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<DIV>One last note regarding Paul Barry's Modulo 2 Binomial Transform.
</DIV>
<DIV> </DIV>
<DIV>If we apply the triangle formed by reading Pascal's triangle mod 2:
</DIV>
<DIV><A
href="http://www.research.att.com/projects/OEIS?Anum=A047999">http://www.research.att.com/projects/OEIS?Anum=A047999</A> </DIV>
<DIV>as a transformation on a sequence A, and apply this transform M times,
</DIV>
<DIV>the formula for this (Mod 2 Binomial transformation)^M of
A is: </DIV>
<DIV> </DIV>
<DIV>(1) a(n) = Sum_{k=0..n} M^A000120(k)*(C(n,k) Mod 2)*A(k) </DIV>
<DIV>where A000120(k) = number of 1's in binary expansion of k. </DIV>
<DIV> </DIV>
<DIV>This can be expressed using PARI code as:</DIV>
<DIV>a(n)=sum(k=0,n,M^subst(Pol(binary(n-k)),x,1)*(binomial(n,k)%2)*A(k))</DIV>
<DIV> </DIV>
<DIV> </DIV>
<DIV>A special case exists when A(n) = B^n for integer B; </DIV>
<DIV>for then (1) is equivalent to the product formula: </DIV>
<DIV> </DIV>
<DIV>
<DIV>(2) a(n) = Product_{k=0..[log_2(n)]} ( B^(2^k) + M )^b(n,k)
</DIV>
<DIV>where b(n,k) = coefficient of 2^k in the binary expansion of
n. </DIV></DIV>
<DIV> </DIV>
<DIV>This can be expressed using PARI code as:</DIV>
<DIV>a(n)=prod(k=0,length(binary(n))-1,(B^(2^k)+M)^bittest(n,k))
<BR> </DIV>
<DIV>The product formula (2) explains the factorization behavior noted
earlier. </DIV>
<DIV> </DIV>
<DIV>(Merry Christmas)^2,</DIV>
<DIV> Paul</DIV></BODY></HTML>