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.

A214964 Least m > 0 such that for every r and s in the set S = {{h*(1+sqrt(5))/2} : h = 1,..,n} of fractional parts, if r < s, then r < k/m < s for some integer k; m is the least separator of S as defined at A001000.

Original entry on oeis.org

2, 3, 4, 5, 6, 8, 8, 10, 10, 13, 13, 13, 16, 16, 16, 21, 21, 21, 21, 21, 28, 30, 30, 30, 34, 34, 34, 34, 34, 34, 34, 34, 34, 43, 45, 50, 50, 50, 50, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 72, 73, 73, 73, 81, 81, 81, 81, 81, 81, 89, 89, 89, 89
Offset: 2

Views

Author

Clark Kimberling, Aug 12 2012

Keywords

Comments

a(n) is the least separator of S, as defined at A001000, which includes a guide to related sequences. - Clark Kimberling, Aug 12 2012

Links

Crossrefs

Cf. A001000, A214961, A214965.

Programs

  • Mathematica
    leastSeparatorShort[seq_, s_] := Module[{n = 1},
While[Or @@ (n #1[[1]] <= s + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[seq, 2, 1], n++]; n];
Table[leastSeparatorShort[Sort[N[FractionalPart[GoldenRatio*Range[n]], 50]], 1], {n, 2, 100}]
(* Peter J. C. Moses, Aug 01 2012 *)

A214965 Least m > 0 such that for every r and s in the set S = {{h*e} : h = 1,..,n} of fractional parts, if r < s, then r < k/m < s for some integer k; m is the least separator of S as defined at A001000.

Original entry on oeis.org

2, 3, 4, 6, 6, 7, 11, 11, 11, 18, 18, 18, 18, 25, 25, 25, 25, 25, 25, 25, 25, 25, 32, 32, 32, 32, 32, 32, 32, 32, 32, 35, 35, 35, 39, 39, 39, 39, 55, 61, 61, 66, 68, 69, 69, 69, 70, 70, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71, 71
Offset: 2

Views

Author

Clark Kimberling, Aug 12 2012

Keywords

Comments

a(n) is the least separator of S, as defined at A001000, which includes a guide to related sequences. - Clark Kimberling, Aug 12 2012

Examples

			Write the sorted fractional parts {h*e}, for h=1..5, as f1,f2,f3,f4,f5.  Then f1 < 2/6 < f2 < 3/6 < f3 < 4/6 < f5 < 5/6 < f6, and no such separation occurs using fractions k/m having m < 6; so a(5) = 6.

Links

Crossrefs

Cf. A001000, A214961, A214964.

Programs

  • Mathematica
    leastSeparatorShort[seq_, s_] := Module[{n = 1},
While[Or @@ (n #1[[1]] <= s + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[seq, 2, 1], n++]; n];
Table[leastSeparatorShort[Sort[N[FractionalPart[E*Range[n]], 50]], 1], {n, 2, 100}]
(* Peter J. C. Moses, Aug 01 2012 *)

A213005 a(0)=1, a(n) = least k > a(n-1) such that k*a(n-1) is a triangular number.

Original entry on oeis.org

1, 3, 5, 9, 17, 33, 45, 72, 143, 152, 303, 420, 451, 603, 952, 1398, 1572, 2408, 3762, 4233, 5880, 6325, 8469, 13384, 20079, 34189, 62769, 82665, 87448, 161037, 287283, 371337, 515745, 533505, 573815, 734484, 737035, 737149, 767505, 825495, 887865, 1136468, 2272935
Offset: 0

Views

Author

Alex Ratushnyak, Aug 03 2012

Keywords

Comments

Corresponding triangular numbers t(n)=a(n)*a(n+1): 3, 15, 45, 153, 561, 1485, 3240, 10296, 21736, 46056, 127260, 189420, 271953, 574056, 1330896, 2197656, 3785376, 9058896, 15924546, 24890040, 37191000, ...

Crossrefs

Cf. A000217, A214961.
Cf. A081976 (a(0)=1, a(n) = least k > a(n-1) such that k*a(n-1) is a Fibonacci number).
Cf. A006882 (a(0)=a(1)=1, a(n) = least k > a(n-1) such that k*a(n-1) is a factorial).
Cf. A079078 (a(0)=1, a(n) = least k > a(n-1) such that k*a(n-1) is a primorial).

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = For[k = a[n-1]+1, True, k++, If[ IntegerQ[ Sqrt[8k*a[n-1]+1] ], Return[k] ] ]; Table[ Print[a[n]]; a[n], {n, 0, 42}] (* Jean-François Alcover, Sep 14 2012 *)
  • Python
    a = 1
for n in range(55):
    print(a, end=',')
    b = k = 0
    while k<=a:
        tn = b*(b+1)//2
        k = 0
        if tn%a==0:
            k = tn // a
        b += 1
    a = k

A214916 a(0) = a(1) = 1, a(n) = n! / a(n-2).

Original entry on oeis.org

1, 1, 2, 6, 12, 20, 60, 252, 672, 1440, 5400, 27720, 88704, 224640, 982800, 5821200, 21288960, 61102080, 300736800, 1990850400, 8089804800, 25662873600, 138940401600, 1007370302400, 4465572249600, 15397724160000, 90311261040000, 707173952284800, 3375972620697600
Offset: 0

Views

Author

Alex Ratushnyak, Aug 03 2012

Keywords

Comments

a(n) is least k > a(n-1) such that k*a(n-2) is a factorial.
Two periodic subsets of these numbers appear in the coefficients of a series involved in a solution of a Riccati-type differential equation addressed by the Bernoulli brothers: z = 1 - x^4/12 + x^8/672 - x^12/88704 + ... = 1 - 2 * x^4/4! + 60 * x^8/8! - 5400 * x^12/12! + ... . See the MathOverflow question. - Tom Copeland, Jan 24 2017

Links

Crossrefs

Cf. A214914, A214915, A214961, A214963, A007662.

Programs

  • Magma
    [1] cat  [n le 2 select n  else  Factorial(n)  div  Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 25 2017
  • Python
    import math
prpr = prev = 1
for n in range(2, 33):
    print(prpr, end=', ')
    cur = math.factorial(n) // prpr
    prpr = prev
    prev = cur

Formula

a(0) = a(1) = 1, for n>=2, a(n) = n! / a(n-2).
a(n) = n!!!! * (n-1)!!!! = A007662(n) * A007662(n-1), for n >= 1. - David Radcliffe, Jun 05 2025

A214915 a(0)=a(1)=1, a(n) = least k > a(n-1) such that k*a(n-2) is a Lucas number.

Original entry on oeis.org

1, 1, 2, 3, 9, 41, 642, 45378, 300467529, 141919565128041, 23889121268674415030529282, 1882004232433803432495089121844337244861378, 15480785260333522531462377514739365882017797041068613651703152050407493402249
Offset: 0

Views

Author

Alex Ratushnyak, Aug 03 2012

Keywords

Crossrefs

Cf. A214914, A214916, A214961, A214963.
Showing 1-5 of 5 results.

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