# Notation A/B/C

N. J. A. Sloane njas at research.att.com
Sat Dec 29 20:09:35 CET 2007

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Seqfans,
Recall the identity:
(2) Sum_{n>=0} log(1 + q^n*x)^n/n!  =  Sum_{n>=0} C(q^n,n)*x^n.

From this, I found the more general statements:

(3) Sum_{n>=0} m^n*log(1 + q^n*x)^n/n!  =  Sum_{n>=0} C(m*q^n,n)*x^n.

(4) Sum_{n>=0} m^n * (1 + q^n*x)^b * log(1 + q^n*x)^n/n!  =
Sum_{n>=0} C(m*q^n + b, n)*x^n.

Identity (4) is very interesting ... I wonder if it leads to other
results?
It certainly can lead to many significant sequences!

What I would really like is for formula (4) to allow the g.f.
A(x,m,b) = Sum_{n>=0} C(m*q^n + b, n)*x^n
to be manipulated to solve some functional equation ...

Any ideas along these lines from anyone?
Paul

On Sat, 29 Dec 2007 00:20:34 -0500 <pauldhanna at juno.com> writes:
Seqfans,
Just found an o.g.f. for A014070(n) = C(2^n,n):

(1) G.f.: A(x) = Sum_{n>=0} log(1 + 2^n*x)^n / n!.

More generally, we have the nontrivial identity:

(2) Sum_{n>=0} log(1 + q^n*x)^n/n!  =  Sum_{n>=0} C(q^n,n)*x^n.

The proof is easy given the following formula for A008275,
the Stirling numbers of the first kind:
E.g.f. for k-th column of A008275 is ln(1+x)^k/k!.

Can any other interesting result be derived from (2)?
Thanks,
Paul
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<BODY bottomMargin=3D0 leftMargin=3D3 topMargin=3D0 rightMargin=3D3>
<DIV>Seqfans, </DIV>
<DIV>     Recall the identity: </DIV>
<DIV>(2) Sum_{n>=3D0} log(1 + q^n*x)^n/n!  =3D  Sum_{n>=3D0=
}=20
C(q^n,n)*x^n. </DIV>
<DIV> </DIV>
<DIV>From this, I found the more general statements: </DIV>
<DIV> </DIV>
<DIV>(3) Sum_{n>=3D0} m^n*log(1 + q^n*x)^n/n!  =3D  Sum_{n>=
=3D0}=20
C(m*q^n,n)*x^n.</DIV>
<DIV> </DIV>
<DIV>
<DIV>(4) Sum_{n>=3D0} m^n * (1 + q^n*x)^b * log(1 + q^n*x)^n/n!  =
=3D =20
</DIV>
<DIV>     Sum_{n>=3D0} C(m*q^n + b, n)*x^n.</DI=
V>
<DIV> </DIV>
<DIV>Identity (4) is very interesting ... I wonder if it leads to other res=
ults?=20
</DIV>
<DIV>It certainly can lead to many significant sequences! </DIV>
<DIV> </DIV>
<DIV>What I would really like is for formula (4) to allow the g.f=
.=20
</DIV>
<DIV>    A(x,m,b) =3D Sum_{n>=3D0} C(m*q^n + b, n)*x=
^n </DIV>
<DIV>to be manipulated to solve some functional equation ... </DIV>
<DIV> </DIV>
<DIV>Any ideas along these lines from anyone?</DIV>
<DIV>      Paul</DIV></DIV>
<DIV>  </DIV>
<DIV> </DIV>
<DIV>On Sat, 29 Dec 2007 00:20:34 -0500 <<A=20
href=3D"mailto:pauldhanna at juno.com">pauldhanna at juno.com</A>> writes:</DI=
V>
<BLOCKQUOTE dir=3Dltr=20
style=3D"PADDING-LEFT: 10px; MARGIN-LEFT: 10px; BORDER-LEFT: #000000 2px so=
lid">
<DIV></DIV>
<DIV>Seqfans, <BR>        Just=20
found an o.g.f. for A014070(n) =3D C(2^n,n): </DIV>
<DIV> </DIV>
<DIV>(1) G.f.: A(x) =3D Sum_{n>=3D0} log(1 + 2^n*x)^n / n!.</DIV>
<DIV>  </DIV>
<DIV>More generally, we have the nontrivial identity: </DIV>
<DIV> </DIV>
<DIV>(2) Sum_{n>=3D0} log(1 + q^n*x)^n/n!  =3D  Sum_{n>=
=3D0}=20
C(q^n,n)*x^n. </DIV>
<DIV> </DIV>
<DIV>The proof is easy given the following formula for A008275, </DI=
V>
<DIV>the Stirling numbers of the first kind: </DIV>
<DIV>   E.g.f. for k-th column of A008275 is ln(1+x)^k/k!.=
=20
</DIV>
<DIV> </DIV>
<DIV>Can any other interesting result be derived from (2)? </DIV>
<DIV>Thanks, </DIV>
<DIV>     Paul </DIV>
<DIV> </DIV></BLOCKQUOTE></BODY></HTML>

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