A346504
G.f. A(x) satisfies: A(x) = 1 + x + x^3 * A(x)^2 / (1 - x).
Original entry on oeis.org
1, 1, 0, 1, 3, 4, 6, 14, 28, 49, 95, 196, 386, 754, 1524, 3102, 6258, 12700, 26032, 53440, 109772, 226457, 468863, 972300, 2020274, 4208530, 8784556, 18365322, 38461110, 80682740, 169501696, 356579216, 751138916, 1584281062, 3345404514, 7072055268, 14965933024, 31702754496
Offset: 0
-
nmax = 37; A[] = 0; Do[A[x] = 1 + x + x^3 A[x]^2/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = a[1] = 1; a[2] = 0; a[n_] := a[n] = a[n - 1] + Sum[a[k] a[n - k - 3], {k, 0, n - 3}]; Table[a[n], {n, 0, 37}]
CoefficientList[Series[(1 - x)*(1 - Sqrt[(1 - x - 4*x^3 - 4*x^4)/(1 - x)]) / (2*x^3), {x, 0, 40}], x] (* Vaclav Kotesovec, Sep 27 2023 *)
A273344
Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k levels. A level in a bargraph is a maximal sequence of two or more adjacent horizontal steps; it is preceded and followed by either an up step or a down step.
Original entry on oeis.org
1, 1, 1, 3, 2, 6, 7, 14, 19, 2, 33, 53, 11, 79, 148, 47, 1, 194, 409, 181, 10, 482, 1137, 639, 69, 1214, 3159, 2166, 360, 6, 3090, 8793, 7110, 1646, 66, 7936, 24515, 22831, 6868, 490, 2, 20544, 68443, 72145, 26893, 2918, 44, 53545, 191367, 225138, 100598, 15085, 486, 140399, 535762, 695798, 363360, 70847, 3825
Offset: 2
Row 4 is 3,2 because the 5 (=A082582(4)) bargraphs of semiperimeter 4 correspond to the compositions [1,1,1], [1,2], [2,1], [2,2], [3] having 1, 0, 0, 1, 0 levels, respectively.
Triangle starts
1;
1,1;
3,2;
6,7;
14,19,2.
- Alois P. Heinz, Rows n = 2..250, flattened
- A. Blecher, C. Brennan, and A. Knopfmacher, Levels in bargraphs, Ars Math. Contemp., 9, 2015, 297-310.
- A. Blecher, C. Brennan, and A. Knopfmacher, Peaks in bargraphs, Trans. Royal Soc. South Africa, 71, No. 1, 2016, 97-103.
- M. Bousquet-Mélou and A. Rechnitzer, The site-perimeter of bargraphs, Adv. in Appl. Math. 31 (2003), 86-112.
-
G := (1-2*z-z^2+2*z^3-2*t*z^3-sqrt((1-z)*(1-3*z-z^2+3*z^3-4*t*z^3+4*z^4 -4*t*z^4-4*z^5+8*t*z^5-4*t^2*z^5)))/(2*z*(1-z+t*z)): Gser := simplify(series(G, z = 0, 25)): for n from 2 to 20 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 2 to 20 do seq(coeff(P[n], t, j), j = 0 .. degree(P[n])) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, y, t, w) option remember; expand(
`if`(n=0, (1-t), `if`(t<0, 0, b(n-1, y+1, 1, 0))+
`if`(t>0 or y<2, 0, b(n, y-1, -1, 0))+ `if`(y<1, 0,
`if`(w=1, z, 1)*b(n-1, y, 0, min(w+1, 2)))))
end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0$3)):
seq(T(n), n=2..20); # Alois P. Heinz, Jun 04 2016
-
b[n_, y_, t_, w_] := b[n, y, t, w] = Expand[If[n == 0, (1 - t), If[t < 0, 0, b[n - 1, y + 1, 1, 0]] + If[t > 0 || y < 2, 0, b[n, y - 1, -1, 0]] + If[y < 1, 0, If[w == 1, z, 1]*b[n - 1, y, 0, Min[w + 1, 2]]]]]; T[n_] := Function[p, Table[Coefficient[p, z, i], {i, 0, Exponent[p, z]}]][b[n, 0, 0, 0]]; Table[T[n], {n, 2, 20}] // Flatten (* Jean-François Alcover, Nov 29 2016 after Alois P. Heinz *)
A273349
Triangle read by rows: T(n,k) is the number of bargraphs of semiperimeter n having k level steps (n>=2,k>=0). A level step in a bargraph is any pair of adjacent horizontal steps at the same height.
