Logo

Greetings from The On-Line Encyclopedia of Integer Sequences!

Hints

Search: id:A157703
Displaying 1-1 of 1 results found. page 1
     Format: long | short | internal | text      Sort: relevance | references | number      Highlight: on | off
%I A157703
%S A157703 1,1,5,5,2,62,152,62,2,91,1652,5957,5957,1652,91,52,5240,77630,342188,
%T A157703 551180,342188,77630,5240,52,12,8549,424921,5629615,28123559,61108544,
%U A157703 61108544,28123559,5629615,424921,8549,12
%N A157703 G.f.s of the z^p coefficients of the polynomials in the GF2 denominators 
               of A156925
%C A157703 The formula for the PDGF2(z;n) polynomials in the GF2 denominators of 
               A156925 can be found below.
%C A157703 The general structure of the GFKT2(z;p) that generate the z^p coefficients 
               of the PDGF2(z; n) polynomials can also be found below. The KT2(z;
               p) polynomials in the nominators of the GFKT2(z;p) have a nice symmetrical 
               structure.
%C A157703 The sequence of the number of terms of the first few KT2(z;p) polynomials 
               is: 1, 1, 2, 5, 6, 9, 12, 13, 16, 19, 22, 23, 26. The first differences 
               follow a simple pattern. The positions of the 1's follow the Lazy 
               Caterer's sequence A000124 with one exception, here a(0) = 0.
%C A157703 A Maple algorithm that generates relevant GFKT2(z;p) information can 
               be found below.
%F A157703 PDGF2(z;n) = product((1-m*z)^(n+1-m),m=1..n) with n = 1, 2, 3, ..
%F A157703 GFKT2(z;p) = (-1)^(p)*(z^q2)*KT2(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, 
               ..
%F A157703 The recurrence relation for the z^p coefficients a(n) is: a(n) = sum((-1)^(k+1)* 
               binomial(3*p + 1, k) *a(n-k), k=1 .. 3*p+1) with p = 0, 1, 2, .. 
               .
%e A157703 Some PDGF2 (z;n) are:
%e A157703 PDGF2(z;n=3) = (1-z)^3*(1-2*z)^2*(1-3*z)
%e A157703 PDGF2(z;n=4) = (1-z)^4*(1-2*z)^3*(1-3*z)^2*(1-4*z)
%e A157703 The first few GFKT2's are:
%e A157703 GFKT2(z;p=0) = 1/(1-z)
%e A157703 GFKT2(z;p=1) = -z/(z-1)^4
%e A157703 GFKT2(z;p=2) = z^2*(5+5*z)/(1-z)^7
%e A157703 Some KT2(z,p) polynomials are:
%e A157703 KT2(z;p=2) = 5+5*z
%e A157703 KT2(z;p=3) = 2+62*z+152*z^2+62*z^3+2*z^4
%e A157703 KT2(z;p=4) = 91+1652*z+5957*z^2+5957*z^3+1652*z^4+91*z^5
%p A157703 p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) 
               = fn,a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-m*z)^(n2+1-m),
               m=1..n2): a(n2):= coeff(fz(n2),z,p): end do: b:=n-> a(n): seq(b(n), 
               n=0..3*p+1); a(n)=fn; a(k)=sort(simplify(fk)); GFKT2(p):=sum((fk)*z^k,
               k=0..infinity); q2:=ldegree((numer(GFKT2(p)))): KT2(p):=sort((-1)^p*simplify((GFKT2(p)*(1-z)^(3*p+1))/
               z^q2),z, ascending);
%Y A157703 Originator sequence A156925
%Y A157703 See A000292 for the z^1 coefficients and A040977 for the z^2 coefficients 
               divided by 5.
%Y A157703 Row sums equal A025035
%Y A157703 Cf. A157702, A157704, A157705
%Y A157703 Sequence in context: A023580 A021648 A128006 this_sequence A158349 A079384 
               A158274
%Y A157703 Adjacent sequences: A157700 A157701 A157702 this_sequence A157704 A157705 
               A157706
%K A157703 easy,nonn,tabf
%O A157703 0,3
%A A157703 Johannes W. Meijer (meijgia(AT)hotmail.com), Mar 07 2009

    
page 1

Search completed in 0.003 seconds

Lookup | Welcome | Find friends | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Puzzles | Hot | Classics
More pages | Superseeker | Maintained by N. J. A. Sloane (njas@research.att.com)

Last modified December 2 11:54 EST 2009. Contains 167921 sequences.


AT&T Labs Research