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.

A265278 Expansion of (x^4+x^3-x^2+x)/(x^3+x^2-3*x+1).

Original entry on oeis.org

0, 1, 2, 6, 16, 40, 98, 238, 576, 1392, 3362, 8118, 19600, 47320, 114242, 275806, 665856, 1607520, 3880898, 9369318, 22619536, 54608392, 131836322, 318281038, 768398400, 1855077840, 4478554082, 10812186006, 26102926096, 63018038200, 152139002498, 367296043198
Offset: 0

Views

Author

Alois P. Heinz, Apr 06 2016

Keywords

Crossrefs

Agrees with A213667 except for initial terms.

Programs

  • Mathematica
    Table[2 Fibonacci[n-1, 2] + LucasL[n-1, 2]/2 + KroneckerDelta[n-1] - 1, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
    LinearRecurrence[{3,-1,-1},{0,1,2,6,16},40] (* Harvey P. Dale, Mar 18 2018 *)
  • PARI
    concat(0, Vec(x*(1-x+x^2+x^3)/((1-x)*(1-2*x-x^2)) + O(x^50))) \\ Colin Barker, Apr 12 2016

Formula

From Colin Barker, Apr 12 2016: (Start)
a(n) = (-2 + (1-sqrt(2))^n + (1+sqrt(2))^n)/2 for n>1.
a(n) = 3*a(n-1)-a(n-2)-a(n-3) for n>4.
(End)
E.g.f.: x + (cosh(sqrt(2)*x) - 1)*exp(x). - Ilya Gutkovskiy, Sep 16 2016

A065113 Sum of the squares of the a(n)-th and the (a(n)+1)st triangular numbers (A000217) is a perfect square.

Original entry on oeis.org

6, 40, 238, 1392, 8118, 47320, 275806, 1607520, 9369318, 54608392, 318281038, 1855077840, 10812186006, 63018038200, 367296043198, 2140758220992, 12477253282758, 72722761475560, 423859315570606, 2470433131948080, 14398739476117878, 83922003724759192
Offset: 1

Views

Author

Robert G. Wilson v, Nov 12 2001

Keywords

Comments

The sequence of square roots of the sum of the squares of the n-th and the (n+1)st triangular numbers is A046176.

Examples

			T6 = 21 and T7 = 28, 21^2 + 28^2 = 441 + 784 = 1225 = 35^2.
		

Crossrefs

Cf. A001652, A002315, A003499 (first differences), A065651.

Programs

  • Mathematica
    CoefficientList[ Series[2*(x - 3)/(-1 + 7x - 7x^2 + x^3), {x, 0, 24} ], x]
    LinearRecurrence[{7,-7,1},{6,40,238},41] (* Harvey P. Dale, Dec 27 2011 *)
  • PARI
    a(n)=-1+subst(poltchebi(abs(n+1))-poltchebi(abs(n)),x,3)/2
    
  • PARI
    Vec(2*x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^40)) \\ Colin Barker, Mar 05 2016

Formula

a(n) = 2*A001652(n) = -1 + A002315(n).
a(n) - a(n-1) = A003499(n).
From Michael Somos, Apr 07 2003: (Start)
G.f.: 2*x*(3-x)/((1-6*x+x^2)*(1-x)).
a(n) = 6*a(n-1) - a(n-2) + 4.
a(-1-n) = -a(n) - 2. (End)
a(1)=6, a(2)=40, a(3)=238, a(n) = 7*a(n-1)-7*a(n-2)+a(n-3). - Harvey P. Dale, Dec 27 2011
a(n)^2 + (a(n)+2)^2 = A075870(n+1)^2 = A165518(n+1). - Joerg Arndt, Feb 15 2012
a(n) = (-2-(3-2*sqrt(2))^n*(-1+sqrt(2))+(1+sqrt(2))*(3+2*sqrt(2))^n)/2. - Colin Barker, Mar 05 2016
From Klaus Purath, Sep 05 2021: (Start)
(a(n+1) - a(n) - a(n-1) + a(n-2))/8 = A005319(n), for n >= 3.
((a(n) - a(n-1))^2)/2 - 2 = A005319(n)^2 = 2*A132592(n), for n>= 2.
a(n) = A265278(2*n+1).
a(n) = A293004(2*n+1).
a(n) = A213667(2*n).
a(n) = Sum_{k=1..n} A003499(k). (End)

A231690 Cardinalities of the sub-operad FF_6 of the operad MFF.

