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<DIV>Seqfans,</DIV>
<DIV> There is a simple answer my own
questions (in prior email). <BR>Since asinh(x) = log(sqrt(1+x^2) + x),
</DIV>
<DIV>then the sum (13) fits the form:</DIV>
<DIV> Sum_{n>=0} log(F(q^n*x))^n/n! = Sum_{n>=0} x^n
[y^n] F(y)^(q^n)</DIV>
<DIV>where F(y) = sqrt(1+y^2) + y .</DIV>
<DIV> </DIV>
<DIV>Thus, </DIV>
<DIV>G(x,q) = Sum_{n>=0} asinh( q^n*x )^n / n! = </DIV>
<DIV> Sum_{n>=0} x^n [y^n] ( sqrt(1+y^2) + y )^(q^n) </DIV>
<DIV> </DIV>
<DIV>When q=2,</DIV>
<DIV>
<DIV>G(x,2) = Sum_{n>=0} asinh( 2^n*x )^n / n! = </DIV>
<DIV> Sum_{n>=0} x^n [y^n] ( sqrt(1+y^2) + y )^(2^n) =
</DIV></DIV>
<DIV> 1 + 2*x + 8*x^2 + 84*x^3 + 2688*x^4 + 276892*x^5 + 94978048*x^6
+...</DIV>
<DIV> </DIV>
<DIV>then to get the sum of (12) we simply take the
bisection: (G(x,2) - G(-x,2))/2. </DIV>
<DIV> </DIV>
<DIV>Thanks, </DIV>
<DIV> Paul </DIV>
<DIV> </DIV>
<BLOCKQUOTE dir=ltr
style="PADDING-LEFT: 10px; MARGIN-LEFT: 10px; BORDER-LEFT: #000000 2px solid">
<DIV>Here is an example using the hyperbolic sine series applied on the
inverse sinh:<BR>(12) G.f.: A(x) = Sum_{n>=0} asinh( 2^(2n+1)*x )^(2n+1) /
(2n+1)! = <BR>2*x + 84*x^3 + 276892*x^5 + 111457917800*x^7 +
<BR>6660816097416169260*x^9 + 66597307693046550483175282456*x^11
+<BR>120167520447600665027319450022840022638104*x^13 +...</DIV>
<DIV> </DIV>
<DIV>Is there a simple formula for the integer coefficients on the right side
of (12)? <BR> </DIV>
<DIV> </DIV>
<DIV>Of course (12) is an example of the more general: </DIV>
<DIV>(13) G.f.: A(x) = Sum_{n>=0} asinh( q^(2n+1)*x )^(2n+1) /
(2n+1)! = ? <BR>yielding some unknown integer series for
all integer q. </DIV>
<DIV> </DIV>
<DIV>Is there a simple formula for the integer coefficients on the right side
of (13)? </DIV>
<DIV>It would be nice if it turned out to be as simple a formula as:
</DIV>
<DIV>(2) Sum_{n>=0} log(1 + q^n*x)^n/n! = Sum_{n>=0}
C(q^n,n)*x^n. </DIV>
<DIV> </DIV></BLOCKQUOTE></BODY></HTML>