[DivisorSigma[0, #]/DivisorSigma[-1, #]] == 2 &]
    Select[Range[500000],Denominator[HarmonicMean[Divisors[#]]]==2&] (* Harvey P. Dale, Apr 06 2023 *)
  • PARI
    isok(m) = my(d=divisors(m)); denominator(#d/sum(k=1, #d, 1/d[k])) == 2; \\ Michel Marcus, Oct 18 2021
    
  • Python
    from sympy import gcd, divisor_sigma
    A348411_list = [n for n in range(1,10**3) if (lambda x, y: 2*gcd(x,y*n)==x)(divisor_sigma(n),divisor_sigma(n,0))] # Chai Wah Wu, Oct 20 2021

A348415 Numbers k such that k and k+1 have the same denominator of the harmonic means of their divisors.

Original entry on oeis.org

12, 88, 180, 266, 321, 604, 4277, 4364, 8632, 15861, 18720, 28461, 47613, 63546, 97412, 98907, 135078, 137333, 154132, 179621, 185776, 192699, 203709, 265489, 284883, 344217, 383466, 517610, 604197, 876469, 1089604, 1277518, 1713865, 1839123, 1893268, 2349390
Offset: 1

Views

Author

Amiram Eldar, Oct 17 2021

Keywords

Comments

Numbers k such that A099378(k) = A099378(k+1).
The common denominators of k and k+1 are 7, 45, 91, 30, 36, 133, 96, 637, ...
Can 3 consecutive numbers have the same denominator of harmonic mean of divisors? There are no such numbers below 10^10.

Examples

			12 is a term since the harmonic means of the divisors of 12 and 13 are 18/7 and 13/7, respectively, and both have the denominator 7.
		

Crossrefs

Similar sequences: A002961, A238380.

Programs

  • Mathematica
    dh[n_] := Denominator[DivisorSigma[0, n]/DivisorSigma[-1, n]]; Select[Range[10^6], dh[#] == dh[# + 1] &]
  • PARI
    f(n) = my(d=divisors(n)); denominator(#d/sum(k=1, #d, 1/d[k])); \\ A099378
    isok(k) = f(k) == f(k+1); \\ Michel Marcus, Oct 20 2021

A348694 a(n) is the least number k such that the numerator of the harmonic mean of the divisors of k is equal to n, or -1 if no such k exists.

Original entry on oeis.org

1, 6, 3, 2, 5, 270, 7, 672, 84, 30, 11, 4, 13, 18620, 420, 24, 17, 12, 19, 10, 21, 22, 23, 30240, 1550, 78, 9, 168, 29, 60, 31, 8, 132, 102, 35, 18, 37, 38, 39, 3360, 41, 3724, 43, 7392, 45, 15456, 47, 1080, 49, 6051500, 153, 26, 53, 540, 55, 56, 57, 174, 59, 90
Offset: 1

Views

Author

Amiram Eldar, Oct 30 2021

Keywords

Examples

			a(2) = 6 since the harmonic mean of the divisors of 6 is 2.
a(3) = 3 since the harmonic mean of the divisors of 3 is 3/2.
		

Crossrefs

Programs

  • Mathematica
    num[n_] := Numerator[DivisorSigma[0, n]/DivisorSigma[-1, n]]; seq[m_] := Module[{s = Table[0, {m}], c = 0, n = 1, i}, While[c < m, i = num[n]; If[i <= m && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[100]
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