Original entry on oeis.org

1, 6, 56, 640, 8158, 111258, 1588544, 23446248, 354855218, 5477342222, 85893429256, 1364577254040, 21916000458014, 355251287893170, 5804407209709312, 95493879511032240, 1580592247322440642, 26301843121772151254, 439764358275666481496, 7384252698468635017936, 124469446338979722639294
Offset: 1

Views

Author

N. J. A. Sloane, Nov 14 2013

Keywords

Crossrefs

Programs

  • Mathematica
    InverseSeries[x(1 - 3x - x^2 + x^3)/(1 + 3x + x^2 - x^3) + O[x]^22] // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Oct 24 2018, from PARI *)
  • PARI
    N=22; x='x+O('x^N);
    Vec(serreverse(Ser(x*(1-3*x-x^2+x^3)/(1+3*x+x^2-x^3))))  \\ Gheorghe Coserea, Jan 13 2017

Formula

A(x) = serreverse(A213667(-x))(-x). - Gheorghe Coserea, Jan 13 2017

Extensions

Offset changed and more terms from Gheorghe Coserea, Jan 13 2017

A213666 Irregular triangle read by rows: T(n,k) is the number of dominating subsets with k vertices of the graph G(n) obtained by taking n copies of the path P_3 and identifying one of their endpoints (a star with n branches of length 2).

Original entry on oeis.org

1, 3, 1, 0, 3, 8, 5, 1, 0, 0, 7, 20, 18, 7, 1, 0, 0, 0, 15, 48, 56, 32, 9, 1, 0, 0, 0, 0, 31, 112, 160, 120, 50, 11, 1, 0, 0, 0, 0, 0, 63, 256, 432, 400, 220, 72, 13, 1, 0, 0, 0, 0, 0, 0, 127, 576, 1120, 1232, 840, 364, 98, 15, 1
Offset: 1

Views

Author

Emeric Deutsch, Jul 01 2012

Keywords

Comments

Rows also give the coefficients of the domination polynomial of the n-helm graph (divided by x, i.e., with initial 0 dropped from rows). - Eric W. Weisstein, May 28 2017
Row n contains 2n + 1 entries (first n-1 of which are 0).
Sum of entries in row n = 2*3^{n-1} - 1 = A048473(n).
Sum of entries in column k = A213667(k).

Examples

			Row 2 is 0,3,8,5,1 because G(2) is the path P_5 abcde; no domination subset of size 1, three of size 2 (ad, bd, be), all subsets of size 3 with the exception of abc and cde are dominating (binomial(5,3)-2=8), all binomial(5,4)=5 subsets of size 4 are dominating, and abcde is dominating.
Triangle starts:
  1, 3, 1;
  0, 3, 8,  5,  1;
  0, 0, 7, 20, 18,  7,  1;
  0, 0, 0, 15, 48, 56, 32, 9, 1;
		

Crossrefs

Programs

  • Maple
    T := proc (n, k) if k = n then 2^n-1 else 2^(2*n-k)*(2*binomial(n, k-n-1) + binomial(n, k-n)) end if end proc: for n to 10 do seq(T(n, k), k = 1 .. 2*n+1) end d; # yields sequence in triangular form
  • Mathematica
    T[n_, n_] := 2^n - 1;
    T[n_, k_] := 2^(2*n - k)*(2*Binomial[n, k - n - 1] + Binomial[n, k - n]);
    Table[T[n, k], {n, 1, 10}, {k, 1, 2*n + 1}] // Flatten (* Jean-François Alcover, Dec 02 2017 *)

Formula

T(n,k) = 2^(2*n-k)*(2*binomial(n,k-n-1)+binomial(n,k-n)) if k > n; T(n,n)=2^n - 1.
The generating polynomial of row n is g[n] = g[n,x] = (1+x)(x*(2+x))^n - x^n (= domination polynomial of the graph G(n)).
Bivariate g.f.: G(x,z) = x*z*(1+x)*(2+x)/(1-2*x*z-x^2*z)-x*z/(1-xz).

