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).

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

Offset: 0

Author

Paul D. Hanna, Dec 13 2008

Keywords

• nonn
• tabf

Comments

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

Examples

			Triangle begins:
[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],
[21,16,16,16,16,13,12,12,12,12,12,11,9,8,8,8,5,5,5,5,4,1,1,1,1,1,1,1],
[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 - - -
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 - - - - - - - -
4: o o o o - - - - - - - -
1: 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.
		

Links

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

Crossrefs

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

Programs

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(""))

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.