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%I #23 Sep 24 2019 09:19:49
%S 1,1,2,1,3,4,1,4,9,9,1,5,16,32,21,1,6,25,78,120,51,1,7,36,155,404,473,
%T 127,1,8,49,271,1025,2208,1925,323,1,9,64,434,2181,7167,12492,8034,
%U 835,1,10,81,652,4116,18583,51945,72589,34188,2188,1,11,100,933,7120,41363,164255,387000,430569,147787,5798
%N T(n,k) = Number of length n+1 nonnegative integer arrays starting and ending with 0 with adjacent elements differing by no more than k.
%C Table starts
%C ...1....1.....1......1.......1.......1........1........1........1.........1
%C ...2....3.....4......5.......6.......7........8........9.......10........11
%C ...4....9....16.....25......36......49.......64.......81......100.......121
%C ...9...32....78....155.....271.....434......652......933.....1285......1716
%C ..21..120...404...1025....2181....4116.....7120....11529....17725.....26136
%C ..51..473..2208...7167...18583...41363....82440...151125...259459....422565
%C .127.1925.12492..51945..164255..431445...991152..2057553..3945655...7098949
%C .323.8034.72589.387000.1493142.4629851.12262470.28832499.61766005.122779448
%H R. H. Hardin, <a href="/A204213/b204213.txt">Table of n, a(n) for n = 1..9999</a>
%F Empirical for rows:
%F T(1,k) = 1
%F T(2,k) = k + 1
%F T(3,k) = k^2 + 2*k + 1
%F T(4,k) = (4/3)*k^3 + (7/2)*k^2 + (19/6)*k + 1
%F T(5,k) = (23/12)*k^4 + (37/6)*k^3 + (91/12)*k^2 + (13/3)*k + 1
%F T(6,k) = (44/15)*k^5 + (133/12)*k^4 + 17*k^3 + (161/12)*k^2 + (167/30)*k + 1
%F T(7,k) = (841/180)*k^6 + (101/5)*k^5 + (1325/36)*k^4 + (73/2)*k^3 + (946/45)*k^2 + (34/5)*k + 1
%F ...
%F G.f. for column k: exp( Sum_{n>=1} L(n,k)*x^n/n ) - 1, where L(n,k) = central coefficient of (1+x+x^2+x^3+...+x^(2*k))^n. - _Paul D. Hanna_, Aug 01 2013
%F T(n,k):=Sum_{i=1..n}((Sum_{j=0..(i*k)/(2*k+1)}(binomial(i,j)*(-1)^j*binomial(-j*(2*k+1)+i*(k+1)-1,i*k-j*(2*k+1))))*T(n-i,k))/n, T(0,k)=1. - _Vladimir Kruchinin_, Apr 06 2017
%e Some solutions for n=5 k=3
%e ..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0
%e ..3....0....1....2....2....2....1....3....0....3....1....2....2....3....2....1
%e ..5....0....4....5....1....1....4....4....2....4....0....4....5....1....0....1
%e ..5....1....2....4....3....0....1....1....3....1....1....2....5....3....2....2
%e ..2....3....1....2....0....2....0....0....1....3....1....2....2....1....2....1
%e ..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0
%t T[n_, k_] := T[n, k] = If[n == 0, 1, Sum[(Sum[Binomial[i, j]*(-1)^j* Binomial[-j*(2*k + 1) + i*(k + 1) - 1, i*k - j*(2*k + 1)], {j, 0, (i*k)/(2*k + 1)}])*T[n - i, k], {i, 1, n}]/n];
%t Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* _Jean-François Alcover_, Sep 24 2019, after _Vladimir Kruchinin_ *)
%o (PARI)
%o {L(n,k)=polcoeff( ( (1-x^(2*k+1))/(1-x) +x*O(x^(k*n)) )^n, k*n)}
%o {T(n,k)=polcoeff(exp(sum(m=1, n, L(m,k)*x^m/m)+x*O(x^n)), n)}
%o for(n=1, 10,for(k=1,10, print1(T(n,k), ", "));print("")) \\ _Paul D. Hanna_, Aug 01 2013
%o (Maxima)
%o T(n,k):=if n=0 then 1 else sum((sum(binomial(i,j)*(-1)^j*binomial(-j*(2*k+1)+i*(k+1)-1,i*k-j*(2*k+1)),j,0,(i*k)/(2*k+1)))*T(n-i,k),i,1,n)/n; /* _Vladimir Kruchinin_, Apr 06 2017 */
%Y Column 1 is A001006; column 2 is A104184; column 3 is A204208.
%Y Row 4 is A051662.
%K nonn,tabl
%O 1,3
%A _R. H. Hardin_, Jan 12 2012
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