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A118340
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Pendular triangle, read by rows, where row n is formed from row n-1 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) + T(n-1,k), for n>=k>0, with T(n,0)=1 and T(n,n)=0^n.
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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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,8
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COMMENTS
| 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.
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FORMULA
| 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.
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EXAMPLE
| 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).
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PROG
| (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)))))}
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CROSSREFS
| Cf. A108447 (central terms), A118341, A118341, A118343; variants: A118344 (Catalan), A118362 (ternary), A118350, A118355.
Sequence in context: A017827 A094266 A071569 * A071921 A003992 A171882
Adjacent sequences: A118337 A118338 A118339 * A118341 A118342 A118343
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KEYWORD
| nonn,tabl
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AUTHOR
| Paul D. Hanna (pauldhanna(AT)juno.com), Apr 25 2006
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