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Re: new differential approach to combinations

  • To: mathgroup at smc.vnet.net
  • Subject: [mg109583] Re: new differential approach to combinations
  • From: Roger Bagula <roger.bagula at gmail.com>
  • Date: Thu, 6 May 2010 04:52:27 -0400 (EDT)
  • References: <hrot06$5sm$1@smc.vnet.net>

Another set of differential combinatorial triangle sequences has been found.
These are based on:
1/(1+t)=Sum[(-t)^n,{n,0,Infinity}] =Sum[f(0,n)*t^n/n!,{n,0,Infinity}]
so that
 f(0,n)=(-1)^n*n!
Adding the t in
t/(1+t)
gives the 1/q expansions as combinatorial.
This kind of infinite differential expansion is closely related to the
Umbral Calculus
and expansion function.


%I A177227
%S A177227 2,2,2,2,2,2,2,3,3,2,2,4,6,4,2,2,5,10,10,5,2,2,6,15,20,15,
%T A177227 6,2,2,7,21,35,35,21,7,2,2,8,28,56,70,56,28,8,2,2,9,36,
%U A177227 84,126,126,84,36,9,2,2,10,45,120,210,252,210,120,45,10
%V A177227
2,2,2,2,-2,2,2,-3,-3,2,2,-4,-6,-4,2,2,-5,-10,-10,-5,2,2,-6,-15,-20,-15,
%W A177227
-6,2,2,-7,-21,-35,-35,-21,-7,2,2,-8,-28,-56,-70,-56,-28,-8,2,2,-9,-36,
%X A177227
-84,-126,-126,-84,-36,-9,2,2,-10,-45,-120,-210,-252,-210,-120,-45,-10
%N A177227 A Combinatorial differential triangle sequence:q=2;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))
%C A177227 The row sums are:
%C A177227 {2, 4, 2, -2, -10, -26, -58, -122, -250, -506, -1018,...}.
%C A177227 The Taylor expansion derivatives of 1/(1+t) are:
f(0,n)=(-1)^n*n!
%C A177227 from:
%C A177227 1/(1+t)=Sum[(-t)^n,{n,0,Infinity}]
%C A177227 Adding the t to get t/(1+t)
%C A177227 gives the symmetrical sequence as combinatorial.
%D A177227 J. Riordan, Combinatorial Identities, Wiley, 1968, p.22-23
%F A177227 q=2;t=1/q;
%F A177227 f(t,n)=d^n/dt^n*(t/(1+t);
%F A177227 c(t.n,m)=(1/(1+t)*f(n,t)/(f(t,m)*f(t,(n-m))
%e A177227 {2},
%e A177227 {2, 2},
%e A177227 {2, -2, 2},
%e A177227 {2, -3, -3, 2},
%e A177227 {2, -4, -6, -4, 2},
%e A177227 {2, -5, -10, -10, -5, 2},
%e A177227 {2, -6, -15, -20, -15, -6, 2},
%e A177227 {2, -7, -21, -35, -35, -21, -7, 2},
%e A177227 {2, -8, -28, -56, -70, -56, -28, -8, 2},
%e A177227 {2, -9, -36, -84, -126, -126, -84, -36, -9, 2},
%e A177227 { 2, -10, -45, -120, -210, -252, -210, -120, -45, -10, 2}
%t A177227 f[t_, n_] := D[t/(1 + t), {t, n}];
%t A177227 a = Table[f[t, n], {n, 0, 20}];
%t A177227 c[t_, n_, m_] = (1/(1 + t))*a[[n + 1]]/(a[[m + 1]]*a[[n - m
+ 1]]);
%t A177227 Table[Flatten[Table[Table[c[1/q, n, m], {m, 0, n}], {n, 0,
10}]], {q, 2, 10}]
%K A177227 sign,tabl
%O A177227 0,1
%A A177227 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), May 05 2010

