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<DIV>Seqfans, <BR> Here is
an accidental observation somewhat related to my question. </DIV>
<DIV>Consider Vladeta Jovovic's nice sequence A060435, where: </DIV>
<DIV>A060435(n) = number of functions f:{1,2,...,n}->{1,2,...,n} with even
cycles only. </DIV>
<DIV> </DIV>
<DIV>It has the rather complicated formula for the n-th term:</DIV>
<DIV> </DIV>
<DIV><FONT face="Courier New">a(n)=(n-1)!*Sum_{k=0..floor((n-2)/2)}
(k+1)/2^(2*k+1)*binomial(2*k+2, k+1)*n^(n-2-2*k)/(n-2-2*k)!.</FONT></DIV>
<DIV> </DIV>
<DIV>(how does Vladeta do that?!). </DIV>
<DIV> </DIV>
<DIV>Now if we square of the e.g.f. of A060435 after we include
the terms </DIV>
<DIV>A060435(0)=1, A060435(1)=0, </DIV>
<DIV>then we find that the NEW sequence (A134095) has a
simple formula: <BR> <BR>a(n) = Sum_{k=0..n} C(n,k) * (n-k)^k *
k^(n-k). <BR> </DIV>
<DIV>and the e.g.f. follows from Vladeta's formula for A060435: </DIV>
<DIV> </DIV>
<DIV>E.g.f.: A(x) = 1/(1 - LambertW(-x)^2 ).</DIV>
<DIV> </DIV>
<DIV>Below I supply the initial terms of A134095 and the PARI code.
</DIV>
<DIV> </DIV>
<DIV>Just one of many accidental coincidences of the OEIS. </DIV>
<DIV> Paul </DIV>
<DIV> </DIV>
<DIV>A134095 begins:
<BR>1,0,2,12,120,1480,22320,396564,8118656,188185680,4871980800,</DIV>
<DIV>139342178140,4363291266048,148470651659928,5455056815237120,</DIV>
<DIV>215238256785814500,9077047768435752960,407449611073696325536,</DIV>
<DIV> <BR>(PARI)<BR>a(n)=sum(k=0,n,(n-k)^k*k^(n-k)*binomial(n,k))<BR>/*
Generated by E.G.F. 1/(1 - LambertW(-x)^2 ):
*/<BR>{a(n)=local(LambertW=-x*sum(k=0,n,(-x)^k*(k+1)^(k-1)/k!)
+x*O(x^n));<BR>n!*polcoeff(1/(1-subst(LambertW,x,-x)^2),n)}<BR> </DIV>
<DIV>END.</DIV>
<DIV>On Sat, 6 Oct 2007 01:38:29 -0400 Paul D. Hanna <<A
href="mailto:pauldhanna@juno.com">pauldhanna@juno.com</A>> writes:<BR>>
Seqfans,<BR>> Consider the sequence:<BR>>
A062817(n) = Sum_{k=0..n} (n-k)^k*k^(n-k), n>=1<BR>> 0, 1, 4, 22,
152, 1251, 11980, 130908, 1607488, ...<BR>> Examples: <BR>> A062817(4) =
3^1*1^3 + 2^2*2^2 + 1^3*3^1 = 22 ;<BR>> A062817(5) = 4^1*1^4 + 3^2*2^3 +
2^3*3^2 + 1^4*4^1 = 152 ;<BR>> A062817(6) = 5^1*1^5 + 4^2*2^4 + 3^3*3^3 +
2^4*4^2 + 1^5*5^1 = 1251 <BR>> ;<BR>> <BR>> This formula
looks similar to a self-convolution ... and, <BR>> surprisingly (to me),
<BR>> it is indeed a self-convolution of an integer sequence: A132608
<BR>> (below). <BR>> <BR>> Can anyone find another (simpler?)
formula or g.f. for either <BR>> A062817 or A132608? <BR>> <BR>>
Also, what is the value of the limit: <BR>>
Limit_{n->infinity} [A062817(n+1)/A062817(n)] / n = c <BR>> <BR>>
At n=12000, c ~ 1.35936744... and is decreasing. <BR>> Does the
limit c = exp(1)/2 ? <BR>> <BR>> Thanks,
<BR>> Paul <BR>>
------------------------------------------ <BR>> A132608 <BR>> <BR>>
1,2,9,58,469,4530,50491,634790,8861043,135750454,2262315973,<BR>>
40726646802,787471241647,16275700505510,358103286781293,<BR>>
8357593147404346,206241859929682177,5366082228239257410<BR>> <BR>>
Self-convolution square-root of A062817 (offset 2); <BR>> thus g.f. A(x)
satisfies: <BR>> A(x)^2 = Sum(n>=2} A062817(n)*x^n,
<BR>> where A062817(n) = Sum_{k=0..n} (n-k)^k*k^(n-k).<BR>> <BR>>
EXAMPLE.<BR>> A(x) = x + 2x^2 + 9x^3 + 58x^4 + 469x^5 + 4530x^6 +...+
a(n)*x^n <BR>> +...<BR>> A(x)^2 = x^2 + 4x^3 + 22x^4 + 152x^5 + 1251x^6
+...+ A062817(n)*x^n <BR>> +...<BR>> <BR>> (PARI) <BR>>
{a(n)=polcoeff((sum(m=2,n+1,sum(k=0,m,(m-k)^k*k^(m-k))*x^m <BR>>
+x*O(x^(n+1))))^(1/2),n)}<BR>> </DIV>
<DIV> </DIV>
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