Original entry on oeis.org
1, 1, 1, 3, 1, 1, 6, 5, 1, 1, 14, 12, 7, 1, 1, 33, 34, 19, 9, 1, 1, 79, 95, 61, 27, 11, 1, 1, 194, 261, 193, 95, 36, 13, 1, 1, 482, 728, 585, 333, 136, 46, 15, 1, 1, 1214, 2022, 1797, 1091, 521, 184, 57, 17, 1, 1, 3090, 5634, 5439, 3629, 1821, 763, 239, 69, 19, 1, 1
Offset: 2
Row 4 is 3,1,1 because the 5 (=A082582(4)) bargraphs of semiperimeter 4 correspond to the compositions [1,1,1], [1,2], [2,1], [2,2], [3] which, clearly, have 2,0,0,1,0 level steps.
Triangle starts
1;
1,1;
3,1,1;
6,5,1,1;
14,12,7,1,1
- A. Blecher, C. Brennan, and A. Knopfmacher, Combinatorial parameters in bargraphs (preprint).
-
G:=((1-t*z-z-2*z^2+t*z^2-sqrt((1-t*z-z-2*z^2+t*z^2)^2-4*z^3))*(1/2))/z: Gser:=simplify(series(G,z=0,21)): for n from 2 to 18 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 2 to 18 do seq(coeff(P[n], t, j), j = 0 .. n-2) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, y, t, w) option remember; expand(
`if`(n=0, (1-t), `if`(t<0, 0, b(n-1, y+1, 1, 0))+
`if`(t>0 or y<2, 0, b(n, y-1, -1, 0))+ `if`(y<1, 0,
`if`(w=1, z, 1)*b(n-1, y, 0, min(w+1, 1)))))
end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(n, 0$3)):
seq(T(n), n=2..18); # Alois P. Heinz, Jun 04 2016
-
b[n_, y_, t_, w_] := b[n, y, t, w] = Expand[If[n == 0, 1 - t, If[t < 0, 0, b[n - 1, y + 1, 1, 0]] + If[t > 0 || y < 2, 0, b[n, y - 1, -1, 0]] + If[y < 1, 0, If[w == 1, z, 1]*b[n - 1, y, 0, Min[w + 1, 1]]]]];
T[n_] := Function [p, Table[Coefficient[p, z, i], {i, 0, Exponent[p, z]}]][ b[n, 0, 0, 0]];
Table[T[n], {n, 2, 18}] // Flatten (* Jean-François Alcover, Jul 29 2016, after Alois P. Heinz *)
A126191
Triangle read by rows: number of 0-1-2 trees (i.e., ordered trees with vertices of outdegrees 0, 1, or 2) with n edges and exactly k vertices that have 2 children, both being leaves (n >= 0, 0 <= k <= floor((n+2)/4)).
Original entry on oeis.org
1, 1, 1, 1, 3, 1, 6, 3, 14, 7, 33, 17, 1, 79, 45, 3, 194, 117, 12, 482, 313, 40, 1214, 843, 129, 2, 3090, 2287, 411, 10, 7936, 6247, 1278, 50, 20544, 17139, 3942, 210, 53545, 47219, 12045, 820, 5, 140399, 130527, 36559, 3052, 35, 370098, 361851, 110388
Offset: 0
-
G:=(1-z-sqrt(1-2*z-3*z^2+4*z^4-4*z^4*t))/2/z^2: Gser:=simplify(series(G,z=0,21)): for n from 0 to 17 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 17 do seq(coeff(P[n],t,j),j=0..floor((n+2)/4)) od; # yields sequence in triangular form
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