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 A204213 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. 9
 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, 127, 1, 8, 49, 271, 1025, 2208, 1925, 323, 1, 9, 64, 434, 2181, 7167, 12492, 8034, 835, 1, 10, 81, 652, 4116, 18583, 51945, 72589, 34188, 2188, 1, 11, 100, 933, 7120, 41363, 164255, 387000, 430569, 147787, 5798 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Table starts ...1....1.....1......1.......1.......1........1........1........1.........1 ...2....3.....4......5.......6.......7........8........9.......10........11 ...4....9....16.....25......36......49.......64.......81......100.......121 ...9...32....78....155.....271.....434......652......933.....1285......1716 ..21..120...404...1025....2181....4116.....7120....11529....17725.....26136 ..51..473..2208...7167...18583...41363....82440...151125...259459....422565 .127.1925.12492..51945..164255..431445...991152..2057553..3945655...7098949 .323.8034.72589.387000.1493142.4629851.12262470.28832499.61766005.122779448 LINKS R. H. Hardin, Table of n, a(n) for n = 1..9999 FORMULA Empirical for rows: T(1,k) = 1 T(2,k) = k + 1 T(3,k) = k^2 + 2*k + 1 T(4,k) = (4/3)*k^3 + (7/2)*k^2 + (19/6)*k + 1 T(5,k) = (23/12)*k^4 + (37/6)*k^3 + (91/12)*k^2 + (13/3)*k + 1 T(6,k) = (44/15)*k^5 + (133/12)*k^4 + 17*k^3 + (161/12)*k^2 + (167/30)*k + 1 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 ... 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 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 EXAMPLE Some solutions for n=5 k=3 ..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0 ..3....0....1....2....2....2....1....3....0....3....1....2....2....3....2....1 ..5....0....4....5....1....1....4....4....2....4....0....4....5....1....0....1 ..5....1....2....4....3....0....1....1....3....1....1....2....5....3....2....2 ..2....3....1....2....0....2....0....0....1....3....1....2....2....1....2....1 ..0....0....0....0....0....0....0....0....0....0....0....0....0....0....0....0 MATHEMATICA 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]; Table[T[n - k + 1, k], {n, 1, 11}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Sep 24 2019, after Vladimir Kruchinin *) PROG (PARI) {L(n, k)=polcoeff( ( (1-x^(2*k+1))/(1-x) +x*O(x^(k*n)) )^n, k*n)} {T(n, k)=polcoeff(exp(sum(m=1, n, L(m, k)*x^m/m)+x*O(x^n)), n)} for(n=1, 10, for(k=1, 10, print1(T(n, k), ", ")); print("")) \\ Paul D. Hanna, Aug 01 2013 (Maxima) 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 */ CROSSREFS Column 1 is A001006; column 2 is A104184; column 3 is A204208. Row 4 is A051662. Sequence in context: A055208 A051128 A137614 * A143326 A327086 A186686 Adjacent sequences:  A204210 A204211 A204212 * A204214 A204215 A204216 KEYWORD nonn,tabl AUTHOR R. H. Hardin, Jan 12 2012 STATUS approved

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Last modified May 31 00:29 EDT 2020. Contains 334747 sequences. (Running on oeis4.)