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.

A097063 Expansion of (1-2*x+3*x^2)/((1+x)*(1-x)^3).

Original entry on oeis.org

1, 0, 3, 4, 9, 12, 19, 24, 33, 40, 51, 60, 73, 84, 99, 112, 129, 144, 163, 180, 201, 220, 243, 264, 289, 312, 339, 364, 393, 420, 451, 480, 513, 544, 579, 612, 649, 684, 723, 760, 801, 840, 883, 924, 969, 1012, 1059, 1104, 1153, 1200, 1251, 1300, 1353, 1404
Offset: 0

Views

Author

Paul Barry, Jul 22 2004

Keywords

Comments

Partial sums of A097062. Pairwise sums are A002061. Binomial transform is essentially A007466.

Links

Crossrefs

A diagonal of A326296.

Programs

Formula

G.f. : (1-2*x+3*x^2)/((1-x^2)(1-x)^2).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
a(n) = Sum_{k=0..n} (k^2-k+1)*(-1)^(n-k).
a(2n) = A058331(n); a(2n+1) = A046092(n). - R. J. Mathar, Oct 27 2008
a(n) = binomial(n+1, 2) - ceiling((n+1)/2) + 2((n+1) mod 2). - Wesley Ivan Hurt, Mar 08 2014
a(n) = 2*floor(n/2) + ceiling((n-1)^2/2). - M. Ryan Julian Jr., Sep 10 2019
a(n) = A326296(n + 1, n) for n > 0. - Andrew Howroyd, Sep 23 2019

A097064 Expansion of (1 - 4*x + 6*x^2)/(1 - 2*x)^2.

Original entry on oeis.org

1, 0, 2, 8, 24, 64, 160, 384, 896, 2048, 4608, 10240, 22528, 49152, 106496, 229376, 491520, 1048576, 2228224, 4718592, 9961472, 20971520, 44040192, 92274688, 192937984, 402653184, 838860800, 1744830464, 3623878656, 7516192768, 15569256448, 32212254720, 66571993088
Offset: 0

Views

Author

Paul Barry, Jul 22 2004

Keywords

Comments

Binomial transform of A097062.

Links

Crossrefs

Essentially the same as A036289.
Cf. A001787, A002162, A016578, A097062.

Programs

  • Mathematica
    CoefficientList[Series[(1-4x+6x^2)/(1-2x)^2,{x,0,30}],x] (* or *) Join[{1},LinearRecurrence[{4,-4},{0,2},30]] (* Harvey P. Dale, May 26 2011 *)

Formula

a(n) = (n-1)*2^(n-1) + 3*0^n/2.
a(n) = 4*a(n-1) - 4*a(n-2), n>2.
a(n) = Sum_{k=0..n} binomial(n, k)*((2k-1)/2 + 3*(-1)^k/2).
a(n+1)/2 = A001787(n).
From Amiram Eldar, Oct 01 2022: (Start)
Sum_{n>=2} 1/a(n) = log(2) (A002162).
Sum_{n>=2} (-1)^n/a(n) = log(3/2) (A016578). (End)
E.g.f.: (3 - exp(2*x)*(1 - 2*x))/2. - Stefano Spezia, Feb 12 2023

A142954 a(n) = 2*n + 3 + 3*(-1)^n.

Original entry on oeis.org

2, 10, 6, 14, 10, 18, 14, 22, 18, 26, 22, 30, 26, 34, 30, 38, 34, 42, 38, 46, 42, 50, 46, 54, 50, 58, 54, 62, 58, 66, 62, 70, 66, 74, 70, 78, 74, 82, 78, 86, 82, 90, 86, 94, 90, 98, 94, 102, 98, 106, 102, 110, 106, 114, 110
Offset: 1

Views

Author

Paul Curtz, Sep 29 2008

Keywords

Comments

First differences of A142717.

Links

Crossrefs

First differences of A214345.

Programs

  • GAP
    List([1..55],n->2*n+3+3*(-1)^n); # Muniru A Asiru, Nov 01 2018
  • Magma
    I:=[2,10,6]; [n le 3 select I[n] else Self(n-1)+Self(n-2)-Self(n-3): n in [1..60]]; // Vincenzo Librandi, Apr 03 2013
  • Maple
    seq(2*n+3+3*(-1)^n,n=1..55); # Muniru A Asiru, Nov 01 2018
  • Mathematica
    Table[2 n + 3 + 3 (-1)^n, {n, 1, 60}] (* Vincenzo Librandi, Apr 03 2013 *)
LinearRecurrence[{1,1,-1},{2,10,6},60] (* Harvey P. Dale, Aug 20 2015 *)
  • Python
    for n in range(1,50): print(2*n+3+3*(-1)**n, end=', ') # Stefano Spezia, Nov 01 2018

Formula

a(2n+1) = A016825(n). a(2n) = A016825(n+1).
From R. J. Mathar, Oct 24 2008: (Start)
G.f.: 2*x*(1+4*x-3*x^2)/((1+x)*(1-x)^2). [corrected by Jason Yuen, Oct 01 2024]
a(n) = a(n-1)+a(n-2)-a(n-3) = 2*A097062(n+2). (End)

Extensions

Edited by R. J. Mathar, Oct 24 2008

A215415 a(2*n) = n, a(4*n+1) = 2*n-1, a(4*n+3) = 2*n+3.

