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 1-10 of 10 results.

A334471 a(n) = Product_{d|n} (A069934(n) / sigma(d)) where A069934(n) = lcm_{d|n} sigma(d).

Original entry on oeis.org

1, 3, 4, 441, 6, 144, 8, 385875, 2704, 324, 12, 12446784, 14, 576, 576, 37418184916875, 18, 197413632, 20, 42007896, 1024, 1296, 24, 38118276000000, 34596, 1764, 35152000, 99574272, 30, 26873856, 32, 1409355934894096875, 2304, 2916, 2304, 1695648500686393344
Offset: 1

Views

Author

Jaroslav Krizek, May 01 2020

Keywords

Examples

			For n = 6; divisors d of 6: {1, 2, 3, 6}; sigma(d): {1, 3, 4, 12}; lcm_{d|6} sigma(d) = 12; a(6) = 12/1 * 12/3 * 12/4 * 12/12 = 144.
		

Crossrefs

Cf. Similar sequence with tau(d): A334470.

Programs

  • Magma
    [&*[ LCM([&+Divisors(d): d in Divisors(n)]) / &+Divisors(d): d in Divisors(n)]: n in [1..100]]
    
  • Mathematica
    a[n_] := (LCM @@ (s = DivisorSigma[1, Divisors[n]]))^Length[s] / Times @@ s; Array[a, 36] (* Amiram Eldar, May 02 2020 *)
  • PARI
    a(n) = {my(d=divisors(n), lcms = lcm(vector(#d, k, sigma(d[k])))); vecprod(vector(#d, k, lcms/sigma(d[k])));} \\ Michel Marcus, May 02 2020

Formula

a(n) = ((lcm_{d|n} sigma(d))^tau(n)) / Product_{d|n} (sigma(d)).
a(n) = A069934(n)^A000005(n) / A206032(n).
a(p) = p + 1 for p = primes (A000040).

A265709 a(n) = numerator of Sum_{d|n} 1/sigma(d).

Original entry on oeis.org

1, 4, 5, 31, 7, 5, 9, 54, 69, 14, 13, 155, 15, 3, 35, 1709, 19, 23, 21, 31, 45, 13, 25, 27, 223, 10, 703, 93, 31, 35, 33, 15536, 65, 38, 21, 713, 39, 7, 75, 9, 43, 15, 45, 403, 161, 25, 49, 1709, 521, 446, 95, 155, 55, 703, 91, 243, 21, 62, 61, 155, 63, 11
Offset: 1

Views

Author

Jaroslav Krizek, Dec 24 2015

Keywords

Comments

a(n) = numerator of Sum_{d|n} 1/A000203(d).
Are there numbers n > 1 such that Sum_{d|n} 1/sigma(d) is an integer?

Examples

			For n = 6; divisors d of 6: {1, 2, 3, 6}; sigma(d): {1, 3, 4, 12}; Sum_{d|6} 1/sigma(d) = 1/1 + 1/3 + 1/4 + 1/12 = 20/12 = 5/3; a(n) = 5.
		

Crossrefs

Programs

  • Magma
    [Numerator(&+[1/SumOfDivisors(d): d in Divisors(n)]): n in [1..1000]]
    
  • Mathematica
    A265709[n_] := Numerator[DivisorSum[n, 1/DivisorSigma[1,#]&]];
    Array[A265709, 100] (* Paolo Xausa, Feb 06 2024 *)
  • PARI
    A265709(n) = numerator(sumdiv(n,d,1/sigma(d))); \\ Antti Karttunen, Nov 19 2017

Formula

a(n) = A265710(n) * Sum_{d|n} 1/sigma(d) = A265708(n) * A265710(n) / A069934(n).
a(1) = 1; a(p) = p + 2 for p = prime.

A265710 a(n) = denominator of Sum_{d|n} 1/sigma(d).

