|
|
A177229
|
|
A combinatorial differential triangle sequence:q=4;t=1/q;f(t,n)=d^n/dt^n*(t/(1+t); c(t.n,m)=(1/(1+t)*f(n,t)/(f(t,m)*f(t,(n-m))
|
|
0
|
|
|
4, 4, 4, 4, -2, 4, 4, -3, -3, 4, 4, -4, -6, -4, 4, 4, -5, -10, -10, -5, 4, 4, -6, -15, -20, -15, -6, 4, 4, -7, -21, -35, -35, -21, -7, 4, 4, -8, -28, -56, -70, -56, -28, -8, 4, 4, -9, -36, -84, -126, -126, -84, -36, -9, 4, 4, -10, -45, -120, -210, -252, -210, -120, -45, -10
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
The name contains an unmatched parenthesis. - Editors, Mar 13 2024
The row sums are:
{4, 8, 6, 2, -6, -22, -54, -118, -246, -502, -1014,...}.
The Taylor expansion derivatives of 1/(1+t) are: f(0,n)=(-1)^n*n!
from:
1/(1+t)=Sum[(-t)^n,{n,0,Infinity}]
Adding the t to get t/(1+t)
gives the symmetrical sequence as combinatorial.
|
|
REFERENCES
|
J. Riordan, Combinatorial Identities, Wiley, 1968, p.22-23
|
|
LINKS
|
|
|
FORMULA
|
q=4;t=1/q;
f(t,n)=d^n/dt^n*(t/(1+t);
c(t.n,m)=(1/(1+t)*f(n,t)/(f(t,m)*f(t,(n-m))
|
|
EXAMPLE
|
{4},
{4, 4},
{4, -2, 4},
{4, -3, -3, 4},
{4, -4, -6, -4, 4},
{4, -5, -10, -10, -5, 4},
{4, -6, -15, -20, -15, -6, 4},
{4, -7, -21, -35, -35, -21, -7, 4},
{4, -8, -28, -56, -70, -56, -28, -8, 4},
{4, -9, -36, -84, -126, -126, -84, -36, -9, 4},
{ 4, -10, -45, -120, -210, -252, -210, -120, -45, -10, 4}
|
|
MATHEMATICA
|
f[t_, n_] := D[t/(1 + t), {t, n}];
a = Table[f[t, n], {n, 0, 20}];
c[t_, n_, m_] = (1/(1 + t))*a[[n + 1]]/(a[[m + 1]]*a[[n - m + 1]]);
Table[Flatten[Table[Table[c[1/q, n, m], {m, 0, n}], {n, 0, 10}]], {q, 2, 10}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|