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