[seqfan] Re: The Sequencer OEIS survey‏‏

[apovolot at gmail.com]

The generating function for A145052 is:

(3 + 4*x+ 4*x^2 - 5*x^3)/(1 - x - 3*x^3 + 3*x^4)

Indeed 

Taylor series expansion of
 (3 + 4*x+ 4*x^2 - 5*x^3)/(1 - x - 3*x^3 + 3*x^4)
at x=0 

Gives 
3+7*x+11*x^2+15*x^3+27*x^4+39*x^5+51*x^6+87*x^7+123*x^8+159*x^9+267*x^10+O(x^11) 

Alexander R. Povolotsky

On Apr 14, 2015, at 7:14 PM, David Corneth <davidacorneth at gmail.com> wrote:
> 
> Hi all,
> 
> For A145052, the first few terms are 3, 7, 11, 15, 27, ...
> Note how the first differences are 4, 4, 4, 12, 12, 12, 36, 36, 36, ...
> based on the first differences, I'd like to propose:
> a(n) = 6*(3^(n\3)/3 - 1) + 3 + 4 * (n mod 3) * 3^(n \ 3)/3. where n\k
> denotes integer division; n\k = floor(n/k).
> 
> So we merely have a geometric sum for the n divisible by 3. For the others
> (n not divisible by 3), add a bit as shown in the formula.
> 
> For A027983, I'd like to conjecture, based on the formula Sent from my iPad
> Kimberling <https://oeis.org/wiki/User:Clark_Kimberling> I presume that T(n,
> k) = Lucas(k+1), and the Lucas numbers that
> a(n) = 2 * a(n - 1) + L(n + 1), where L(m) is a Lucas number, as in A000032.
> 
> I feel a bit reluctant to put the conjectures in the sequence as they are
> being discussed here.
> 
> Best,
> David
> 
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