A201780
Riordan array ((1-x)^2/(1-2x), x/(1-2x)).
Original entry on oeis.org
1, 0, 1, 1, 2, 1, 2, 5, 4, 1, 4, 12, 13, 6, 1, 8, 28, 38, 25, 8, 1, 16, 64, 104, 88, 41, 10, 1, 32, 144, 272, 280, 170, 61, 12, 1, 64, 320, 688, 832, 620, 292, 85, 14, 1, 128, 704, 1696, 2352, 2072, 1204, 462, 113, 16, 1
Offset: 0
Triangle begins:
1;
0, 1;
1, 2, 1;
2, 5, 4, 1;
4, 12, 13, 6, 1;
8, 28, 38, 25, 8, 1;
- Benjamin Braun, W. K. Hough, Matching and Independence Complexes Related to Small Grids, arXiv preprint arXiv:1606.01204 [math.CO], 2016.
- Wesley K. Hough, On Independence, Matching, and Homomorphism Complexes, (2017), Theses and Dissertations--Mathematics, 42.
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CoefficientList[#, y]& /@ CoefficientList[(1-x)^2/(1-(y+2)*x) + O[x]^10, x] // Flatten (* Jean-François Alcover, Nov 03 2018 *)
A076616
Number of permutations of {1,2,...,n} that result in a binary search tree (when elements of the permutation are inserted in that order) of height n-1 (i.e., the second largest possible height).
Original entry on oeis.org
0, 0, 0, 2, 16, 64, 208, 608, 1664, 4352, 11008, 27136, 65536, 155648, 364544, 843776, 1933312, 4390912, 9895936, 22151168, 49283072, 109051904, 240123904, 526385152, 1149239296, 2499805184, 5419040768, 11710496768, 25232932864, 54223962112, 116232552448
Offset: 0
a(3) = 2 because only the permutations (2,1,3) and (2,3,1) result in a search tree of height 2 (notice we count empty external nodes in determining the height). The largest such trees are of height 3.
-
a:= n-> max(-(<<0|1|0>, <0|0|1>, <8|-12|6>>^n. <<1/2, 1, 1>>)[1$2], 0):
seq(a(n), n=0..40); # Alois P. Heinz, Sep 20 2011
-
Join[{0, 0, 0}, LinearRecurrence[{6, -12, 8}, {2, 16, 64}, 40]] (* Jean-François Alcover, Jan 09 2025 *)
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concat(vector(3), Vec(2*x^3*(1+2*x-4*x^2)/(1-2*x)^3 + O(x^50))) \\ Colin Barker, May 16 2016
A109436
Triangle of numbers: row n gives the elements along the subdiagonal of A109435 that connects 2^n with (n+2)*2^(n-1).
Original entry on oeis.org
0, 0, 1, 1, 2, 3, 4, 7, 8, 8, 15, 19, 20, 16, 31, 43, 47, 48, 32, 63, 94, 107, 111, 112, 64, 127, 201, 238, 251, 255, 256, 128, 255, 423, 520, 558, 571, 575, 576, 256, 511, 880, 1121, 1224, 1262, 1275, 1279, 1280, 512, 1023, 1815, 2391, 2656, 2760, 2798, 2811
Offset: 0
The triangle A109435 begins
1;
2, 1;
4, 3, 1;
8, 7, 3, 1;
16, 15, 8, 3, 1;
32, 31, 19, 8, 3, 1;
64, 63, 43, 20, 8, 3, 1;
128, 127, 94, 47, 20, 8, 3, 1;
If we read this triangle starting at 2^n in its first column along its n-th subdiagonal up to the first occurrence of (n+2)*2^(n-1), we get row n of the current triangle, which begins:
0, 0;
1, 1;
2, 3;
4, 7, 8;
8, 15, 19, 20;
16, 31, 43, 47, 48;
32, 63, 94, 107, 111, 112;
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T[n_, m_] := Length[ Select[ StringPosition[ #, StringDrop[ ToString[10^m], 1]] & /@ Table[ ToString[ FromDigits[ IntegerDigits[i, 2]]], {i, 2^n, 2^(n + 1) - 1}], # != {} &]]; Flatten[ Table[ T[n + i, i], {n, 0, 9}, {i, 0, n}]]
A127718
A007318 * A002260 as infinite lower triangular matrices; A002260 = [1; 1,2; 1,2,3; ...].
