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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A334420 Numbers m such that sigma(d)/tau(d) is an integer for all divisors d of m.

Original entry on oeis.org

1, 3, 5, 7, 11, 13, 15, 17, 19, 21, 23, 29, 31, 33, 35, 37, 39, 41, 43, 47, 49, 51, 53, 55, 57, 59, 61, 65, 67, 69, 71, 73, 77, 79, 83, 85, 87, 89, 91, 93, 95, 97, 101, 103, 105, 107, 109, 111, 113, 115, 119, 123, 127, 129, 131, 133, 137, 139, 141, 143, 145
Offset: 1

Views

Author

Jaroslav Krizek, Apr 29 2020

Keywords

Comments

Sequences of numbers m from this sequence with k such divisors for 1 < k < 6:
k = 2: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ... (A065091 - odd primes).
k = 3: 49, 169, 361, 961, 1369, 1849, 3721, 4489, 5329, 6241, 9409, ...
k = 4: 15, 21, 33, 35, 39, 51, 55, 57, 65, 69, 77, 85, 87, 91, 93, ...
k = 5: 923521, 13845841, 519885601, 1073283121, 1982119441, ...
See A334421 for sequence of the smallest numbers m with k such divisors.
All divisors of a member of the sequence are members of the sequence. - Robert Israel, May 01 2020
Numbers for which all divisors are in A003601. - Michel Marcus, May 02 2020

Examples

			Number 15 with divisors 1, 3, 5 and 15 is a term because sigma(1)/tau(1) = 1/1 = 1, sigma(3)/tau(3) = 4/2 = 2, sigma(5)/tau(5) = 6/2 = 3, sigma(15)/tau(15) = 24/4 = 6.

Links

Crossrefs

Subsequence of A306639.
Cf. A000005 (tau), A000203 (sigma), A003601, A324499, A324500, A334421.
Includes A056911.

Programs

  • Magma
    [m: m in [1..10^6] | &+[SumOfDivisors(d) mod NumberOfDivisors(d): d in Divisors(m)] eq 0];
  • Maple
    filter:= n -> andmap(d -> numtheory:-sigma(d) mod numtheory:-tau(d)=0, numtheory:-divisors(n)):
select(filter, [$1..200]); # Robert Israel, May 01 2020
  • Mathematica
    divQ[n_] := Divisible[DivisorSigma[1, n], DivisorSigma[0, n]]; Select[Range[150], AllTrue[Divisors[#], divQ] &] (* Amiram Eldar, Apr 29 2020 *)
  • PARI
    isok(m) = fordiv(m, d, if (sigma(d) % numdiv(d), return (0))); return(1); \\ Michel Marcus, Apr 29 2020

Formula

A324500(a(n)) = 1.

A334421 a(n) is the smallest number m with n divisors d such that sigma(d)/tau(d) is an integer for all divisors d.

Original entry on oeis.org

1, 3, 49, 15, 923521, 147, 88245939632761, 105, 8281, 2770563, 4345096786921664259621718196367601, 735, 9024585590445680759701490904755712009585829774768244676951841, 264737818898283, 45252529, 1155
Offset: 1

Views

Author

Jaroslav Krizek, May 01 2020

Keywords

Comments

Sequences of numbers m with n such divisors for 1 < n < 6:
n = 2: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, ... (A065091 - odd primes).
n = 3: 49, 169, 361, 961, 1369, 1849, 3721, 4489, 5329, 6241, 9409, ...
n = 4: 15, 21, 33, 35, 39, 51, 55, 57, 65, 69, 77, 85, 87, 91, 93, ...
n = 5: 923521, 13845841, 519885601, 1073283121, 1982119441, ...
Other terms: a(18) = 24843, a(24) = 8085, a(32) = 15015.
a(p) = (k*p# + 1)^(p-1) for some k > 0 and p# is the product of primes <= k. - David A. Corneth, May 01 2020 [Proof: a(p) must be of the form q^(p-1), where q is a prime. Thus r | (q^r - 1)/(q - 1) if r <= p is prime. Suppose that q - 1 is not divisible by r, then 0 == q^r - 1 == q - 1 (mod r), a contradiction! Therefore, q - 1 is divisible by any primes r <= p. In conclusion, q = k*p# + 1 for some k > 0. - Jinyuan Wang, May 02 2020]
a(17) = 4084081^16 and a(19) = 106696591^18 are too large to be included. - Amiram Eldar, May 02 2020

Examples

			a(4) = 15 because number 15 with divisors 1, 3, 5 and 15 is the smallest number with 4 such divisors: sigma(1)/tau(1) = 1/1 = 1, sigma(3)/tau(3) = 4/2 = 2, sigma(5)/tau(5) = 6/2 = 3, sigma(15)/tau(15) = 24/4 = 6.

Crossrefs

Cf. A000005 (tau), A000203 (sigma), A003601, A306639, A324499, A324500, A334420.

