A227978
a(0)=1, a(1)=2; for n>1, a(n) = n*(2^n+4)/4.
Original entry on oeis.org
1, 2, 4, 9, 20, 45, 102, 231, 520, 1161, 2570, 5643, 12300, 26637, 57358, 122895, 262160, 557073, 1179666, 2490387, 5242900, 11010069, 23068694, 48234519, 100663320, 209715225, 436207642, 905969691, 1879048220, 3892314141, 8053063710, 16642998303
Offset: 0
-
[1,2] cat [n*(2^n+4)/4: n in [2..40]]; // Bruno Berselli, Oct 11 2013
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Join[{1, 2}, Table[n (2^n + 4)/4, {n, 2, 40}]] (* Bruno Berselli, Oct 11 2013 *)
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a(n) = if (n == 0, 1, if (n == 1, 2, n*(2^n+4)/4)); \\ Michel Marcus, Oct 11 2013
A283260
Product of the products of elements of all the nonempty subsets of the sets of numbers from 1 to n.
Original entry on oeis.org
1, 4, 1296, 110075314176, 1848842588950364160000000000000000, 27204445973673520186989292010512474445356561349778469394841600000000000000000000000000000000
Offset: 1
Rows with nonempty subsets of the sets of numbers from 1 to n:
{1};
{1}, {2}, {1, 2};
{1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3};
...
Rows with the products of elements of these subsets:
1;
1, 2, 2;
1, 2, 3, 2, 3, 6, 6;
...
a(1) = (1), a(2) = (1*2*2) = 4, a(3) = (1*2*3*2*3*6*6) = 1296, ...
A349706
Square array T(n,k) = Sum_{j=0..k} binomial(k,j) * j^n for n and k >= 0, read by ascending antidiagonals.
Original entry on oeis.org
1, 0, 2, 0, 1, 4, 0, 1, 4, 8, 0, 1, 6, 12, 16, 0, 1, 10, 24, 32, 32, 0, 1, 18, 54, 80, 80, 64, 0, 1, 34, 132, 224, 240, 192, 128, 0, 1, 66, 342, 680, 800, 672, 448, 256, 0, 1, 130, 924, 2192, 2880, 2592, 1792, 1024, 512, 0, 1, 258, 2574, 7400, 11000, 10752, 7840, 4608, 2304, 1024
Offset: 0
Square array begins:
1 2 4 8 16 32
0 1 4 12 32 80
0 1 6 24 80 240
0 1 10 54 224 800
0 1 18 132 680 2880
0 1 34 342 2192 11000
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
- Renate Golombek, Aufgabe 1088, El. Math., 49 (1994), 126-127.
- Simsek Yilmaz, New families of special numbers for computing negative order Euler numbers and related numbers and polynomials, Applicable Analysis and Discrete Mathematics 2018 Volume 12, Issue 1, Pages: 1-35. See B(n,k).
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T[n_, k_] := Sum[Binomial[k, j] * If[j == n == 0, 1, j^n], {j, 0, k}]; Table[T[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 26 2021 *)
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T(n,k) = sum(j=0, k, binomial(k,j)*j^n);
A364688
Number of 8-cycles in the hypercube graph Q_n.
Original entry on oeis.org
0, 0, 0, 6, 696, 6720, 39840, 184800, 736512, 2644992, 8801280, 27624960, 82790400, 238977024, 668688384, 1822679040, 4858183680, 12700876800, 32647938048, 82682707968, 206650736640, 510425825280, 1247438438400, 3019527684096, 7245593051136, 17248655769600
Offset: 0
-
Table[Length[FindCycle[HypercubeGraph[n], {8}, All]], {n, 0, 9}]
Table[2^(n - 4) n (n - 1) (n - 2) (27 n - 79), {n, 0, 20}]
Table[3 2^(n - 3) Binomial[n, 3] (27 n - 79), {n, 0, 20}]
LinearRecurrence[{10, -40, 80, -80, 32}, {0, 0, 0, 6, 696}, 20]
CoefficientList[Series[6 x^3 (1 + 106 x)/(1 - 2 x)^5, {x, 0, 20}], x]
A372873
Triangle read by rows: T(n,k) is the number of flattened Catalan words of length n with exactly k runs of descents.
