A393385 draft (necklace graph): should a(0) be 1 or 3?

[Robert McKone]

Hello SeqFan,

For A393385 - OEIS "number of minimal edge covers in the n-necklace graph", there is a question about what to do with a(0).

For n >= 1, the counts begin:

7, 69, 571, 4729, 39227

A transfer-matrix/trace derivation leads to the g.f.:

(3 - 14*x - 10*x^2)/(1 - 7*x - 10*x^2 - 6*x^3)

If you try and interpret n = 0 combinatorially as the "empty necklace" (the empty graph), there is exactly 1 minimal edge cover (the empty set) that corresponds to the normalised g.f.:

(1 + 10*x^2 + 12*x^3)/(1 - 7*x - 10*x^2 - 6*x^3)

which differs from the trace g.f. by a constant (normalised = trace - 2), and therefore leaves all coefficients for n >= 1 unchanged.

The draft discussion is here:
https://oeis.org/draft/A393385

Question:  For cyclically-constructed graph families that are only canonically defined for n >= 1, is it better to (a) keep the recurrence/trace continuation a(0) = 3, (b) normalised so that a(0) = 1 matches the "empty object" convection, or (c) set OFFSET 1 and avoid n = 0 entirely?

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Thanks,

Robert P. P. McKone

Game Designer | Mathematician | Electrical Engineer

LinkedIn:  https://www.linkedin.com/in/robert-mckone/

[Max Alekseyev]

I would suggest (b), that is, make the sequence term a(0) to match the given combinatorial definition. Correspondingly, the g.f. would need to changed to 

(3 - 14*x - 10*x^2)/(1 - 7*x - 10*x^2 - 6*x^3) - 2

= (1 + 10*x^2 + 12*x^3)/(1 - 7*x - 10*x^2 - 6*x^3)

Regards,

Max

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