Original entry on oeis.org

1, 3, 4, 21, 6, 3, 8, 35, 52, 9, 12, 84, 14, 2, 24, 1085, 18, 13, 20, 18, 32, 9, 24, 14, 186, 7, 520, 56, 30, 18, 32, 9765, 48, 27, 16, 364, 38, 5, 56, 5, 42, 8, 44, 252, 104, 18, 48, 868, 456, 279, 72, 98, 54, 390, 72, 140, 16, 45, 60, 72, 62, 8, 416, 1240155
Offset: 1

Views

Author

Jaroslav Krizek, Dec 24 2015

Keywords

Comments

a(n) = denominator of Sum_{d|n} 1/A000203(d).
Are there numbers n > 1 such that Sum_{d|n} 1/sigma(d) is an integer?
a(n) = 2 for n = 14, 244, 494, 45994. Are there any others? - Robert Israel, Apr 02 2017

Examples

			For n = 6; divisors d of 6: {1, 2, 3, 6}; sigma(d): {1, 3, 4, 12}; Sum_{d|6} 1/sigma(d) = 1/1 + 1/3 + 1/4 + 1/12 = 20/12 = 5/3; a(n) = 3.
		

Crossrefs

Programs

  • Maple
    f:= n -> denom(add(1/numtheory:-sigma(d), d = numtheory:-divisors(n))):
    map(f, [$1..200]); # Robert Israel, Apr 02 2017
  • Mathematica
    Table[Denominator[Plus@@(1/DivisorSigma[1, Divisors[n]])], {n, 70}] (* Alonso del Arte, Dec 24 2015 *)
  • PARI
    a(n) = denominator(sumdiv(n, d, 1/sigma(d))); \\ Michel Marcus, Feb 06 2024

Formula

a(1) = 1; a(p) = p + 1 for p = prime.
a(n) = A265709(n) / (Sum_{d|n} 1/sigma(d)) = A265709(n) * A069934(n) / A265708(n).

A265708 a(n) = lcm_{d|n} sigma(d) * Sum_{d|n} 1/sigma(d), where sigma(d) represents the sum of divisors of d (A000203(d)).

Original entry on oeis.org

1, 4, 5, 31, 7, 20, 9, 162, 69, 28, 13, 155, 15, 36, 35, 5127, 19, 276, 21, 217, 45, 52, 25, 810, 223, 60, 703, 279, 31, 140, 33, 15536, 65, 76, 63, 2139, 39, 84, 75, 1134, 43, 180, 45, 403, 483, 100, 49, 25635, 521, 892, 95, 465, 55, 2812, 91, 1458, 105, 124
Offset: 1

Views

Author

Jaroslav Krizek, Dec 24 2015

Keywords

Examples

			For n = 6; divisors d of 6: {1, 2, 3, 6}; sigma(d): {1, 3, 4, 12}; lcm_{d|6} sigma(d) = 12; a(6) = 12/1 + 12/3 + 12/4 + 12/12 = 20.
		

Crossrefs

Programs

  • Magma
    [&+[1/SumOfDivisors(d): d in Divisors(n)] * LCM([SumOfDivisors(d): d in Divisors(n)]): n in [1..100]];
    
  • Mathematica
    a[n_] := LCM @@ DivisorSigma[1, Divisors[n]] * DivisorSum[n, 1/DivisorSigma[1, #] &]; Array[a, 100] (* Amiram Eldar, Dec 09 2022 *)
  • PARI
    A069934(n) = my(d = divisors(n)); lcm(vector(#d, k, sigma(d[k])));
    A265708(n) = (A069934(n) * sumdiv(n,d,1/sigma(d))); \\ Antti Karttunen, Nov 19 2017

Formula

a(n) = A069934(n) * Sum_{d|n} 1/A000203(d) = A265709(n) * A069934(n) / A265710(n).
Multiplicative with a(p^e) = (1/1 + ..., + 1/sigma(p^(e-1)) + 1/sigma(p^(e))) * lcm{1, ..., sigma(p^(e-1)), sigma(p^(e))}.

A265711 Numbers n such that floor(Sum_{d|n} 1 / sigma(d)) = 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 73
Offset: 1

Views

Author

Jaroslav Krizek, Dec 25 2015

Keywords

Comments

Numbers n such that A265710(n) = floor(A265708(n) / A069934(n)) = floor(A265709(n) / A265710(n)) = 1.
See A265714(n) = the smallest number k such that floor(Sum_{d|k} 1/sigma(d)) = n.

Examples

			6 is a term because floor(Sum_{d|6} 1/sigma(d)) = floor(1/1 + 1/3 + 1/4 + 1/12) = floor(5/3) = 1.
		