A281261 Triangle T(n,k) read by rows: coefficients of polynomials P_n(t) defined in Formula section.

Original entry on oeis.org

1, 2, 2, 1, 5, 2, 5, 9, 2, 1, 15, 14, 2, 7, 35, 20, 2, 1, 28, 70, 27, 2, 9, 84, 126, 35, 2, 1, 45, 210, 210, 44, 2, 11, 165, 462, 330, 54, 2, 1, 66, 495, 924, 495, 65, 2, 13, 286, 1287, 1716, 715, 77, 2, 1, 91, 1001, 3003, 3003, 1001, 90, 2, 15, 455, 3003, 6435, 5005, 1365, 104, 2, 1, 120, 1820, 8008, 12870, 8008, 1820, 119, 2
Offset: 1

Views

Author

Gheorghe Coserea, Jan 18 2017

Keywords

Comments

Row n>1 contains floor((n+3)/2) terms.

Examples

			A(x;t) = x + (2*t+2)*x^2 + (t^2+5*t+2)*x^3 + (5*t^2+9*t+2)*x^4 + ...
Triangle starts:
n\k  [1]      [2]      [3]      [4]      [5]      [6]      [7]      [8]
[1]  1;
[2]  2,       2;
[3]  1,       5,       2;
[4]  5,       9,       2;
[5]  1,       15,      14,      2;
[6]  7,       35,      20,      2;
[7]  1,       28,      70,      27,      2;
[8]  9,       84,      126,     35,      2;
[9]  1,       45,      210,     210,     44,      2;
[10] 11,      165,     462,     330,     54,      2;
[11] 1,       66,      495,     924,     495,     65,      2;
[12] 13,      286,     1287,    1716,    715,     77,      2;
[13] 1,       91,      1001,    3003,    3003,    1001,    90,      2;
[14] 15,      455,     3003,    6435,    5005,    1365,    104,     2;
[15] ...
		

Programs

  • Mathematica
    Reverse[CoefficientList[#, t]]& /@ CoefficientList[x*((1-t)*x^3 + (t^2 - 2*t - 1)*x^2 + (2*t - 1)*x + 1)/((t-1)*x^3 + (3-t)*x^2 - 3*x + 1) + O[x]^16, x] // Rest // Flatten (* Jean-François Alcover, Feb 18 2019 *)
  • PARI
    N=16; x='x+O('x^N); concat(apply(p->Vec(p),  Vec(Ser(x*((1-t)*x^3 + (t^2-2*t-1)*x^2 + (2*t-1)*x + 1)/((t-1)*x^3 + (3-t)*x^2 - 3*x + 1)))))
    
  • PARI
    N = 14; concat(1, concat(vector(N, n, Vec(substpol(((1+t)^(n+2) + (1-t)^(n+2))/2 - t^2 + 1, t^2, t)))))

Formula

A(x;t) = Sum{n>=1} P_n(t)*x^n = x*((1-t)*x^3 + (t^2-2*t-1)*x^2 + (2*t-1)*x + 1)/((t-1)*x^3 + (3-t)*x^2 - 3*x + 1).
A278457(x;t) = serreverse(A(-x;t))(-x).
A151821(n) = P_n(1), A213667(n) = P_n(2).
P_n(t^2) = ((1+t)^(n+1) + (1-t)^(n+1))/2 - t^2 + 1, for n>1.
Showing 1-5 of 5 results.