A136219 Number of terms in rows of irregular triangle A136218.

1, 3, 7, 13, 22, 33, 47, 64, 84, 106, 131, 159, 190, 224, 261, 301, 343, 388, 436, 487, 541, 598, 658, 721, 787, 856, 928, 1003, 1081, 1162, 1245, 1331, 1420, 1512, 1607, 1705, 1806, 1910, 2017, 2127, 2240, 2356, 2475, 2597, 2722, 2850, 2981, 3115, 3252

Offset: 0

Author

Paul D. Hanna, Dec 23 2007

Keywords

• nonn

Comments

A136218 is a triangle where row n+1 is generated from row n by first inserting zeros in row n at positions {[m*(m+7)/6], m>=0} and then taking partial sums, starting with a '1' in row 0.

Examples

			G.f. A(x) = (1+x+x^2)/(1-x)^3 - (x+x^2+x^3+x^5+x^9+x^16+x^30+x^58+...)/(1-x)^2.
		

Crossrefs

Cf. A136218; A136169.

Programs

PARI

{a(n)=local(A,X=x+x*O(x^n),bd=#binary(2*n), B=(1 - x^2*(1+x+x^2)/(1-x^3+x*O(x^bd))^2 )/(1-2*x)); A=(1+x+x^2)/(1-X)^3 - sum(k=0,bd,x^polcoeff(B,k))/(1-X)^2;polcoeff(A,n)}

Formula

G.f. A(x) = (1+x+x^2)/(1-x)^3 - [Sum_{n>=0} x^b(n)]/(1-x)^2 where exponents b(n) = A136169(n) satisfy: b(n) = 2*b(n-1) - [(n+1)/3] for n>0 with b(0)=1 and the g.f. of the exponents is B(z) = [1 - z^2*(1+z+z^2)/(1-z^3)^2]/(1-2*z).