Programs

  • Magma
    [Min([m: m in[1..1000] | &+[&+Divisors(d) mod #Divisors(d): d in Divisors(m)] eq 0 and #Divisors(m) eq k]): k in [1..4]]

Extensions

a(13)-a(14) from Amiram Eldar, May 02 2020

A324500 a(n) = denominator of Sum_{d|n} sigma(d)/tau(d) where sigma(k) = the sum of the divisors of k (A000203) and tau(k) = the number of divisors of k (A000005).

Original entry on oeis.org

1, 2, 1, 6, 1, 2, 1, 12, 3, 1, 1, 2, 1, 2, 1, 60, 1, 3, 1, 3, 1, 2, 1, 4, 3, 1, 3, 6, 1, 1, 1, 60, 1, 1, 1, 9, 1, 2, 1, 3, 1, 2, 1, 6, 3, 2, 1, 20, 1, 6, 1, 3, 1, 3, 1, 12, 1, 1, 1, 1, 1, 2, 3, 420, 1, 2, 1, 3, 1, 1, 1, 18, 1, 1, 1, 6, 1, 1, 1, 15, 15, 1, 1, 2
Offset: 1

Views

Author

Jaroslav Krizek, Mar 02 2019

Keywords

Comments

Sum_{d|n} sigma(d)/tau(d) > 1 for all n > 1.
Sum_{d|n} sigma(d)/tau(d) = n only for numbers n = 1, 3, 10 and 30.

Examples

			Sum_{d|n} sigma(d)/tau(d) for n >= 1: 1, 5/2, 3, 29/6, 4, 15/2, 5, 103/12, 22/3, 10, 7, 29/2, 8, 25/2, 12, 887/60, ...
For n=4; Sum_{d|4} sigma(d)/tau(d) = sigma(1)/tau(1) + sigma(2)/tau(2) + sigma(4)/tau(4) = 1/1 + 3/2 + 7/3 = 29/6;  a(4) = 6.

Links

Crossrefs

Cf. A000005, A000203, A323779, A323780, A323781, A324499 (numerators), A306639.

Programs

  • Magma
    [Denominator(&+[SumOfDivisors(d) / NumberOfDivisors(d): d in Divisors(n)]): n in [1..100]]
  • Mathematica
    Table[Denominator[Sum[DivisorSigma[1, k]/DivisorSigma[0, k], {k, Divisors[n]}]], {n, 1, 100}] (* G. C. Greubel, Mar 04 2019 *)
  • PARI
    a(n) = denominator(sumdiv(n, d, sigma(d)/numdiv(d))); \\ Michel Marcus, Mar 03 2019
  • Sage
    [sum(sigma(k,1)/sigma(k,0) for k in n.divisors() ).denominator() for n in (1..100)] # G. C. Greubel, Mar 04 2019

Formula

a(p) = 1 for odd primes p.
a(n) = 1 for numbers in A306639.

A306639 Numbers m such that Sum_{d|m} (sigma(d)/tau(d)) is an integer h where sigma(k) = the sum of the divisors of k (A000203) and tau(k) = the number of divisors of k (A000005).

Original entry on oeis.org

1, 3, 5, 7, 10, 11, 13, 15, 17, 19, 21, 23, 26, 29, 30, 31, 33, 34, 35, 37, 39, 41, 43, 47, 49, 51, 53, 55, 57, 58, 59, 60, 61, 65, 67, 69, 70, 71, 73, 74, 75, 77, 78, 79, 82, 83, 85, 87, 89, 91, 93, 95, 97, 98, 101, 102, 103, 105, 106, 107, 109, 110, 111, 113
Offset: 1

Views

Author

Jaroslav Krizek, Mar 02 2019

Keywords

Comments

Sum_{d|n} sigma(d)/tau(d) > 1 for all n > 1.
Sum_{d|n} sigma(d)/tau(d) = n only for numbers 1, 3, 10 and 30.
Odd primes are terms.
Corresponding values of integers h: 1, 3, 4, 5, 10, 7, 8, 12, 10, 11, 15, 13, 20, 16, 30, 17, 21, 25, 20, 20, 24, ...

Examples

			Sum_{d|n} sigma(d)/tau(d) for n >= 1: 1, 5/2, 3, 29/6, 4, 15/2, 5, 103/12, 22/3, 10, 7, 29/2, 8, 25/2, 12, 887/60, ...
10 is a term because Sum_{d|10} (sigma(d)/tau(d)) = sigma(1)/tau(1) + sigma(2)/tau(2) + sigma(5)/tau(5) + sigma(10)/tau(10) = 1/1 + 3/2 + 6/2 + 18/4 = 10 (integer).

Crossrefs

Cf. A000005, A000203, A324499, A324500.

Programs

  • Magma
    [n: n in [1..1000] | Denominator(&+[SumOfDivisors(d) / NumberOfDivisors(d): d in Divisors(n)]) eq 1]
  • PARI
    isok(n) = frac(sumdiv(n, d, sigma(d)/numdiv(d))) == 0; \\ Michel Marcus, Mar 03 2019

Formula

A324500(a(n)) = 1.
Showing 1-4 of 4 results.

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