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 1-5 of 5 results.

A054482 Denominator of lowest terms fraction from division of a highly composite number (1) by its predecessor.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 3, 4, 1, 2, 3, 2, 1, 6, 2, 3, 2, 1, 2, 3, 2, 3, 4, 10, 11, 9, 10, 2, 3, 2, 3, 4, 5, 2, 9, 5, 12, 2, 3, 2, 3, 4, 5, 2, 9, 5, 4, 3, 5, 4, 2, 15, 17, 12, 5, 2, 9, 5, 4, 3, 5, 4, 2, 15, 19, 2, 15, 2, 19, 18, 5, 4, 3, 5, 4, 2, 3, 2, 3, 2, 6, 35, 92, 21, 23, 35, 92, 21, 2, 23, 21, 2
Offset: 2

Views

Author

Henry Bottomley, Mar 31 2000

Keywords

Comments

Successive fractions are 2/1,2/1,3/2,2/1,2/1,3/2,4/3,5/4,2/1,3/2,4/3,3/2,2/1 a(n) is frequently A054483(n)-1, though not for example for n=26.

Examples

			a(7)=2 since A002182(7)=36, A002182(6)=24 and 36/24=3/2 in lowest terms.
		

Crossrefs

Programs

  • Mathematica
    s = Import["https://oeis.org/A002182/b002182.txt", "Data"][[All, -1]]; Array[Denominator[s[[# + 1]]/s[[#]]] &, 120] (* Michael De Vlieger, Oct 11 2023 *)

Formula

a(n) = A002182(n-1)/A054481(n).

Extensions

More terms from Jud McCranie, Jun 11 2005

A054483 Numerator of lowest terms fraction from division of a highly composite number (1) by its predecessor.

Original entry on oeis.org

2, 2, 3, 2, 2, 3, 4, 5, 2, 3, 4, 3, 2, 7, 3, 4, 3, 2, 3, 4, 3, 4, 5, 11, 18, 10, 11, 3, 4, 3, 4, 5, 6, 3, 10, 6, 13, 3, 4, 3, 4, 5, 6, 3, 10, 6, 5, 4, 6, 5, 3, 17, 20, 17, 6, 3, 10, 6, 5, 4, 6, 5, 3, 19, 20, 3, 19, 3, 20, 19, 6, 5, 4, 6, 5, 3, 4, 3, 4, 3, 7, 46, 105, 23, 28, 46, 105, 23, 3, 28, 23, 3
Offset: 2

Views

Author

Henry Bottomley, Mar 31 2000

Keywords

Comments

Successive fractions are 2/1,2/1,3/2,2/1,2/1,3/2,4/3,5/4,2/1,3/2,4/3,3/2,2/1 a(n) is frequently A054482(n)+1, though not for example for n=26.

Examples

			a(7)=3 since A002182(7)=36, A002182(6)=24 and 36/24=3/2 in lowest terms.
		

Crossrefs

Programs

  • Mathematica
    s = Import["https://oeis.org/A002182/b002182.txt", "Data"][[All, -1]]; Array[Numerator[s[[# + 1]]/s[[#]]] &, 120] (* Michael De Vlieger, Oct 11 2023 *)

Formula

a(n) = A002182(n)/A054481(n).

Extensions

More terms from Jud McCranie, Jun 11 2005

A262501 First differences of A002182 (highly composite numbers, definition 1).

Original entry on oeis.org

1, 2, 2, 6, 12, 12, 12, 12, 60, 60, 60, 120, 360, 120, 420, 420, 840, 2520, 2520, 2520, 5040, 5040, 5040, 2520, 17640, 5040, 5040, 27720, 27720, 55440, 55440, 55440, 55440, 166320, 55440, 110880, 55440, 360360, 360360, 720720, 720720, 720720, 720720, 2162160, 720720, 1441440, 2162160, 3603600, 2882880, 4324320, 10810800, 4324320, 6486480, 18018000
Offset: 1

Views

Author

Antti Karttunen, Sep 24 2015

Keywords

Crossrefs

Cf. also A054481 (gives the gcds of successive terms of A002182, from which this sequence differs for the first time at n=25).

Programs

Formula

a(n) = A002182(n+1) - A002182(n).

A328520 GCD of terms in A002182 that have n prime factors counted with multiplicity.

Original entry on oeis.org

1, 2, 2, 12, 12, 12, 60, 120, 2520, 2520, 2520, 55440, 55440, 720720, 720720, 12252240, 36756720, 698377680, 3491888400, 80313433200, 160626866400, 160626866400, 9316358251200, 288807105787200, 2021649740510400, 74801040398884800, 74801040398884800, 3066842656354276800, 131874234223233902400
Offset: 0

Views

Author

David A. Corneth, Jan 04 2020

Keywords

Comments

If for every term t > 1 in A002182 there exists a term in A002182 of the form t/p for some prime p|t (Cf. A328523) then (1): a(n+1) is a multiple of a(n) for n >= 1 and (2): this sequence can be used to find terms to A199337 and A330737.
Proof of (1): Let Generation(n) be the terms in A002182 with n prime factors counted with multiplicity. Then a(n) = GCD(Generation(n)). As each term in Generation(n) is of the form a(n) * t for some t and each term in Generation(n + 1) is p * g for some g in Generation(n), a(n + 1) is a multiple of a(n).
Proof of (2): As a(n + 1) is a multiple of a(n) we have that if m | a(n), we also have m | a(n + k), k >= 0 hence the largest number not divisible by m can have at most n - 1 prime factors counted with multiplicity.
Given Generation(n) we can find all candidates for Generation(n + 1) and from there on find terms to A002182, A199337 and A330737.

Examples

			The terms in A002182 with n = 4 prime divisors counted with multiplicity are 24, 36 and 60. Their GCD is 12 hence a(4) = 12.
Furthermore, If for every term t > 1 in A002182 there exists a term in A002182 of the form t/p for some prime p|t then we have that each term with more than 4 prime divisors counted with multiplicity is a multiple of at least one of 24, 36 or 60 hence is divisible by 12.
		

Crossrefs

Programs

  • Mathematica
    (* First, load the function f at A025487, then: *)
    Block[{s = Union@ Flatten@ f@ 20, t}, t = DivisorSigma[0, s]; s = Map[s[[FirstPosition[t, #][[1]] ]] &, Union@ FoldList[Max, t]]; t = PrimeOmega[s]; Drop[Array[GCD @@ s[[Position[t, #][[All, 1]] ]] &, Max@ t + 1, 0], -3] ] (* Michael De Vlieger, Jan 12 2020 *)

A053878 Difference between numerator and denominator of lowest terms fraction from division of a highly composite number (1) by its predecessor.

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, 7, 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, 3, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 13, 2, 5, 11, 13, 2, 1, 5, 2, 1, 1, 1, 1, 3, 1, 1, 1, 3, 1, 1, 1
Offset: 2

Views

Author

Henry Bottomley, Mar 31 2000

Keywords

Comments

Successive fractions are 2/1, 2/1, 3/2, 2/1, 2/1, 3/2, 4/3, 5/4, 2/1, 3/2, 4/3, 3/2, 2/1.
This sequence is not multiplicative.

Examples

			a(7)=1 since A002182(7)=36, A002182(6)=24, 36/24=3/2 in lowest terms and 3-2=1.
		

Crossrefs

Cf. A002182.

Formula

a(n) = A054483(n)-A054482(n) =(A002182(n)-A002182(n-1))/A054481(n).

Extensions

More terms from Jud McCranie, Jun 09 2005
Showing 1-5 of 5 results.