cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A200785 T(n,k) is the number of arrays of n+2 elements from {0,1,...,k} with no two consecutive ascents.

Original entry on oeis.org

8, 26, 16, 60, 75, 32, 115, 225, 216, 64, 196, 530, 840, 622, 128, 308, 1071, 2425, 3136, 1791, 256, 456, 1946, 5796, 11100, 11704, 5157, 512, 645, 3270, 12152, 31395, 50775, 43681, 14849, 1024, 880, 5175, 23136, 75992, 169884, 232275, 163020, 42756, 2048
Offset: 1

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Author

R. H. Hardin Nov 22 2011

Keywords

Comments

All the conjectured formulas are true, and follow from the Burstein-Mansour paper. - N. J. A. Sloane, May 21 2013

Examples

			Table starts
....8.....26......60.......115.......196........308.........456.........645
...16.....75.....225.......530......1071.......1946........3270........5175
...32....216.....840......2425......5796......12152.......23136.......40905
...64....622....3136.....11100.....31395......75992......164004......324087
..128...1791...11704.....50775....169884.....474566.....1160616.....2562633
..256...5157...43681....232275....919413....2964416.....8216484....20273247
..512..14849..163020...1062500...4975322...18514405....58154912...160338680
.1024..42756..608400...4860250..26924106..115637431...411637168..1268210421
.2048.123111.2270580..22232375.145698840..722234149..2913595712.10030582998
.4096.354484.8473921.101698250.788446400.4510869636.20622837480.79335475611
Some arrays for n=4, k=3:
..0....1....0....0....1....0....3....3....0....1....3....0....2....2....2....2
..3....0....2....2....0....2....0....0....3....1....0....0....0....3....3....3
..2....3....2....2....2....2....3....3....1....0....1....0....2....1....3....3
..1....0....2....1....0....0....2....2....2....2....1....2....2....0....0....2
..0....3....0....0....1....2....1....2....0....0....3....2....0....3....1....3
..3....3....0....3....0....2....3....2....0....3....0....0....2....2....1....3
		

Crossrefs

Column 1 is A000079
Column 2 is A076264
Column 3 is A072335
Row 1 is A002413
Cf. A200781.

Formula

T(n-2,k) = \sum_{L=0}^n (-1)^L / L! * \sum_{M=0}^{min(L,[(n-L)/2])} binomial(n-L-M,M) * M! * (k+1)^(n-L-2*M) B_{L,M}(x_1,x_2,...), where B_{L,M}() are Bell polynomials, x_i = binomial(k+1,i+2) * i! * f(i), i=1,2,..., and f(i) has period of length 6: [0,1,1,0,-1,-1] (i.e., f(0)=0, f(1)=1, etc.). This formula implies that for a fixed n, T(n,k) is a polynomial in k, which is easy to compute. - Max Alekseyev, Dec 12 2011
Empirical formulas for columns:
k=1: a(n) = 2*a(n-1)
k=2: a(n) = 3*a(n-1) -a(n-3)
k=3: a(n) = 4*a(n-1) -4*a(n-3) +a(n-4)
k=4: a(n) = 5*a(n-1) -10*a(n-3) +5*a(n-4)
k=5: a(n) = 6*a(n-1) -20*a(n-3) +15*a(n-4) -a(n-6)
k=6: a(n) = 7*a(n-1) -35*a(n-3) +35*a(n-4) -7*a(n-6) +a(n-7)
k=7: a(n) = 8*a(n-1) -56*a(n-3) +70*a(n-4) -28*a(n-6) +8*a(n-7)
Empirical recurrence for general column k:
0 = sum{i=0..floor(k/3) (binomial(k+1,3*i+1)*T(n-(3*i+1),k))} - sum{i=0..floor((k+1)/3) (binomial(k+1,3*i)*T(n-3*i,k))}
Formulae for rows:
T(1,k) = (5/6)*k^3 + 3*k^2 + (19/6)*k + 1
T(2,k) = (17/24)*k^4 + (43/12)*k^3 + (151/24)*k^2 + (53/12)*k + 1
T(3,k) = (7/12)*k^5 + (47/12)*k^4 + (39/4)*k^3 + (133/12)*k^2 + (17/3)*k + 1
T(4,k) = (349/720)*k^6 + (321/80)*k^5 + (1883/144)*k^4 + (1013/48)*k^3 + (3139/180)*k^2 + (413/60)*k + 1
T(5,k) = (2017/5040)*k^7 + (1427/360)*k^6 + (5759/360)*k^5 + (607/18)*k^4 + (28459/720)*k^3 + (9113/360)*k^2 + (848/105)*k + 1
T(6,k) = (6679/20160)*k^8 + (4799/1260)*k^7 + (26449/1440)*k^6 + (2162/45)*k^5 + (212153/2880)*k^4 + (6019/90)*k^3 + (174571/5040)*k^2 + (3893/420)*k + 1
T(7,k) = (99377/362880)*k^9 + (48247/13440)*k^8 + (243673/12096)*k^7 + (60529/960)*k^6 + (2076437/17280)*k^5 + (274529/1920)*k^4 + (952027/9072)*k^3 + (152461/3360)*k^2 + (26399/2520)*k + 1

A225682 Triangle read by rows: T(n,k) (0 <= k <= n) = chi(k)*binomial(n,k), where chi(k) = 1,-1,0 according as k == 0,1,2 mod 3.

Original entry on oeis.org

1, 1, -1, 1, -2, 0, 1, -3, 0, 1, 1, -4, 0, 4, -1, 1, -5, 0, 10, -5, 0, 1, -6, 0, 20, -15, 0, 1, 1, -7, 0, 35, -35, 0, 7, -1, 1, -8, 0, 56, -70, 0, 28, -8, 0, 1, -9, 0, 84, -126, 0, 84, -36, 0, 1, 1, -10, 0, 120, -210, 0, 210, -120, 0, 10, -1, 1, -11, 0, 165, -330, 0, 462, -330, 0, 55, -11, 0, 1, -12, 0, 220, -495, 0, 924, -792, 0, 220, -66, 0, 1
Offset: 0

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Author

Keywords

Comments

Corresponding to row n of this triangle, define a generating function G_n(x) = 1/(Sum_{k=0..n} T(n,k)*x^k).
Then G_n(x) is the g.f. for the number of words of length n over an alphabet of size n which do not contain any strictly decreasing factor (consecutive subword) of length 3.
For example, G_2, G_3, G_4, G_5, G_6 are g.f.'s for A000079, A076264, A072335, A200781, A200782.

Examples

			Triangle begins:
[1],
[1, -1],
[1, -2, 0],
[1, -3, 0, 1],
[1, -4, 0, 4, -1],
[1, -5, 0, 10, -5, 0],
[1, -6, 0, 20, -15, 0, 1],
[1, -7, 0, 35, -35, 0, 7, -1],
[1, -8, 0, 56, -70, 0, 28, -8, 0],
...
		

Crossrefs

Programs

  • Maple
    f:=proc(n) local k,s;
    s:=k->if k mod 3 = 0 then 1 elif k mod 3 = 1 then -1 else 0; fi;
    [seq(s(k)*binomial(n,k),k=0..n)];
    end;
    [seq(f(n),n=0..12)];
  • Mathematica
    chi[k_] := Switch[Mod[k, 3], 0, 1, 1, -1, 2, 0]; t[n_, k_] := chi[k]*Binomial[n, k]; Table[t[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 14 2014 *)
Showing 1-2 of 2 results.