%I A177228
%S A177228 3,3,3,3,2,3,3,3,3,3,3,4,6,4,3,3,5,10,10,5,3,3,6,15,20,15,
%T A177228 6,3,3,7,21,35,35,21,7,3,3,8,28,56,70,56,28,8,3,3,9,36,
%U A177228 84,126,126,84,36,9,3,3,10,45,120,210,252,210,120,45,10
%V A177228
3,3,3,3,-2,3,3,-3,-3,3,3,-4,-6,-4,3,3,-5,-10,-10,-5,3,3,-6,-15,-20,-15,
%W A177228
-6,3,3,-7,-21,-35,-35,-21,-7,3,3,-8,-28,-56,-70,-56,-28,-8,3,3,-9,-36,
%X A177228
-84,-126,-126,-84,-36,-9,3,3,-10,-45,-120,-210,-252,-210,-120,-45,-10
%N A177228 A Combinatorial differential triangle sequence:q=3;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))
%C A177228 The row sums are:
%C A177228 {3, 6, 4, 0, -8, -24, -56, -120, -248, -504, -1016,...}.
%C A177228 The Taylor expansion derivatives of 1/(1+t) are:
f(0,n)=(-1)^n*n!
%C A177228 from:
%C A177228 1/(1+t)=Sum[(-t)^n,{n,0,Infinity}]
%C A177228 Adding the t to get t/(1+t)
%C A177228 gives the symmetrical sequence as combinatorial.
%D A177228 J. Riordan, Combinatorial Identities, Wiley, 1968, p.22-23
%F A177228 q=3;t=1/q;
%F A177228 f(t,n)=d^n/dt^n*(t/(1+t);
%F A177228 c(t.n,m)=(1/(1+t)*f(n,t)/(f(t,m)*f(t,(n-m))
%e A177228 {3},
%e A177228 {3, 3},
%e A177228 {3, -2, 3},
%e A177228 {3, -3, -3, 3},
%e A177228 {3, -4, -6, -4, 3},
%e A177228 {3, -5, -10, -10, -5, 3},
%e A177228 {3, -6, -15, -20, -15, -6, 3},
%e A177228 {3, -7, -21, -35, -35, -21, -7, 3},
%e A177228 {3, -8, -28, -56, -70, -56, -28, -8, 3},
%e A177228 {3, -9, -36, -84, -126, -126, -84, -36, -9, 3},
%e A177228 { 3, -10, -45, -120, -210, -252, -210, -120, -45, -10, 3}
%t A177228 f[t_, n_] := D[t/(1 + t), {t, n}];
%t A177228 a = Table[f[t, n], {n, 0, 20}];
%t A177228 c[t_, n_, m_] = (1/(1 + t))*a[[n + 1]]/(a[[m + 1]]*a[[n - m
+ 1]]);
%t A177228 Table[Flatten[Table[Table[c[1/q, n, m], {m, 0, n}], {n, 0,
10}]], {q, 2, 10}]
%K A177228 sign,tabl
%O A177228 0,1
%A A177228 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), May 05 2010


%I A177229
%S A177229 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,
%T A177229 6,4,4,7,21,35,35,21,7,4,4,8,28,56,70,56,28,8,4,4,9,36,
%U A177229 84,126,126,84,36,9,4,4,10,45,120,210,252,210,120,45,10
%V A177229
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,
%W A177229
-6,4,4,-7,-21,-35,-35,-21,-7,4,4,-8,-28,-56,-70,-56,-28,-8,4,4,-9,-36,
%X A177229
-84,-126,-126,-84,-36,-9,4,4,-10,-45,-120,-210,-252,-210,-120,-45,-10
%N 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))
%C A177229 The row sums are:
%C A177229 {4, 8, 6, 2, -6, -22, -54, -118, -246, -502, -1014,...}.
%C A177229 The Taylor expansion derivatives of 1/(1+t) are:
f(0,n)=(-1)^n*n!
%C A177229 from:
%C A177229 1/(1+t)=Sum[(-t)^n,{n,0,Infinity}]
%C A177229 Adding the t to get t/(1+t)
%C A177229 gives the symmetrical sequence as combinatorial.
%D A177229 J. Riordan, Combinatorial Identities, Wiley, 1968, p.22-23
%F A177229 q=4;t=1/q;
%F A177229 f(t,n)=d^n/dt^n*(t/(1+t);
%F A177229 c(t.n,m)=(1/(1+t)*f(n,t)/(f(t,m)*f(t,(n-m))
%e A177229 {4},
%e A177229 {4, 4},
%e A177229 {4, -2, 4},
%e A177229 {4, -3, -3, 4},
%e A177229 {4, -4, -6, -4, 4},
%e A177229 {4, -5, -10, -10, -5, 4},
%e A177229 {4, -6, -15, -20, -15, -6, 4},
%e A177229 {4, -7, -21, -35, -35, -21, -7, 4},
%e A177229 {4, -8, -28, -56, -70, -56, -28, -8, 4},
%e A177229 {4, -9, -36, -84, -126, -126, -84, -36, -9, 4},
%e A177229 { 4, -10, -45, -120, -210, -252, -210, -120, -45, -10, 4}
%t A177229 f[t_, n_] := D[t/(1 + t), {t, n}];
%t A177229 a = Table[f[t, n], {n, 0, 20}];
%t A177229 c[t_, n_, m_] = (1/(1 + t))*a[[n + 1]]/(a[[m + 1]]*a[[n - m
+ 1]]);
%t A177229 Table[Flatten[Table[Table[c[1/q, n, m], {m, 0, n}], {n, 0,
10}]], {q, 2, 10}]
%K A177229 sign,tabl
%O A177229 0,1
%A A177229 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), May 05 2010



--
Respectfully, Roger L. Bagula
11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
http://www.google.com/profiles/Roger.Bagula
alternative email: roger.bagula at gmail.com





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