OFFSET
1,1
COMMENTS
Column 4 of A267245.
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..210
Robert Israel, Maple-assisted proof of empirical recurrence
Index entries for linear recurrences with constant coefficients, signature (24, -246, 1420, -5121, 12084, -18944, 19536, -12720, 4736, -768).
FORMULA
Empirical: a(n) = 24*a(n-1) -246*a(n-2) +1420*a(n-3) -5121*a(n-4) +12084*a(n-5) -18944*a(n-6) +19536*a(n-7) -12720*a(n-8) +4736*a(n-9) -768*a(n-10).
Empirical formula verified (see link). - Robert Israel, Sep 08 2019
EXAMPLE
Some solutions for n=4
..0..0..0..0....0..0..0..0....0..0..1..1....0..0..1..1....0..0..0..1
..0..0..0..0....0..0..0..1....0..0..1..1....0..1..0..1....0..1..1..0
..0..1..1..1....0..1..1..0....0..1..1..1....1..0..1..0....0..1..1..1
..1..0..1..1....0..1..1..0....1..0..1..1....1..0..1..0....0..1..1..1
MAPLE
states:= select(proc(x) (x[1]=x[2] or x[5]=1) and (x[2]=x[3] or x[6]=1) and (x[3]=x[4] or x[7]=1) end proc, [seq(seq(seq(seq(seq(seq(seq([a, b, c, d, e, f, g], g=0..1), f=0..1), e=0..1), d=0..1), c=0..1), b=0..1), a=0..1)]):
T:= Matrix(54, 54, proc(i, j) local k;
if add(states[j, k]-states[i, k], k=1..4) > 0 then return 0 fi;
if states[j, 5]>states[i, 5] or states[j, 6]>states[i, 6] or states[j, 7]>states[i, 7] then return 0 fi;
if states[i, 1]>=states[i, 2] and states[j, 5]<> states[i, 5] then return 0 fi;
if states[i, 2]>=states[i, 3] and states[j, 6]<> states[i, 6] then return 0 fi;
if states[i, 3]>=states[i, 4] and states[j, 7]<> states[i, 7] then return 0 fi;
1
end proc):
U:= Vector(54, 1):
E[0]:= Vector(54): E[0][1]:= 1:
for k from 1 to 25 do E[k]:= T . E[k-1] od:
seq(U^%T . E[j], j=1..25); # Robert Israel, Sep 08 2019
MATHEMATICA
LinearRecurrence[{24, -246, 1420, -5121, 12084, -18944, 19536, -12720, 4736, -768}, {5, 22, 105, 567, 3351, 20676, 129129, 804817, 4982759, 30629206, 187121865}, 25] (* Jean-François Alcover, Oct 25 2022, after Robert Israel *)
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Jan 12 2016
STATUS
approved