A241619 T(n,k)=Number of length n+2 0..k arrays with no consecutive three elements summing to more than k

4, 10, 6, 20, 20, 9, 35, 50, 40, 13, 56, 105, 125, 76, 19, 84, 196, 315, 295, 147, 28, 120, 336, 686, 889, 711, 287, 41, 165, 540, 1344, 2254, 2567, 1730, 556, 60, 220, 825, 2430, 5040, 7586, 7483, 4175, 1077, 88, 286, 1210, 4125, 10242, 19374, 25774, 21631, 10077

Offset

1,1

Comments

Table starts

...4...10....20.....35......56.......84......120.......165.......220........286

...6...20....50....105.....196......336......540.......825......1210.......1716

...9...40...125....315.....686.....1344.....2430......4125......6655......10296

..13...76...295....889....2254.....5040....10242.....19305.....34243......57772

..19..147...711...2567....7586....19374....44274.....92697....180829.....332761

..28..287..1730...7483...25774....75180...193194....449295....963886....1934647

..41..556..4175..21631...86828...289248...835812...2159025...5093737...11151140

..60.1077.10077..62547..292621..1113348..3617703..10380183..26932543...64309245

..88.2091.24377.181255..988303..4294574.15692003..50011289.142701909..371651553

.129.4057.58928.524877.3335451.16553380.68014233.240772037.755538278.2146210209

Links

R. H. Hardin, Table of n, a(n) for n = 1..10011

Formula

Empirical for column k, apparently a recurrence of order (k+1)*(k+2)/2:

k=1: a(n) = a(n-1) +a(n-3)

k=2: a(n) = a(n-1) +a(n-2) +2*a(n-3) -a(n-5) -a(n-6)

k=3: a(n) = 2*a(n-1) +4*a(n-3) -3*a(n-4) -a(n-5) -3*a(n-6) +2*a(n-7) +a(n-9) -a(n-10)

k=4: [order 15]

k=5: [order 21]

k=6: [order 28]

k=7: [order 36]

k=8: [order 45]

k=9: [order 55]

k=10: [order 66]

k=11: [order 78]

k=12: [order 91]

Empirical for row n, apparently a polynomial of degree n+2:

n=1: a(n) = (1/6)*n^3 + 1*n^2 + (11/6)*n + 1

n=2: a(n) = (1/12)*n^4 + (2/3)*n^3 + (23/12)*n^2 + (7/3)*n + 1

n=3: a(n) = (1/24)*n^5 + (5/12)*n^4 + (13/8)*n^3 + (37/12)*n^2 + (17/6)*n + 1

n=4: [polynomial of degree 6]

n=5: [polynomial of degree 7]

n=6: [polynomial of degree 8]

n=7: [polynomial of degree 9]

From Robert Israel, Sep 04 2019: (Start)

Column k satisfies a recurrence of order (k+1)*(k+2)/2, since a(n)=e^T T^n e where T is a (k+1)*(k+2)/2 matrix and e the vector of all 1's (see proofs at A241615 and A241618).

Row n is the Ehrhart polynomial of degree n+2 corresponding to the polytope {(x(1),...,x(n+2)): all x(i)>=0, x(i)+x(i+1)+x(i+2)<=1 for i=1..n}, whose vertices have all entries in {0,1}. (End)

Example

Some solutions for n=5 k=4
..1....0....2....1....0....2....0....1....0....0....0....2....1....0....1....2
..0....0....1....3....0....0....4....2....1....0....1....1....3....0....0....1
..0....3....0....0....0....1....0....1....3....0....0....0....0....2....1....0
..0....0....0....1....2....0....0....0....0....0....1....0....1....1....1....2
..2....0....3....0....0....2....0....0....0....0....2....1....0....0....0....0
..0....1....0....1....2....1....2....0....1....0....0....1....1....0....3....1
..0....0....0....1....0....0....2....4....2....2....0....0....2....0....0....0

Maple

for m from 1 to 12 do
  r:= [seq(seq([i, j], j=0..m-i), i=0..m)];
  T[m]:= Matrix((m+1)*(m+2)/2, (m+1)*(m+2)/2, proc(i, j) if r[i][1]=r[j][2] and r[i][1]+r[i][2]+r[j][1]<=m then 1 else 0 fi end proc):
  U[m, 0]:= Vector((m+1)*(m+2)/2, 1);
od:
R:= NULL:
for i from 2 to 12 do
  for j from 1 to i-1 do
    U[i-j, j]:= T[i-j] . U[i-j, j-1];
    R:= R, convert(U[i-j, j], `+`)
od od:
R; # Robert Israel, Sep 04 2019

Crossrefs

Column 1 is A000930(n+4)

Row 1 is A000292(n+1)

Row 2 is A002415(n+2)

Row 3 is A006414

Row 4 is A114244

Sequence in context: A003564 A347116 A205016 * A129531 A298264 A014476

Adjacent sequences: A241616 A241617 A241618 * A241620 A241621 A241622

Keyword

nonn,tabl

Author

R. H. Hardin, Apr 26 2014

Status

approved