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 1-3 of 3 results.

A168317 Eigensequence of triangle A168316.

Original entry on oeis.org

1, 1, 3, 6, 16, 39, 103, 263, 690, 1791, 4693, 12247, 32073, 83869, 219598, 574658, 1504540, 3938272, 10310703
Offset: 1

Views

Author

Gary W. Adamson, Nov 22 2009

Keywords

Comments

Conjectured convergent of a(n)/a(n-1) = phi^2 = 2.6180339... E.g.: a(19)/a(18) = 10310703/3938272 = 2.6180779....

Crossrefs

Formula

Equals lim_{n->oo} M^2, the left shifted vector considered as a sequence where M = triangle A168316 as an infinite lower triangular matrix, then shifted down one row, and inserting a "1" at top.

A168318 Triangle read by rows, A168316 * its diagonalized eigensequence, A168317.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 1, 2, 3, 0, 1, 3, 6, 6, 0, 0, 2, 9, 12, 16, 0, 0, 2, 12, 18, 32, 39, 0, 0, 1, 9, 24, 48, 78, 103, 0, 0, 1, 9, 30, 64, 117, 206, 263, 0, 0, 0, 6, 24, 80, 156, 309, 526, 690, 0, 0, 0, 6, 24, 96, 195, 412, 789, 1380, 1791
Offset: 1

Views

Author

Gary W. Adamson, Nov 22 2009

Keywords

Comments

Row sums = A168317: (1, 1, 3, 6, 16, 39, 103, 263,...).
Rightmost diagonal = A168317 prefaced with a 1.
Sum of n-th row terms = rightmost term of next row.

Examples

			First few rows of the triangle =
1;
0, 1;
0, 2, 1;
0, 1, 2, 3;
0, 1, 3, 6, 6;
0, 0, 2, 9, 12, 16;
0, 0, 2, 12, 18, 32, 39;
0, 0, 1, 9, 24, 48, 78, 103;
0, 0, 1, 9, 30, 64, 117, 206, 263;
0, 0, 0, 6, 24, 80, 156, 309, 526, 690;
0, 0, 0, 6, 24, 96, 195, 412, 789, 1380, 1791;
0, 0, 0, 3, 18, 80, 234, 515, 1052, 2070, 3582, 4693;
0, 0, 0, 3, 18, 80, 273, 618, 1315, 2760, 5373, 9386, 12247;
0, 0, 0, 0, 12, 64, 234, 721, 1578, 3450, 7164, 14079, 24494, 32073;
...
		

Crossrefs

Formula

Triangle read by rows, M*Q. M = A168316, Q = an infinite lower triangular matrix
with A168317 prefaced with a 1; (1, 1, 1, 3, 6, 16, 39, 103,...) as the right
diagonal and the rest zeros.

A129819 Antidiagonal sums of triangular array T: T(j,k) = (k+1)/2 for odd k, T(j,k) = 0 for k = 0, T(j,k) = j+1-k/2 for even k > 0; 0 <= k <= j.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 7, 8, 12, 14, 19, 21, 27, 30, 37, 40, 48, 52, 61, 65, 75, 80, 91, 96, 108, 114, 127, 133, 147, 154, 169, 176, 192, 200, 217, 225, 243, 252, 271, 280, 300, 310, 331, 341, 363, 374, 397, 408, 432, 444, 469, 481, 507, 520, 547, 560, 588, 602, 631
Offset: 0

Views

Author

Paul Curtz, May 20 2007

Keywords

Comments

Interleaving of A077043 and A006578.
First differences are in A124072.
If the values of the second, fourth, sixth, ... column are replaced by the corresponding negative values, the antidiagonal sums of the resulting triangular array are 0, 0, 1, 1, -1, -2, -1, -2, -6, -8, -7, -9, ... .
Row sums of triangle A168316 = (1, 1, 3, 4, 7, 8, 12, ...). - Gary W. Adamson, Nov 22 2009

Examples

			First seven rows of T are
  0;
  0, 1;
  0, 1, 2;
  0, 1, 3, 2;
  0, 1, 4, 2, 3;
  0, 1, 5, 2, 4, 3;
  0, 1, 6, 2, 5, 3, 4;.
		

Crossrefs

Programs

  • Magma
    m:=59; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do for k:=2 to j do if k mod 2 eq 0 then M[j, k]:= k div 2; else M[j, k]:=j-(k div 2); end if; end for; end for; [ &+[ M[j-k+1, k]: k in [1..(j+1) div 2] ]: j in [1..m] ]; // Klaus Brockhaus, Jul 16 2007
    
  • Magma
    A129819:= func< n | Floor(((n-1)*(3*n+1) +(2*n+5)*((n+1) mod 2))/16) >;
    [A129819(n): n in [0..70]]; // G. C. Greubel, Sep 19 2024
    
  • Mathematica
    CoefficientList[Series[x^2*(1+x^2+x^3)/((1-x)*(1-x^2)*(1-x^4)), {x, 0, 70}], x] (* G. C. Greubel, Sep 19 2024 *)
  • PARI
    {vector(59, n, (n-2+n%2)*(n+n%2)/8+floor((n-2-n%2)^2/16))} \\ Klaus Brockhaus, Jul 16 2007
    
  • SageMath
    def A129819(n): return ((n-1)*(3*n+1) + (2*n+5)*((n+1)%2))//16
    [A129819(n) for n in range(71)] # G. C. Greubel, Sep 19 2024

Formula

a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-4) - a(n-5) - a(n-6) + a(n-7) for n > 6, with a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 1, a(4) = 3, a(5) = 4, a(6) = 7.
G.f.: x^2*(1+x^2+x^3)/((1-x)^3*(1+x)^2*(1+x^2)).
a(n) = (3/16)*(n+2)*(n+1) - (5/8)*(n+1) + 7/32 + (3/32)*(-1)^n + (1/16)*(n+1)*(-1)^n - (1/8)*cos(n*Pi/2) + (1/8)*sin(n*Pi/2). - Richard Choulet, Nov 27 2008

Extensions

Edited and extended by Klaus Brockhaus, Jul 16 2007
Showing 1-3 of 3 results.