Original entry on oeis.org
1, 0, 2, 0, 1, 4, 0, 0, 6, 8, 0, 0, 1, 24, 16, 0, 0, 0, 10, 80, 32, 0, 0, 0, 1, 60, 240, 64, 0, 0, 0, 0, 14, 280, 672, 128, 0, 0, 0, 0, 1, 112, 1120, 1792, 256, 0, 0, 0, 0, 0, 18, 672, 4032, 4608, 512, 0, 0, 0, 0, 0, 1, 180, 3360, 13440, 11520, 1024
Offset: 1
The triangle begins:
1;
0, 2;
0, 1, 4;
0, 0, 6, 8;
0, 0, 1, 24, 16;
0, 0, 0, 10, 80, 32;
0, 0, 0, 1, 60, 240, 64;
0, 0, 0, 0, 14, 280, 672, 128;
0, 0, 0, 0, 1, 112, 1120, 1792, 256;
...
T(4,3) = 6 since there 6 flattened Catalan words of length 4 with 3 runs of descents: 0010, 0100, 0101, 0110, 0120, and 0121.
- Jean-Luc Baril, Pamela E. Harris, and José L. Ramírez, Flattened Catalan Words, arXiv:2405.05357 [math.CO], 2024. See p. 11.
-
T[n_,k_]:=SeriesCoefficient[x*y*(1-2*x*y)/(1-4*x*y-x^2*y+4x^2*y^2),{x,0,n},{y,0,k}]; Table[T[n,k],{n,11},{k,n}]//Flatten (* or *)
T[n_,k_]:=2^(2*k-n-1)*Binomial[n-1, 2*(n-k)]; Table[T[n,k],{n,11},{k,n}]//Flatten
A157933
Triangle T(i,j) such that Sum_{j=0..i} T(i,j)*x(i,j)/2^i = Sum_{k=0..i, j=0..k} x(k,j), if x(k-1,j) = (x(k,j) + x(k,j+1))/2.
Original entry on oeis.org
1, 3, 3, 7, 10, 7, 15, 25, 25, 15, 31, 56, 66, 56, 31, 63, 119, 154, 154, 119, 63, 127, 246, 337, 372, 337, 246, 127, 255, 501, 711, 837, 837, 711, 501, 255, 511, 1012, 1468, 1804, 1930, 1804, 1468, 1012, 511, 1023, 2035, 2992, 3784, 4246, 4246, 3784, 2992, 2035, 1023
Offset: 0
To get the 3rd row of the triangle, consider the pyramid
f
d e
a b c
where d=(a+b)/2, e=(b+c)/2, f=(d+e)/2. Then a+b+c+d+e+f=(7a+10b+7c)/2^2, which yields the row (7,10,7).
Triangle begins:
1,
3, 3;
7, 10, 7;
15, 25, 25, 15;
31, 56, 66, 56, 31;
63, 119, 154, 154, 119, 63;
...
A181372
Square array read by antidiagonals: a(p,n) is the number of inversions in all p-ary words of length n on {0,1,2,...,p-1} (p>=2, n>=2).
Original entry on oeis.org
1, 3, 6, 6, 27, 24, 10, 72, 162, 80, 15, 150, 576, 810, 240, 21, 270, 1500, 3840, 3645, 672, 28, 441, 3240, 12500, 23040, 15309, 1792, 36, 672, 6174, 32400, 93750, 129024, 61236, 4608, 45, 972, 10752, 72030, 291600, 656250, 688128, 236196, 11520
Offset: 2
a(2,3)=6 because the words 010, 100, 101, 110 have 1, 2, 1, 2 inversions, respectively, while 000, 001, 011 and 111 have none.
a(3,2)=3 because each of the words 10, 20, 21 has 1 inversion, while 00, 01, 02, 11, 12, 22 have none.
