|
| |
|
|
A213500
|
|
Rectangular array T(n,k): (row n) = b**c, where b(h) = h, c(h) = h + n - 1, n >= 1, h >= 1, and ** = convolution.
|
|
89
|
|
|
|
1, 4, 2, 10, 7, 3, 20, 16, 10, 4, 35, 30, 22, 13, 5, 56, 50, 40, 28, 16, 6, 84, 77, 65, 50, 34, 19, 7, 120, 112, 98, 80, 60, 40, 22, 8, 165, 156, 140, 119, 95, 70, 46, 25, 9, 220, 210, 192, 168, 140, 110, 80, 52, 28, 10, 286, 275, 255, 228, 196, 161, 125, 90
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
Row 1: (1,2,3,...)**(1,2,3,...) = A000292.
Row 2: (1,2,3,...)**(2,3,4,...) = A005581.
Row 3: (1,2,3,...)**(3,4,5,...) = A006503.
Row 4: (1,2,3,...)**(4,5,6,...) = A060488.
Row 5: (1,2,3,...)**(5,6,7,...) = A096941.
Row 6: (1,2,3,...)**(6,7,8,...) = A096957.
...
In general, the convolution of two infinite sequences is defined from the convolution of two n-tuples: let X(n) = (x(1),...,x(n)) and Y(n)=(y(1),...,y(n)); then X(n)**Y(n) = x(1)*y(n)+x(2)*y(n-1)+...+x(n)*y(1); this sum is the n-th term in the convolution of infinite sequences:(x(1),...,x(n),...)**(y(1),...,y(n),...), for all n>=1.
...
In the following guide to related arrays and sequences, row n of each array T(n,k) is the convolution b**c of the sequences b(h) and c(h+n-1). The principal diagonal is given by T(n,n) and the n-th antidiagonal sum by S(n). In some cases, T(n,n) or S(n) differs in offset from the listed sequence.
b(h)........ c(h)........ T(n,k) .. T(n,n) .. S(n)
...
Suppose that u = (u(n)) and v = (v(n)) are sequences having generating functions U(x) and V(x), respectively. Then the convolution u**v has generating function U(x)*V(x). Accordingly, if u and v are homogeneous linear recurrence sequences, then every row of the convolution array T satisfies the same homogeneous linear recurrence equation, which can be easily obtained from the denominator of U(x)*V(x). Also, every column of T has the same homogeneous linear recurrence as v.
|
|
|
LINKS
|
|
|
|
FORMULA
|
T(n,k) = 4*T(n,k-1) - 6*T(n,k-2) + 4*T(n,k-3) - T(n,k-4).
T(n,k) = 2*T(n-1,k) - T(n-2,k).
G.f. for row n: x*(n - (n - 1)*x)/(1 - x)^4.
|
|
|
EXAMPLE
|
Northwest corner (the array is read by southwest falling antidiagonals):
1, 4, 10, 20, 35, 56, 84, ...
2, 7, 16, 30, 50, 77, 112, ...
3, 10, 22, 40, 65, 98, 140, ...
4, 13, 28, 50, 80, 119, 168, ...
5, 16, 34, 60, 95, 140, 196, ...
6, 19, 40, 70, 110, 161, 224, ...
T(6,1) = (1)**(6) = 6;
T(6,2) = (1,2)**(6,7) = 1*7+2*6 = 19;
T(6,3) = (1,2,3)**(6,7,8) = 1*8+2*7+3*6 = 40.
|
|
|
MATHEMATICA
|
b[n_] := n; c[n_] := n
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}] (* A213500 *)
|
|
|
PROG
|
(PARI)
t(n, k) = sum(i=0, k - 1, (k - i) * (n + i));
tabl(nn) = {for(n=1, nn, for(k=1, n, print1(t(k, n - k + 1), ", "); ); print(); ); };
(Python)
def t(n, k): return sum((k - i) * (n + i) for i in range(k))
for n in range(1, 13):
print([t(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Mar 26 2017
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
