0,8

Definitions. A pendular triangle is a triangle in which row n is

generated from the pendular sums of row n-1. Pendular sums of a row

are partial sums taken in back-and-forth order, starting with the

left-most term, jumping to the right-most term, back to the left-most

unused term, then forward to the right-most unused term, etc.

In each pass, the partial sum is placed in the new row direcly under

the term most recently used in the sum. Continue in this way until

all the terms of the prior row have been used, and then complete the

new row by appending a zero at the end. Pendular sums are so named

because the process resembles a swinging pendulum that slows down

on each pass until it eventually comes to rest in the center.

In the simplest case, pendular triangles obey the recurrence:

if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else

T(n,k) = T(n,n-1-k) + p*T(n-1,k), for n>=k>0,

with T(n,0)=1 and T(n,n)=0^n, for any number p.

In which case the g.f. G=G(x) of the central terms satisfies:

G = 1 - p*x*G + p*x*G^2 + x*G^3.

More generally, a pendular triangle is defined by the recurrence:

if n > 2k, T(n,k) = T(n,n-k) + T(n-1,k), else

T(n,k) = T(n,n-1-k) + Sum_{j>=1} p(j)*T(n-1,k-1+j),

for n>=k>0, with T(n,0)=1 and T(n,n)=0^n.

Remarkably, the g.f. G=G(x) of the central terms satisfies:

G = 1 + x*G^3 + Sum_{j>=1} p(j)*x^j*[G^(2*j) - G^(2*j-1)].

Further, the g.f. of the m-th lower semi-diagonal equals G(x)^(m+1)

for m>=0, where the m-th semi-diagonal consists of those terms

located at m rows below the central terms.

For variants of pendular triangles, the main diagonal may be nonzero,

but then the g.f.s of the semi-diagonals are more complex.

T(2*n+m,n) = [A108447^(m+1)](n), i.e., the m-th lower semi-diagonal forms the self-convolution (m+1)-power of A108447; compare semi-diagonals to the diagonals of convolution triangle A118343.

Row 6 equals the pendular sums of row 5:

[1, 4, 9, 5, 1, 0], where the sums proceed as follows:

[1,__,__,__,__,__], T(6,0) = T(5,0) = 1;

[1,__,__,__,__, 1], T(6,5) = T(6,0) + T(5,5) = 1 + 0 = 1;

[1, 5,__,__,__, 1], T(6,1) = T(6,5) + T(5,1) = 1 + 4 = 5;

[1, 5,__,__, 6, 1], T(6,4) = T(6,1) + T(5,4) = 5 + 1 = 6;

[1, 5,15,__, 6, 1], T(6,2) = T(6,4) + T(5,2) = 6 + 9 = 15;

[1, 5,15,20, 6, 1], T(6,3) = T(6,2) + T(5,3) = 15 + 5 = 20;

[1, 5,15,20, 6, 1, 0] finally, append a zero to obtain row 6.

Triangle begins:

1;

1, 0;

1, 1, 0;

1, 2, 1, 0;

1, 3, 4, 1, 0;

1, 4, 9, 5, 1, 0;

1, 5, 15, 20, 6, 1, 0;

1, 6, 22, 48, 28, 7, 1, 0;

1, 7, 30, 85, 113, 37, 8, 1, 0;

1, 8, 39, 132, 282, 169, 47, 9, 1, 0;

1, 9, 49, 190, 519, 688, 237, 58, 10, 1, 0;

1, 10, 60, 260, 837, 1762, 1074, 318, 70, 11, 1, 0;

1, 11, 72, 343, 1250, 3330, 4404, 1568, 413, 83, 12, 1, 0; ...

Central terms are T(2*n,n) = A108447(n);

semi-diagonals form successive self-convolutions of the

central terms:

T(2*n+1,n) = A118341(n) = [A108447^2](n),

T(2*n+2,n) = A118342(n) = [A108447^3](n).

(PARI) {T(n, k)=if(n<k|k<0, 0, if(k==0, 1, if(n==k, 0, if(n>2*k, T(n-1, k)+T(n, n-k), T(n-1, k)+T(n, n-1-k)))))}

Cf. A108447 (central terms), A118341, A118341, A118343; variants: A118344 (Catalan), A118362 (ternary), A118350, A118355.

Sequence in context: A017827 A094266 A071569 this_sequence A071921 A003992 A118345

Adjacent sequences: A118337 A118338 A118339 this_sequence A118341 A118342 A118343

nonn,tabl

Paul D. Hanna (pauldhanna(AT)juno.com), Apr 25 2006