[Fwd: coefficients of Moonshine (McKay-Thompson) series]

[N. J. A. Sloane]

Equation 1, below, also fails for n = 23974881469864850330128892 and
other values.

I suggest a modified version:

 a(n) = floor(n/(2 - exp(gamma)) - 1/2 + exp(gamma)/(24n))

Offhand, I do not know of any values for which my modification fails.
But if there are such values, then a more accurate argument of the
floor function could be employed...

David W. Cantrell


----- Original Message ----- 
From: "N. J. A. Sloane" <njas at research.att.com>
To: <seqfan at ext.jussieu.fr>
Cc: <njas at research.att.com>
Sent: Friday, August 10, 2007 06:40
Subject: Conjectured formula for A004081 from Jon Schoenfield


>
> Jon Schoenfield, <jonscho at hiwaay.net>, who is not presently a member
> of this
> list, suggested that I post the following message, which
> seems very interesting. - Neil
>
>>From: "Jon Schoenfield" <jonscho at hiwaay.net>
>>To: <njas at research.att.com>
>>Subject: Equation that works for A004081 -- with only 1 exception??
>>Date: Thu, 9 Aug 2007 22:08:41 -0500
>>
>>Neil,
>>
>>I've spent a while on this ... too long, given my current
>>responsibilities
>>at home and at work <sigh>, which is why I'm putting it away for
>>now, and
>>turning it over to you in case you or the SeqFan list might find it
>>interesting.
>>
>>Regarding A004081, i.e., "a(n) = n-th positive integer such that
>>only one
>>integer lies between exp(s(m)) and exp(s(m+1)), where s(m) = 1 + 1/2
>>+ 1/3 +
>>. . . + 1/m," I've discovered that the following equation (I'll call
>>it
>>"Equation 1"),
>>
>>       a(n) = int(n / (2 - exp(Gamma)) - 0.5),
>>
>>gives the correct result for A004081(n) for all n < 11577.  At
>>n=11577, it
>>gives a(11577)=int(52879.999997856) = 52879, but the correct result
>>is 52880
>>since exp(s(52880)) = 94184.000000934 and exp(s(52881)) =
>>94185.781073352.
>>>From additional tests, I've concluded that, except at the single
>>>anomalous
>>case n = 11577, Equation 1 gives the correct value for A004081(n)
>>for all n
>>up to at least 2x10^13, and probably far beyond that.
>>
>>It seems to me that the best (and probably only necessary) values of
>>n to
>>examine for a failure of Equation 1 to give the correct value for
>>A004081(n)
>>are those values where n / (2 - exp(Gamma)) - 0.5 falls very close
>>to an
>>integer.  It looks as though these values of n can be obtained by
>>dividing
>>by two any even denominator that appears in one of the convergents
>>or
>>semiconvergents of the continued fraction for 1 / (2 - exp(Gamma)),
>>where
>>Gamma is Euler's constant; doing this, I get the values n = 1, 8,
>>15, 22,
>>59, 96, 443, 2562, 5567, 8572, 11577, 222968, 457513, 692058, ....
>>If, for
>>each of these values of n (and many that follow in that sequence!),
>>we
>>evaluate
>>
>>                        x = n / (2 - exp(Gamma)) - 0.5
>>and compute
>>                        m = round(x)
>>and then
>>                        p = exp(s(m)) = m * exp(Gamma) *
>> exp(1/(2*m)) /
>>exp(1/(12*m^2)) * exp(1/(120*m^4)) / exp(1/(252*m^6)) * ...
>>and
>>                        q = round(p)
>>
>>then I think Equation 1 gives the correct value for A004081(n) if
>>and only
>>if x and p are BOTH below or BOTH above their respective nearest
>>integers.
>>(The value given by Equation 1 at n = 11577 is not the correct value
>>for
>>A004081(11577); x = 52879.999997856 is less than m = 52880, but p =
>>94184.000000934 is greater than q = 94184.)
>>
>>I'm very curious to know whether n = 11577 is the only case where
>>Equation 1
>>doesn't give the correct result for A004081(n).
>>
>>Best wishes,
>>
>>-- Jon
>>
>>




Dear Seqfans,

Do you know some reference about the so-called "first child
- next sibling" bijection from ordered trees to binary
trees?

Thanks.
Emeric

P.S. Sorry, in my previous message I faile to complete the



Dear seqfans,

Do you know some reference about the so-called "first child
- next sibling" bijection from ordered trees to binary
trees?

Thanks. 
Emeric