Square array starts:
1,6,24,80,240, ...;
3,27,162,810,3645, ... ;
6,72,576, 3840,23040, ... ;
10,150,1500,12500, 93750,...;
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a := proc (p, n) options operator, arrow: (1/4)*n*(n-1)*(p-1)*p^(n-1) end proc: for p from 2 to 8 do seq(a(p, n), n = 2 .. 8) end do; # yields the first seven rows and seven columns of the square array
A276985
Triangle read by rows: T(n,k) = number of k-dimensional elements in an n-dimensional cross-polytope, n>=1, 0<=k
Original entry on oeis.org
2, 4, 4, 6, 12, 8, 8, 24, 32, 16, 10, 40, 80, 80, 32, 12, 60, 160, 240, 192, 64, 14, 84, 280, 560, 672, 448, 128, 16, 112, 448, 1120, 1792, 1792, 1024, 256, 18, 144, 672, 2016, 4032, 5376, 4608, 2304, 512, 20, 180, 960, 3360, 8064, 13440, 15360, 11520, 5120
Offset: 1
T(4, 1..4) = 8, 24, 32, 16, because the 16-cell has 8 0-faces (vertices), 24 1-faces (edges), 32 2-faces (faces) and 16 3-faces (cells).
Triangle starts
2
4, 4
6, 12, 8
8, 24, 32, 16
10, 40, 80, 80, 32
12, 60, 160, 240, 192, 64
14, 84, 280, 560, 672, 448, 128
16, 112, 448, 1120, 1792, 1792, 1024, 256
18, 144, 672, 2016, 4032, 5376, 4608, 2304, 512
20, 180, 960, 3360, 8064, 13440, 15360, 11520, 5120, 1024
- H. S. M. Coxeter, Regular Polytopes, Third Edition, Dover Publications, 1973, ISBN 9780486141589.
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Table[2^(k + 1) Binomial[n, k + 1], {n, 10}, {k, 0, n - 1}] // Flatten (* Michael De Vlieger, Sep 25 2016 *)
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T(n, k) = 2^(k+1)*binomial(n, k+1)
trianglerows(n) = for(x=1, n, for(y=0, x-1, print1(T(x, y), ", ")); print(""))
trianglerows(10) \\ print initial 10 rows of triangle
A381457
Integers encoding the recursive structure of a bitonic sorter network of n elements in their binary expansion.
Original entry on oeis.org
0, 2, 5, 2762, 22325, 175338, 1405781, 187796017958058, 12026023042822997, 768764969792360106, 49208135664973067605, 3148398007269257431722, 201504853864147844281685, 12895366188400224861219498, 825310989999256684769531221
Offset: 1
n | k | i | Comparator network (1 = \ and 0 = /). | a(n)
---+----+---------------------+---------------------------------------+----------------
2 | 2 | 10 | \/ | 2
---+----+---------------------+---------------------------------------+----------------
4 | 2 | 1010 | \/ \/ | 2762
| 4 | 1100 | \ / |
| 4 | 1010 | \/ \/ |
---+----+---------------------+---------------------------------------+----------------
8 | 2 | 1010 1010 | \/ \/ \/ \/ | 187796017958058
| 4 | 1100 1100 | \ / \ / |
| 4 | 1010 1010 | \/ \/ \/ \/ |
| 8 | 1111 0000 | \ / |
| 8 | 1100 1100 | \ / \ / |
| 8 | 1010 1010 | \/ \/ \/ \/
For n = 5 the network is similar to the network for n = 4 plus a column of ones, which is also a network contained in the network for n = 8.
5 | 2 | 1010 1 | \/ \/ \ | 22325
| 4 | 1100 1 | \ / \ |
| 4 | 1010 1 | \/ \/ \ |
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def a(n):
e, k = 0, 2
while k <= n:
j = k >> 1
while j > 0:
for i in range(n):
e = (e << 1) | ((i & j) == 0)
j >>= 1
k <<= 1
return e
print([a(n) for n in range(1,16)])
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