Original entry on oeis.org
1, 2, 2, 4, 6, 3, 8, 14, 12, 4, 16, 30, 33, 20, 5, 32, 62, 78, 64, 30, 6, 64, 126, 171, 168, 110, 42, 7, 128, 254, 360, 396, 320, 174, 56, 8, 256, 510, 741, 876, 815, 558, 259, 72, 9, 512, 1022, 1506, 1864, 1910, 1536, 910, 368, 90, 10, 1024, 2046, 3039, 3872, 4240
Offset: 1
First few rows of the triangle:
1;
2, 2;
4, 6, 3;
8, 14, 12, 4;
16, 30, 33, 20, 5;
32, 62, 78, 64, 30, 6;
64, 126, 171, 168, 110, 42, 7;
...
-
A007318 := proc(n,k) binomial(n,k) ; end: A002260 := proc(n,k) if k <= n then k; else 0 ; fi ; end: A127718 := proc(n,k) add( A007318(n-1,i-1)*A002260(i,k),i=1..n) ; end: for n from 1 to 15 do for k from 1 to n do printf("%d,",A127718(n,k)) ; od: od: # R. J. Mathar, Oct 02 2007
A332496
Triangular array T(n,k): the number of not necessarily proper colorings of the complete graph on n unlabeled vertices minus an edge using exactly k colors.
Original entry on oeis.org
0, 1, 1, 1, 4, 3, 1, 7, 12, 6, 1, 10, 27, 28, 10, 1, 13, 48, 76, 55, 15, 1, 16, 75, 160, 175, 96, 21, 1, 19, 108, 290, 425, 351, 154, 28, 1, 22, 147, 476, 875, 966, 637, 232, 36, 1, 25, 192, 728, 1610, 2226, 1960, 1072, 333, 45, 1, 28, 243, 1056, 2730, 4536, 4998, 3648, 1701, 460, 55
Offset: 1
T(n,n) = C(n,2) because we need to select the two colors that color the vertices of the removed edge from the n available colors. The remaining vertices are not distinguishable.
Triangle T(n,k) begins:
0;
1, 1;
1, 4, 3;
1, 7, 12, 6;
1, 10, 27, 28, 10;
1, 13, 48, 76, 55, 15;
1, 16, 75, 160, 175, 96, 21;
1, 19, 108, 290, 425, 351, 154, 28;
1, 22, 147, 476, 875, 966, 637, 232, 36;
...
T(4,2) = 7 because when n = 4 there are two unordered pairs of vertices, each of which can be colored in 3 ways using a maximum of 2 colors giving 9 colorings. From this, the two coloring using only one of the two colors needs to be subtracted, so T(4,2) = 9 - 2 = 7. - _Andrew Howroyd_, Feb 15 2020
- E. Palmer and F. Harary, Graphical Enumeration, Academic Press, 1973.
Row sums give 2 *
A084851(n-2) for n>1.
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T:= (n, k)-> `if`(n=1, 0, (k!/2/(n-2)!)*add(abs(Stirling1(n-2, p))
*(Stirling2(p+1, k)+Stirling2(p+2, k)), p=0..n-2)):
seq(seq(T(n, k), k=1..n), n=1..12); # Alois P. Heinz, Feb 14 2020
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T(n,k) = {if(n<2,0,(k!/2/(n-2)!)*sum(p=0,n-2,abs(stirling(n-2,p,1))* (stirling(p+1, k,2)+stirling(p+2, k,2))))};
for(n=1,11,print(" ");for(k=1,n,print1(T(n,k),", "))) \\ Hugo Pfoertner, Feb 14 2020
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