[seqfan] Re: The Sequencer OEIS survey‏‏

[David Corneth]

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 from Clark
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