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 1-4 of 4 results.

A139570 a(n) = 2*n*(n+3).

Original entry on oeis.org

0, 8, 20, 36, 56, 80, 108, 140, 176, 216, 260, 308, 360, 416, 476, 540, 608, 680, 756, 836, 920, 1008, 1100, 1196, 1296, 1400, 1508, 1620, 1736, 1856, 1980, 2108, 2240, 2376, 2516, 2660, 2808, 2960, 3116, 3276, 3440, 3608, 3780, 3956, 4136, 4320, 4508, 4700, 4896
Offset: 0

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Author

Omar E. Pol, May 19 2008

Keywords

Comments

Numbers n such that 2*n + 9 is a square. - Vincenzo Librandi, Nov 24 2010
a(n) appears also as the fourth member of the quartet [p0(n), p1(n), p2(n), a(n)] of the square of [n, n+1, n+2, n+3] in the Clifford algebra Cl_2 for n >= 0. p0(n) = -A147973(n+3), p1(n) = A046092(n), and p2(n) = A054000(n+1). See a comment on A147973, also with a reference. - Wolfdieter Lang, Oct 15 2014

Crossrefs

Programs

Formula

a(n) = 2*A028552(n) = 2*n^2 + 6*n = n*(2*n+6).
a(n) = a(n-1) + 4*n + 4 (with a(0)=0). - Vincenzo Librandi, Nov 24 2010
From Paul Curtz, Mar 27 2011: (Start)
a(n) = A022998(n)*A022998(n+3).
a(n) = 4*A000096(n). (End)
G.f.: 4*x*(2 - x)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 31 2011
From Amiram Eldar, Dec 23 2022: (Start)
Sum_{n>=1} 1/a(n) = 11/36.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/3 - 5/36. (End)
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: 2*exp(x)*x*(4 + x).
a(n) = n*A020739(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A009056 Numbers >= 3.

Original entry on oeis.org

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, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77
Offset: 1

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Author

Keywords

Comments

Same as Pisot sequences E(3,4), P(3,4), T(3,4). See A008776 for definitions of Pisot sequences.
Non-Fermat-exponents, positive integers n such that there are no solutions in positive integers of the equation a^n + b^n = c^n. - Tanya Khovanova, Jul 09 2011
Sums of twin primes. - Charles R Greathouse IV, Jun 21 2012

Crossrefs

Programs

Formula

a(n) = n + 2.
From R. J. Mathar, May 26 2008: (Start)
O.g.f.: x*(3-2*x)/(1-x)^2.
a(n) = A009005(n-1), n > 2. (End)
From Elmo R. Oliveira, Oct 31 2024: (Start)
E.g.f.: exp(x)*(x + 2) - 2.
a(n) = 2*a(n-1) - a(n-2) for n > 2.
a(n) = A055998(n) - A055998(n-1) = A020739(n-1)/2. (End)

A020744 Pisot sequences P(8,10), T(8,10).

Original entry on oeis.org

8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 128, 130, 132, 134, 136, 138
Offset: 0

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Author

Keywords

Comments

Conjecturally, even sums of four primes. - Charles R Greathouse IV, Feb 16 2012

Crossrefs

Subsequence of A005843, A020739. See A008776 for definitions of Pisot sequences.

Programs

  • Mathematica
    LinearRecurrence[{2,-1},{8,10},70] (* Harvey P. Dale, Jul 19 2015 *)
    P[x_, y_, z_] := Block[{a}, a[0] = x; a[1] = y; a[n_] := a[n] = Ceiling[a[n - 1]^2/a[n - 2] - 1/2]; Table[a[n], {n, 0, z}]]; P[8, 10, 65] (* or *)
    T[x_, y_, z_] := Block[{a}, a[0] = x; a[1] = y; a[n_] := a[n] = Floor[a[n - 1]^2/a[n - 2]]; Table[a[n], {n, 0, z}]]; T[8, 10, 65] (* Michael De Vlieger, Aug 08 2016 *)
  • PARI
    a(n)=2*n+8 \\ Charles R Greathouse IV, Feb 16 2012
    
  • PARI
    pisotP(nmax, a1, a2) = {
      a=vector(nmax); a[1]=a1; a[2]=a2;
      for(n=3, nmax, a[n] = ceil(a[n-1]^2/a[n-2]-1/2));
      a
    }
    pisotP(50, 8, 10) \\ Colin Barker, Aug 08 2016

Formula

a(n) = 2*n + 8.
a(n) = 2*a(n-1) - a(n-2).
From Elmo R. Oliveira, Oct 30 2024: (Start)
G.f.: 2*(4 - 3*x)/(1 - x)^2.
E.g.f.: 2*(4 + x)*exp(x).
a(n) = 2*A020705(n) = A028563(n+1) - A028563(n). (End)

A316571 a(1) = 1; for n > 1: a(n) = smallest number such that (Sum_{k=1..n} a(k)) is divisible by n - 1.

Original entry on oeis.org

1, 2, 3, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124
Offset: 1

Views

Author

Jaroslav Krizek, Aug 20 2018

Keywords

Comments

This is the lexicographically earliest increasing sequence such that n-1 divides the sum of the first n terms.
Sequence b(n) of the sums of the first n+1 terms, Sum_{k=1..n+1} a(k): 3, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, ...
Sequence c(n) of quotients when a(n) is calculated, (Sum_{k=1..n+1} a(k) ) / n is 3, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, ...

Examples

			a(1) = 1 because 1 divides the sum of the first 2 (i.e., 1 + 1) terms (a(1) + a(2)) for whatever term a(2) > a(1).
a(2) = 2 because 2 is the smallest number > a(1) and 2 divides the sum of the first 3 (i.e., 2 + 1) terms (a(1) + a(2) + a(3)) for whatever term a(3) > a(2) such that 2 divides the sum a(1) + a(2) + a(3); the smallest number > a(2) with this property for a(3) is 3.
a(3) = 3.
a(4) = 6 because 6 is the smallest number > a(3) such that 3 divides the sum of the first 4 (i.e., 3 + 1) terms.
a(5) = 8 because 8 is the smallest number > a(4) such that 4 divides the sum of the first 5 (i.e., 4 + 1) terms.
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 + 2 x^3 - x^4)/(-1 + x)^2 , {x, 0, 50}], x] (* Stefano Spezia, Sep 16 2018 *)
    Join[{1, 2, 3}, LinearRecurrence[{2, -1}, {6, 8}, 60]] (* Jean-François Alcover, Dec 30 2018 *)

Formula

a(1)=1, a(2)=2, a(3)=3, a(n)=2(n-1) for n >= 4.
a(n) = A005843(n-1) for n >= 4. - Antti Karttunen, Sep 16 2018
G.f.: (1 + 2*x^3 - x^4)/(-1 + x)^2. - Stefano Spezia, Sep 16 2018
a(n) = A020739(n-4) = 2*(n - 4) + 6 for n >= 4. - Georg Fischer, Jan 19 2019

Extensions

Definition clarified by Georg Fischer and Alois P. Heinz, Jan 19 2019
Showing 1-4 of 4 results.