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.

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A152546 Sum of squared terms in rows of the Padovan-Fibonacci triangle A152545.

Original entry on oeis.org

1, 1, 2, 5, 9, 18, 39, 77, 160, 331, 675, 1394, 2871, 5899, 12168, 25047, 51585, 106316, 219047, 451361, 930512, 1917713, 3953407, 8150624, 16804773, 34649369, 71453844, 147346015, 303878941, 626728864, 1292649339, 2666247721
Offset: 0

Views

Author

Paul D. Hanna, Dec 13 2008

Keywords

Crossrefs

Cf. A152545.

A152547 Triangle, read by rows, derived from Pascal's triangle (see g.f. and example for generating methods).

Original entry on oeis.org

1, 2, 3, 1, 4, 2, 2, 5, 3, 3, 3, 1, 1, 6, 4, 4, 4, 4, 2, 2, 2, 2, 2, 7, 5, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 8, 6, 6, 6, 6, 6, 6, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 9, 7, 7, 7, 7, 7, 7, 7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
Offset: 0

Views

Author

Paul D. Hanna, Dec 14 2008

Keywords

Examples

			The number of terms in row n is C(n,[n/2]).
Triangle begins:
[1],
[2],
[3,1],
[4,2,2],
[5,3,3,3,1,1],
[6,4,4,4,4,2,2,2,2,2],
[7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1],
[8,6,6,6,6,6,6,4,4,4,4,4,4,4,4,4,4,4,4,4,4,2,2,2,2,2,2,2,2,2,2,2,2,2,2],
[9,7,7,7,7,7,7,7,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1],
...
ILLUSTRATION OF GENERATING METHOD.
Row n is derived from the binomial coefficients in the following way.
Place markers in an array so that the number of contiguous markers
in row k is C(n,k) and then count the markers along columns.
For example, row 6 of this triangle is generated from C(6,k) like so:
------------------------------------------
1: o - - - - - - - - - - - - - - - - - - -
6: o o o o o o - - - - - - - - - - - - - -
15:o o o o o o o o o o o o o o o - - - - -
20:o o o o o o o o o o o o o o o o o o o o
15:o o o o o o o o o o o o o o o - - - - -
6: o o o o o o - - - - - - - - - - - - - -
1: o - - - - - - - - - - - - - - - - - - -
------------------------------------------
Counting the markers along the columns gives row 6 of this triangle:
[7,5,5,5,5,5,3,3,3,3,3,3,3,3,3,1,1,1,1,1].
Continuing in this way generates all the rows of this triangle.
...
Number of repeated terms in each row of this triangle forms A008315:
1;
1;
1, 1;
1, 2;
1, 3, 2;
1, 4, 5;
1, 5, 9, 5;
1, 6, 14, 14;
1, 7, 20, 28, 14;...
		

Crossrefs

Cf. A152548 (row squared sums), A008315; A152545.

Programs

  • PARI
    {T(n,k)=polcoeff(sum(j=0,n,(x^binomial(n,j) - 1)/(x-1)),k)}
    for(n=0,10, for(k=0, binomial(n,n\2)-1, print1(T(n,k),","));print(""))

Formula

G.f. of row n: Sum_{k=0..n} (x^binomial(n,k) - 1)/(x-1) = Sum_{k=0..binomial(n,n\2)-1} T(n,k)*x^k.
A152548(n) = Sum_{k=0..C(n,[n/2])-1} T(n,k)^2 = Sum_{k=0..[(n+1)/2]} C(n+1, k)*(n+1-2k)^3/(n+1).
Showing 1-2 of 2 results.