Original entry on oeis.org

0, -1, 1, 3, 2, 1, 3, 5, 4, 3, 5, 7, 6, 5, 7, 9, 8, 7, 9, 11, 10, 9, 11, 13, 12, 11, 13, 15, 14, 13, 15, 17, 16, 15, 17, 19, 18, 17, 19, 21, 20, 19, 21, 23, 22, 21, 23, 25, 24, 23, 25, 27, 26, 25, 27, 29, 28, 27, 29, 31, 30, 29, 31, 33, 32, 31, 33, 35, 34, 33, 35, 37
Offset: 0

Views

Author

Paul Curtz, Aug 09 2012

Keywords

Comments

a(n) and higher order differences in further rows:
0, -1, 1, 3, 2, 1,
-1, 2, 2, -1, -1, -2, A134430(n).
3, 0, -3, 0, 3, 0,
-3, -3, 3, 3, -3, -3,
0, 6, 0, -6, 0, 6,
6, -6, -6, 6, 6, -6.
a(n) is the binomial transform of 0, -1, 3, -3, 0, 6, -12, 12, 0, -24, 48, -48, 0, 96..., essentially negated A134813.
By definition, all differences a(n+k)-a(n) are periodic sequences with period length 4. For k=1, 3 and 4 these are A134430, A021307 and A007395, for example.

Links

Crossrefs

Quadrisections: A005843, A060747, A005408, A144396.

Programs

  • Mathematica
    Flatten[Table[{2n, 2n - 1, 2n + 1, 2n + 3}, {n, 0, 19}]] (* Alonso del Arte, Aug 09 2012 *)
  • PARI
    a(n) = ((-3*I)*((-I)^n-I^n)+2*n)/4 \\ Colin Barker, Oct 19 2015
  • PARI
    concat(0, Vec(-x*(1-3*x+x^2)/((x^2+1)*(x-1)^2) + O(x^100))) \\ Colin Barker, Oct 19 2015

Formula

a(2*n) = n, a(2*n+1) = A097062(n+1).
a(n) = (A214297(n+1) - A214297(n-1))/2.
a(3*n) =3*A004525(n).
a(n) = +2*a(n-1) -2*a(n-2) +2*a(n-3) -a(n-4).
G.f. -x*(1-3*x+x^2) / ( (x^2+1)*(x-1)^2 ). - R. J. Mathar, Aug 11 2012
a(n) = ((-3*I)*((-I)^n-I^n)+2*n)/4. - Colin Barker, Oct 19 2015

A237274 a(n) = A236283(n) mod 9.

Original entry on oeis.org

2, 1, 4, 5, 1, 4, 2, 7, 7, 5, 7, 7, 2, 4, 1, 5, 4, 1, 2, 1, 4, 5, 1, 4, 2, 7, 7, 5, 7, 7, 2, 4, 1, 5, 4, 1, 2, 1, 4, 5, 1, 4, 2, 7, 7, 5, 7, 7, 2, 4, 1, 5, 4, 1, 2, 1, 4, 5, 1, 4, 2, 7, 7, 5, 7, 7, 2, 4, 1, 5, 4, 1
Offset: 0

Views

Author

Paul Curtz, Feb 05 2014

Keywords

Comments

(Conjecture) This has period 18: repeat 2, 1, 4, 5, 1, 4, 2, 7, 7, 5, 7, 7, 2, 4, 1, 5, 4, 1.
The first 19 terms and the following 17 are palindromes.
The sorted terms in the conjectured period are 1, 1, 1, 1, 2, 2, 2, 4, 4, 4, 4, 5, 5, 5, 7, 7, 7, 7.
Via the extended differences of A236283(n+1) and A236283(n+18) - A236283(n) which is A008600(n+9)=162, 180,... ,it is easy to see that A236283(0)=2.
A236283(-n) = A236283(n).
A236283(n) difference table:
2, 1, 4, 5, 10, 13, 20, 25, 34, 41,...
-1, 3, 1, 5, 3, 7, 5, 9, 7, 11,... = A097062(n+1)
4, -2, 4, -2, 4, -2, 4, -2, 4, -2,...
-6, 6, -6, 6, -6, 6, -6, 6, -6, 6,... .
A097062(n+1) mod 9 = (a(n+1) -a(n)) mod 9 =
period 18: repeat 8, 3, 1, 5, 3, 7, 5, 0, 7, 2, 0, 4, 2, 6, 4, 8, 6, 1 =b(n). b(n) + b(18-n)= 9, 9, 9, 9, 9, 9, 9, 0, 9.
Ordered b(n)=
period 18: repeat 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8.

Formula

a(n) = A236283(n) mod 9.
Showing 1-5 of 5 results.

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