new differential approach to combinations

• To: mathgroup at smc.vnet.net
• Subject: [mg109542] new differential approach to combinations
• From: Roger Bagula <roger.bagula at gmail.com>
• Date: Tue, 4 May 2010 06:30:25 -0400 (EDT)

```It is easy to prove that the Taylor expansion of  ( 0<t<1)
f(t)=1/(1-t)
has
derivatives ,such that:
f(t,n)=D[1/(1-t),{t,n}]
1/(1-t)=Sum[t^n,{n,0,.Infinity}]=Sum[f(0,n)*t^n/n!,{n,0,.Infinity}]
so that
f(0,n)=n!
That derivation is the basis for this new differential approach
to combination.
The function that makes these triangle sequences is
part of a Cantor rational pair:
1/(1-t)+(-t)/(1+t)=1
The substitution of
q=1/t
allows these q-form like triangle sequence to exist.
The section in Riordan that inspired this approach
is the one on generating functions
where he gives the finite version of the
first two derivatives. It was my idea to use the derivatives
as combinatorial functions.

%I A177097
%S A177097
1,1,1,1,0,1,1,1,1,1,1,2,4,2,1,1,3,8,8,3,1,1,4,13,18,13,4,1,1,5,19,33,
%T A177097
33,19,5,1,1,6,26,54,68,54,26,6,1,1,7,34,82,124,124,82,34,7,1,1,8,43,
%U A177097 118,208,250,208,118,43,8,1
%N A177097 A functional differential triangle function;f(t,n)=D[t/(1 -
t), {t, n}]; c(t,n,m)=(1/(1 - t))*f(t,n )/(f(t/m)*f(t,n-m)) - 1/t + 1
such that t=1/q;q=3 %C A177097 The sequence is base on the 1/(1-t)
having Taylor expansion derivatives of n!.
%C A177097 q=2 is A132044.
%C A177097 Row sums are:
%C A177097 {1, 2, 2, 4, 10, 24, 54, 116, 242, 496, 1006,...}. %F
A177097 f(t,n)=D[t/(1 - t), {t, n}];
%F A177097 c(t,n,m)=(1/(1 - t))*f(t,n )/(f(t/m)*f(t,n-m)) - 1/t + 1 %F
A177097 such that t=1/q;q=3 %e A177097 {1}, %e A177097 {1, 1}, %e
A177097 {1, 0, 1}, %e A177097 {1, 1, 1, 1}, %e A177097 {1, 2, 4, 2,
1}, %e A177097 {1, 3, 8, 8, 3, 1}, %e A177097 {1, 4, 13, 18, 13, 4,
1}, %e A177097 {1, 5, 19, 33, 33, 19, 5, 1}, %e A177097 {1, 6, 26, 54,
68, 54, 26, 6, 1}, %e A177097 {1, 7, 34, 82, 124, 124, 82, 34, 7, 1},
%e A177097 {1, 8, 43, 118, 208, 250, 208, 118, 43, 8, 1} %t A177097
f[t_, n_] := D[t/(1 - t), {t, n}];
%t A177097 a = Table[f[t, n], {n, 0, 20}];
%t A177097 c[t_, n_, m_] = (1/(1 - t))*a[[n + 1]]/(a[[m + 1]]*a[[n - m
+ 1]]) - 1/t + 1;
%t A177097 Table[Flatten[Table[Table[c[1/q, n, m], {m, 0, n}], {n, 0,
10}]], {q, 2, 10}] %Y A177097 Cf. A132044 %K A177097 nonn
%O A177097 0,12
%A A177097 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), May 02 2010

%I A177098
%S A177098
1,1,1,1,1,1,1,0,0,1,1,1,3,1,1,1,2,7,7,2,1,1,3,12,17,12,3,1,1,4,18,32,
%T A177098
32,18,4,1,1,5,25,53,67,53,25,5,1,1,6,33,81,123,123,81,33,6,1,1,7,42,
%U A177098 117,207,249,207,117,42,7,1
%V A177098
1,1,1,1,-1,1,1,0,0,1,1,1,3,1,1,1,2,7,7,2,1,1,3,12,17,12,3,1,1,4,18,32,
%W A177098
32,18,4,1,1,5,25,53,67,53,25,5,1,1,6,33,81,123,123,81,33,6,1,1,7,42,
%X A177098 117,207,249,207,117,42,7,1
%N A177098 A functional differential triangle function;f(t,n)=D[t/(1 -
t), {t, n}]; c(t,n,m)=(1/(1 - t))*f(t,n )/(f(t/m)*f(t,n-m)) - 1/t + 1
such that t=1/q;q=4 %C A177098 The sequence is base on the 1/(1-t)
having Taylor expansion derivatives of n!.