Crossrefs

Programs

  • Magma
    [n: n in [1..1000] | Floor(&+[1/SumOfDivisors(d): d in Divisors(n)]) eq 1]
    
  • Mathematica
    Select[Range@ 73, Floor[Sum[1/DivisorSigma[1, d], {d, Divisors@ #}]] == 1 &] (* Michael De Vlieger, Dec 31 2015 *)
  • PARI
    isok(n) = floor(sumdiv(n, d, 1/sigma(d))) == 1; \\ Michel Marcus, Dec 27 2015

A265712 Numbers n such that floor(Sum_{d|n} 1 / sigma(d)) = 2.

Original entry on oeis.org

60, 72, 84, 90, 120, 144, 168, 180, 210, 216, 240, 252, 264, 270, 280, 288, 300, 312, 324, 330, 336, 360, 378, 384, 390, 396, 408, 420, 432, 450, 456, 462, 468, 480, 504, 510, 528, 540, 546, 552, 560, 570, 576, 588, 600, 612, 624, 630, 648, 660, 672, 684, 690
Offset: 1

Views

Author

Jaroslav Krizek, Dec 25 2015

Keywords

Comments

Numbers n such that A265710(n) = floor(A265708(n) / A069934(n)) = floor(A265709(n) / A265710(n)) = 2.
See A265714(n) = the smallest number k such that floor(Sum_{d|k} 1/sigma(d)) = n.

Examples

			60 is a term because floor(Sum_{d|60} 1/sigma(d)) = floor(155/72) = 2.
		

Crossrefs

Programs

  • Magma
    [n: n in [1..1000] | Floor(&+[1/SumOfDivisors(d): d in Divisors(n)]) eq 2]
    
  • Mathematica
    Select[Range@ 690, Floor[Sum[1/DivisorSigma[1, d], {d, Divisors@ #}]] == 2 &] (* Michael De Vlieger, Dec 31 2015 *)
  • PARI
    isok(n) = floor(sumdiv(n, d, 1/sigma(d))) == 2; \\ Michel Marcus, Dec 27 2015

A265713 Numbers k such that floor(Sum_{d|k} 1 / sigma(d)) = 3.

Original entry on oeis.org

110880, 166320, 221760, 277200, 327600, 332640, 360360, 388080, 393120, 415800, 443520, 471240, 480480, 491400, 498960, 526680, 540540, 554400, 556920, 582120, 589680, 600600, 622440, 637560, 655200, 665280, 693000, 720720, 776160, 786240, 803880, 831600
Offset: 1

Views

Author

Jaroslav Krizek, Dec 25 2015

Keywords

Comments

Numbers k such that A265710(k) = floor(A265708(k) / A069934(k)) = floor(A265709(k) / A265710(k)) = 3.
See A265714(n) = the smallest number k such that floor(Sum_{d|k} 1/sigma(d)) = n.

Examples

			110880 is a term because floor(Sum_{d|110880} 1/sigma(d)) = floor(22333/7440) = 3.
		

Crossrefs

Programs

  • Magma
    [n: n in [1..1000000] | Floor(&+[1/SumOfDivisors(d): d in Divisors(n)]) eq 3]
    
  • Mathematica
    Select[Range[10^5, 9*10^5], Floor[Sum[1/DivisorSigma[1, d], {d, Divisors@ #}]] == 3 &] (* Michael De Vlieger, Dec 31 2015 *)
  • PARI
    isok(n) = floor(sumdiv(n, d, 1/sigma(d))) == 3; \\ Michel Marcus, Dec 27 2015

A266227 a(n) = floor(Sum_{d|n} 1/sigma(d)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1
Offset: 1

Views

Author

Jaroslav Krizek, Dec 24 2015

Keywords

Comments

Sequence of numbers n such that floor(Sum_{d|n} 1/sigma(d)) = k for k = 1, 2, 3:
k = 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ... (A265711);
k = 2: 60, 72, 84, 90, 120, 144, 168, 180, 210, 216, 240, 252, ... (A265712);
k = 3: 110880, 166320, 221760, 277200, 327600, 332640, 360360, ... (A265713).

Examples

			For n = 6; a(6) = floor(Sum_{d|6} 1/sigma(d)) = floor(1/1 + 1/3 + 1/4 + 1/12) = floor(5/3) = 1.
		

Crossrefs

Programs

  • Magma
    [Floor(&+[1/SumOfDivisors(d): d in Divisors(n)]): n in [1..100]]
    
  • Mathematica
    A266227[n_] := Floor[DivisorSum[n, 1/DivisorSigma[1, #]&]];
    Array[A266227, 100] (* Paolo Xausa, Feb 06 2024 *)
  • PARI
    A266227(n) = { my(s=sumdiv(n,d,1/sigma(d))); (numerator(s) \ denominator(s)); }; \\ Antti Karttunen, Nov 19 2017

Formula

a(n) = floor(Sum_{d|n} 1/A000203(d)).
a(n) = floor(A265708(n) / A069934(n)) = floor(A265709(n) / A265710(n)).

A265714 a(n) = the smallest number k such that floor(Sum_{d|k} 1/sigma(d)) = n.

Original entry on oeis.org

1, 60, 110880, 4658179125600, 950542574818669103079134726400, 204614292026733833316841991529248485168966921782532186656980932752000
Offset: 1

Views

Author

Jaroslav Krizek, Dec 24 2015

Keywords

Comments

If a(4) exists, it must be bigger than 5*10^6.
Are there numbers n > 1 such that Sum_{d|n} 1/sigma(d) is an integer?
Conjecture: a(n) is also the smallest number k such that Sum_{d|k} 1/sigma(d) >= n.
Sequence of numbers n such that floor(Sum_{d|n} 1/sigma(d)) = k for k = 1, 2, 3:
k = 1: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ... (A265711);
k = 2: 60, 72, 84, 90, 120, 144, 168, 180, 210, 216, 240, 252, ... (A265712);
k = 3: 110880, 166320, 221760, 277200, 327600, 332640, 360360, ... (A265713).
From Robert Israel, Dec 24 2015: (Start)
Note that g(k) = Sum_{d | k} 1/sigma(d) is multiplicative, with g(p) = 1 + 1/(p+1) for prime p.
Since Product_p (1+1/(p+1)) diverges, there are certainly numbers k with floor(g(k)) = n, including some squarefree numbers.
Conjectured values: a(4) = 4658179125600,
a(5) = 1188178218523336378848918408000,
a(6) = 5354073974699535305124032111682002028587967786642925550857667740344000.
These do have the correct value of floor(g(k)), but may not be the lowest possible.
(End)
Probably, a(7) = 1058687979...2471360000 = 349# * 23# * 7# * 5# * 3#^2 * 2#^3. - Hiroaki Yamanouchi, Dec 31 2015

Examples

			For n = 2; a(2) = 60 because 60 is the smallest number with floor (Sum_{d|60} 1/sigma(d)) = floor(155/72) = 2.
		

Crossrefs

Programs

  • Magma
    a:=1; S:=[a]; for n in [2..3] do k:=0; flag:= true; while flag do k+:=1; if &+[1/SumOfDivisors(d): d in Divisors(k)] ge n then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;

Extensions

a(4)-a(6) from Hiroaki Yamanouchi, Dec 31 2015

A266228 Numbers n such that Sum_{d|n} 1/sigma(d) > Sum_{d|m} 1/sigma(d) for all m < n.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 36, 48, 60, 120, 180, 240, 360, 420, 720, 840, 1260, 1680, 2520, 5040, 10080, 13860, 15120, 18480, 25200, 27720, 55440, 110880, 166320, 277200, 332640, 360360, 720720, 1441440, 2162160, 3603600, 4324320, 6126120, 7207200, 10810800, 12252240, 21621600, 24504480, 36756720, 61261200, 73513440
Offset: 1

Views

Author

Jaroslav Krizek, Dec 24 2015

Keywords

Comments

Where record values of Sum_{d|n} 1/sigma(d) occur.

Examples

			For n = 4; a(4) = 6 because 6 is the smallest number such that Sum_{d|a(4)} 1/sigma(d) = Sum_{d|6} 1/sigma(d) = 5/3 > Sum_{d|a(3)} 1/sigma(d) = Sum_{d|4} 1/sigma(d) = 31/21.
		

Crossrefs

Programs

  • Magma
    a:=1; S:=[a]; for n in [2..25] do k:=0; flag:= true; while flag do k+:=1; if &+[1/SumOfDivisors(d): d in Divisors(a)] lt &+[1/SumOfDivisors(d): d in Divisors(k)] then Append(~S, k); a:=k; flag:=false; end if; end while; end for; S;
    
  • PARI
    b(n)={sumdiv(n, d, 1/sigma(d));}
    { my(m=0); for(n=1, 1e6, if(b(n)>m, m=b(n); print1(n, ", "))) } \\ Andrew Howroyd, Nov 11 2018

Extensions

a(35)-a(47) from Andrew Howroyd, Nov 11 2018
Showing 1-10 of 10 results.