A152546 -id:A152546

Search: a152546 -id:a152546

Displaying 1-1 of 1 result found.

page 1

A152545

Padovan-Fibonacci triangle, read by rows, where the first column equals the Padovan spiral numbers (A134816), while the row sums equal the Fibonacci numbers (A000045).

+10
3

1, 1, 1, 1, 2, 1, 2, 2, 1, 3, 2, 2, 1, 4, 3, 3, 2, 1, 5, 4, 4, 3, 3, 1, 1, 7, 5, 5, 5, 4, 3, 3, 1, 1, 9, 7, 7, 7, 5, 5, 5, 4, 3, 1, 1, 1, 12, 9, 9, 9, 8, 7, 7, 7, 5, 4, 4, 4, 1, 1, 1, 1, 16, 12, 12, 12, 12, 9, 9, 9, 8, 8, 8, 7, 5, 4, 4, 4, 1, 1, 1, 1, 1, 21, 16, 16, 16, 16, 13, 12, 12, 12, 12, 12, 11, 9, 8

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,5

COMMENTS

The number of terms in each row equal the Padovan spiral numbers (A134816, with offset).

LINKS

Paul D. Hanna, Table of n, a(n) for n = 0..261

FORMULA

G.f. for row n: Sum_{k=0..A000931(n+5)-1} (x^{T(n-1,k)+T(n-2,k)} - 1)/(x-1) = Sum_{k=0..A000931(n+6)-1} T(n,k)*x^k for n>1 with T(0,0)=T(1,0)=1, where A000931 is the Padovan sequence.

EXAMPLE

Triangle begins:

[1],

[1,1],

[2,1],

[2,2,1],

[3,2,2,1],

[4,3,3,2,1],

[5,4,4,3,3,1,1],

[7,5,5,5,4,3,3,1,1],

[9,7,7,7,5,5,5,4,3,1,1,1],

[12,9,9,9,8,7,7,7,5,4,4,4,1,1,1,1],

[16,12,12,12,12,9,9,9,8,8,8,7,5,4,4,4,1,1,1,1,1],

[21,16,16,16,16,13,12,12,12,12,12,11,9,8,8,8,5,5,5,5,4,1,1,1,1,1,1,1],

[28,21,21,21,21,20,16,16,16,16,16,16,13,13,12,12,12,12,11,11,8,6,5,5,5,5,5,5,1,1,1,1,1,1,1,1,1],

[37,28,28,28,28,28,22,21,21,21,21,21,20,20,20,20,18,16,16,16,14,13,12,12,12,12,12,11,6,6,6,6,6,5,5,5,5,1,1,1,1,1,1,1,1,1,1,1,1],

[49,37,37,37,37,37,33,28,28,28,28,28,28,28,28,28,27,22,22,21,21,21,20,20,20,20,20,18,17,17,16,16,14,12,12,12,12,7,6,6,6,6,6,6,6,6,6,6,5,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1],

...

ILLUSTRATION OF RECURRENCE.

Start out with row 0 and row 1 consisting of a single '1'.

To obtain any given row of this irregular triangle, first

sum the prior two rows term-by-term; for rows 7 and 8 we get:

[5,4,4,3,3,1,1] + [7,5,5,5,4,3,3,1,1] = [12,9,9,8,7,4,4,1,1].

Place markers in an array so that the number of contiguous markers

in each row correspond to the term-by-term sums like so:

--------------------------

12:o o o o o o o o o o o o

9: o o o o o o o o o - - -

8: o o o o o o o o - - - -

7: o o o o o o o - - - - -

4: o o o o - - - - - - - -

1: o - - - - - - - - - - -

--------------------------

Then count the markers by columns to obtain the desired row;

here, the number of markers in each column yields row 9:

[9,7,7,7,5,5,5,4,3,1,1,1].

Continuing in this way generates all the rows of this triangle.

PROG

(PARI) {T(n, k)=local(G000931=(1-x^2)/(1-x^2-x^3+x*O(x^(n+6)))); if(n<0, 0, if(n<2&k==0, 1, polcoeff(sum(j=0, polcoeff(G000931, n+5)-1, (x^(T(n-1, j)+T(n-2, j)) - 1)/(x-1)), k) ))};

/* To print, use Padovan g.f. to get the number of terms in row n: */

for(n=0, 10, for(k=0, polcoeff((1-x^2)/(1-x^2-x^3+x*O(x^(n+6))), n+6)-1, print1(T(n, k), ", ")); print(""))

CROSSREFS

Cf. A134816, A000045, A000931; A152546 (row squared sums).

KEYWORD

nonn,tabf

AUTHOR

Paul D. Hanna, Dec 13 2008

STATUS

approved

page 1

Search completed in 0.005 seconds