%C A177098 q=2 is A132044.
%C A177098 Row sums are:
%C A177098 {1, 2, 1, 2, 7, 20, 49, 110, 235, 488, 997,...}. %F A177098
f(t,n)=D[t/(1 - t), {t, n}];
%F A177098 c(t,n,m)=(1/(1 - t))*f(t,n )/(f(t/m)*f(t,n-m)) - 1/t + 1 %F
A177098 such that t=1/q;q=4. %e A177098 {1}, %e A177098 {1, 1}, %e
A177098 {1, -1, 1}, %e A177098 {1, 0, 0, 1}, %e A177098 {1, 1, 3, 1,
1}, %e A177098 {1, 2, 7, 7, 2, 1}, %e A177098 {1, 3, 12, 17, 12, 3,
1}, %e A177098 {1, 4, 18, 32, 32, 18, 4, 1}, %e A177098 {1, 5, 25, 53,
67, 53, 25, 5, 1}, %e A177098 {1, 6, 33, 81, 123, 123, 81, 33, 6, 1},
%e A177098 {1, 7, 42, 117, 207, 249, 207, 117, 42, 7, 1} %t A177098
f[t_, n_] := D[t/(1 - t), {t, n}];
%t A177098 a = Table[f[t, n], {n, 0, 20}];
%t A177098 c[t_, n_, m_] = (1/(1 - t))*a[[n + 1]]/(a[[m + 1]]*a[[n - m
+ 1]]) - 1/t + 1;
%t A177098 Table[Flatten[Table[Table[c[1/q, n, m], {m, 0, n}], {n, 0,
10}]], {q, 2, 10}] %Y A177098 Cf. A132044 %K A177098 sign,tabl
%O A177098 0,13
%A A177098 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), May 02 2010

%I A177099
%S A177099
1,1,1,1,2,1,1,1,1,1,1,0,2,0,1,1,1,6,6,1,1,1,2,11,16,11,2,1,1,3,17,
%T A177099
31,31,17,3,1,1,4,24,52,66,52,24,4,1,1,5,32,80,122,122,80,32,5,1,1,6,41,
%U A177099 116,206,248,206,116,41,6,1
%V A177099
1,1,1,1,-2,1,1,-1,-1,1,1,0,2,0,1,1,1,6,6,1,1,1,2,11,16,11,2,1,1,3,17,
%W A177099
31,31,17,3,1,1,4,24,52,66,52,24,4,1,1,5,32,80,122,122,80,32,5,1,1,6,41,
%X A177099 116,206,248,206,116,41,6,1
%N A177099 A functional differential triangle function;f(t,n)=D[t/(1 -
t), {t, n}]; c(t,n,m)=(1/(1 - t))*f(t,n )/(f(t/m)*f(t,n-m)) - 1/t + 1
such that t=1/q;q=5 %C A177099 The sequence is base on the 1/(1-t)
having Taylor expansion derivatives of n!.
%C A177099 q=2 is A132044.
%C A177099 Row sums are:
%C A177099 {1, 2, 0, 0, 4, 16, 44, 104, 228, 480, 988,...}. %F A177099
f(t,n)=D[t/(1 - t), {t, n}];
%F A177099 c(t,n,m)=(1/(1 - t))*f(t,n )/(f(t/m)*f(t,n-m)) - 1/t + 1 %F
A177099 such that t=1/q;q=5. %e A177099 {1}, %e A177099 {1, 1}, %e
A177099 {1, -2, 1}, %e A177099 {1, -1, -1, 1}, %e A177099 {1, 0, 2, 0,
1}, %e A177099 {1, 1, 6, 6, 1, 1}, %e A177099 {1, 2, 11, 16, 11, 2,
1}, %e A177099 {1, 3, 17, 31, 31, 17, 3, 1}, %e A177099 {1, 4, 24, 52,
66, 52, 24, 4, 1}, %e A177099 {1, 5, 32, 80, 122, 122, 80, 32, 5, 1},
%e A177099 {1, 6, 41, 116, 206, 248, 206, 116, 41, 6, 1} %t A177099
f[t_, n_] := D[t/(1 - t), {t, n}];
%t A177099 a = Table[f[t, n], {n, 0, 20}];
%t A177099 c[t_, n_, m_] = (1/(1 - t))*a[[n + 1]]/(a[[m + 1]]*a[[n - m
+ 1]]) - 1/t + 1;
%t A177099 Table[Flatten[Table[Table[c[1/q, n, m], {m, 0, n}], {n, 0,
10}]], {q, 2, 10}] %Y A177099 Cf. A132044 %K A177099 sign,tabl
%O A177099 0,5
%A A177099 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), May 02 2010

--
Respectfully, Roger L. Bagula
11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :