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%I #8 Sep 04 2019 10:28:48
%S 4,10,6,20,20,9,35,50,40,13,56,105,125,76,19,84,196,315,295,147,28,
%T 120,336,686,889,711,287,41,165,540,1344,2254,2567,1730,556,60,220,
%U 825,2430,5040,7586,7483,4175,1077,88,286,1210,4125,10242,19374,25774,21631,10077
%N T(n,k)=Number of length n+2 0..k arrays with no consecutive three elements summing to more than k
%C Table starts
%C ...4...10....20.....35......56.......84......120.......165.......220........286
%C ...6...20....50....105.....196......336......540.......825......1210.......1716
%C ...9...40...125....315.....686.....1344.....2430......4125......6655......10296
%C ..13...76...295....889....2254.....5040....10242.....19305.....34243......57772
%C ..19..147...711...2567....7586....19374....44274.....92697....180829.....332761
%C ..28..287..1730...7483...25774....75180...193194....449295....963886....1934647
%C ..41..556..4175..21631...86828...289248...835812...2159025...5093737...11151140
%C ..60.1077.10077..62547..292621..1113348..3617703..10380183..26932543...64309245
%C ..88.2091.24377.181255..988303..4294574.15692003..50011289.142701909..371651553
%C .129.4057.58928.524877.3335451.16553380.68014233.240772037.755538278.2146210209
%H R. H. Hardin, <a href="/A241619/b241619.txt">Table of n, a(n) for n = 1..10011</a>
%F Empirical for column k, apparently a recurrence of order (k+1)*(k+2)/2:
%F k=1: a(n) = a(n-1) +a(n-3)
%F k=2: a(n) = a(n-1) +a(n-2) +2*a(n-3) -a(n-5) -a(n-6)
%F 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)
%F k=4: [order 15]
%F k=5: [order 21]
%F k=6: [order 28]
%F k=7: [order 36]
%F k=8: [order 45]
%F k=9: [order 55]
%F k=10: [order 66]
%F k=11: [order 78]
%F k=12: [order 91]
%F Empirical for row n, apparently a polynomial of degree n+2:
%F n=1: a(n) = (1/6)*n^3 + 1*n^2 + (11/6)*n + 1
%F n=2: a(n) = (1/12)*n^4 + (2/3)*n^3 + (23/12)*n^2 + (7/3)*n + 1
%F n=3: a(n) = (1/24)*n^5 + (5/12)*n^4 + (13/8)*n^3 + (37/12)*n^2 + (17/6)*n + 1
%F n=4: [polynomial of degree 6]
%F n=5: [polynomial of degree 7]
%F n=6: [polynomial of degree 8]
%F n=7: [polynomial of degree 9]
%F From _Robert Israel_, Sep 04 2019: (Start)
%F 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).
%F 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)
%e Some solutions for n=5 k=4
%e ..1....0....2....1....0....2....0....1....0....0....0....2....1....0....1....2
%e ..0....0....1....3....0....0....4....2....1....0....1....1....3....0....0....1
%e ..0....3....0....0....0....1....0....1....3....0....0....0....0....2....1....0
%e ..0....0....0....1....2....0....0....0....0....0....1....0....1....1....1....2
%e ..2....0....3....0....0....2....0....0....0....0....2....1....0....0....0....0
%e ..0....1....0....1....2....1....2....0....1....0....0....1....1....0....3....1
%e ..0....0....0....1....0....0....2....4....2....2....0....0....2....0....0....0
%p for m from 1 to 12 do
%p r:= [seq(seq([i,j],j=0..m-i),i=0..m)];
%p 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):
%p U[m,0]:= Vector((m+1)*(m+2)/2,1);
%p od:
%p R:= NULL:
%p for i from 2 to 12 do
%p for j from 1 to i-1 do
%p U[i-j,j]:= T[i-j] . U[i-j,j-1];
%p R:= R, convert(U[i-j,j],`+`)
%p od od:
%p R; # _Robert Israel_, Sep 04 2019
%Y Column 1 is A000930(n+4)
%Y Row 1 is A000292(n+1)
%Y Row 2 is A002415(n+2)
%Y Row 3 is A006414
%Y Row 4 is A114244
%K nonn,tabl
%O 1,1
%A _R. H. Hardin_, Apr 26 2014
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