bibkey: barker2019a327692 authors: Colin Barker year: 2019 title: “OEIS A327692, Number of length-n phone numbers that can be dialed by a chess knight on a 0-9 keypad that starts on any number and takes n-1 steps” doi: null url: https://oeis.org/A327692 claim: “Conjectures from Colin Barker, Oct 01 2019: (Start) G.f.: 2x(5 + 10x - 7x^2 - 8x^3 + 2x^4) / (1 - 6x^2 + 4x^4). a(n) = 6a(n-2) - 4a(n-4) for n>6. (End)” strata_touched:
- D5/S3/Combinatorics/KnightDiallerRecurrence license: citation-only triage: anchor
OEIS A327692
The entry counts the digit sequences of length n that a chess knight can dial
on the keypad
1 2 3
4 5 6
7 8 9
* 0 #
starting anywhere and taking n - 1 knight steps, the blank cells * and #
not being dialable. Its terms begin
10, 20, 46, 104, 240, 544, 1256, 2848, 6576, 14912, ….
Verified locator
- URL: https://oeis.org/A327692
- Locator: FORMULA, “Conjectures from Colin Barker, Oct 01 2019: (Start) G.f.: 2x(5 + 10x - 7x^2 - 8x^3 + 2x^4) / (1 - 6x^2 + 4x^4). a(n) = 6a(n-2) - 4a(n-4) for n>6. (End)”
- Revision read: #41, Apr 22 2024. The formula line still carries the word
Conjecturesand records no proof.
Reading of the statement
The recurrence asserted is a(n) = 6*a(n-2) - 4*a(n-4), for n > 6 in Barker’s
own range. A later comment by Francesca Arici, dated Apr 17 2024, adds that the
recurrence also holds at n = 6, and sketches a route: the count is the grand
sum of a power of the adjacency matrix, that matrix is diagonalisable over the
reals with one zero eigenvalue, and the assertion then “reduces to checking an
algebraic condition on the nonzero eigenvalues”. The sketch stops there; the
algebraic condition is not checked and the entry was not reclassified.
Natural subtraction would silently change the statement, so the recorded form is
additive: writing dial n for the count of sequences with n knight steps, so
that dial n is the entry’s a(n+1), the assertion is
dial (n+5) + 4 * dial (n+1) = 6 * dial (n+3) for every n.
That is exactly a(m) = 6*a(m-2) - 4*a(m-4) for every m >= 6, which is
Barker’s conjecture together with the extension to m = 6 that Arici reports.
Scope of the recorded answer
The assertion holds for every n, and the reason is a single dead key.
Let u n be the vector whose entry at digit d is the number of dialable
sequences of n steps starting at d, so u 0 is all ones, u (n+1) is the
knight transfer step applied to u n, and dial n is the sum of the entries of
u n. The transfer step is linear, so it carries the identity forward: if
u (n+5) + 4 * u (n+1) = 6 * u (n+3) holds entrywise at one place, it holds at
the next.
The base case is an evaluation. In matrix language the residual of the recurrence applied to the all-ones vector is not zero,
(A^4 - 6A^2 + 4I) · 1 = (0,0,0,0,0,4,0,0,0,0),
the whole residual sitting on the digit 5; one further step kills it, because
5 is the one key a knight can never leave, both of its knight images being the
blank cells. Concretely
u 1 = (2,2,2,2,3,0,3,2,2,2)
u 3 = (12,10,10,10,16,0,16,10,10,10)
u 5 = (64,52,52,52,84,0,84,52,52,52)
and u 5 + 4 * u 1 = (72,60,60,60,96,0,96,60,60,60) = 6 * u 3.
The same reading explains the exponent bookkeeping exactly. Summing the residual
over all starting digits gives 4 at m = 5, so the recurrence fails there:
a(5) + 4*a(1) = 280 against 6*a(3) = 276. It first holds at m = 6, which is
the range the entry’s later comment reports.
No eigenvalues, no diagonalisability and no real spectrum enter, so the route is shorter than the one the entry sketches.
Bounded prior-resolution evidence
The entry was read in full at revision #41 and records no proof: the formula line
still says Conjectures, and the later comment ends at a reduction. Its
cross-references A280594 and A169696 carry no proof of this recurrence. Searches
for a published proof return knight-dialler programming material, which counts
paths from a fixed start rather than the grand sum and does not address the
closed recurrence. Citation indices were not exhaustively reachable, so this is a
bounded negative finding and no worldwide priority claim is made.