A262612
Triangle read by rows T(n,k) in which column k lists the partial sums of the k-th column of triangle A236104.
Original entry on oeis.org
1, 5, 14, 1, 30, 2, 55, 6, 91, 10, 1, 140, 19, 2, 204, 28, 3, 285, 44, 7, 385, 60, 11, 1, 506, 85, 15, 2, 650, 110, 24, 3, 819, 146, 33, 4, 1015, 182, 42, 8, 1240, 231, 58, 12, 1, 1496, 280, 74, 16, 2, 1785, 344, 90, 20, 3, 2109, 408, 115, 29, 4, 2470, 489, 140, 38, 5, 2870, 570, 165, 47, 9, 3311, 670, 201, 56, 13, 1
Offset: 1
Triangle begins:
1;
5;
14, 1;
30, 2;
55, 6;
91, 10, 1;
140, 19, 2;
204, 28, 3;
285, 44, 7;
385, 60, 11, 1;
506, 85, 15, 2;
650, 110, 24, 3;
819, 146, 33, 4;
1015, 182, 42, 8;
1240, 231, 58, 12, 1;
1496, 280, 74, 16, 2;
1785, 344, 90, 20, 3;
2109, 408, 115, 29, 4;
2470, 489, 140, 38, 5;
2870, 570, 165, 47, 9;
3311, 670, 201, 56, 13, 1;
3795, 770, 237, 72, 17, 2;
4324, 891, 273, 88, 21, 3;
4900, 1012, 322, 104, 25, 4;
...
For n = 6 we have that A175254(6) = [1] + [1 + 3] + [1 + 3 + 4] + [1 + 3 + 4 + 7] + [1 + 3 + 4 + 7 + 6] + [1 + 3 + 4 + 7 + 6 + 12] = 1 + 4 + 8 + 15 + 21 + 33 = 82. On the other hand the alternating sum of the 6th row of the triangle is 91 - 10 + 1 = 82, equaling A175254(6).
Cf.
A000203,
A000217,
A003056,
A024916,
A175254,
A196020,
A235791,
A236104,
A237048,
A237591,
A237593,
A237270,
A237271,
A245092,
A261699,
A262626.
A287195
Independence and clique covering number of the n-triangular honeycomb acute knight graph.
Original entry on oeis.org
1, 3, 3, 5, 9, 9, 12, 18, 18, 22, 30, 30, 35, 45, 45, 51, 63, 63, 70, 84, 84, 92, 108, 108, 117, 135, 135, 145, 165, 165, 176, 198, 198, 210, 234, 234, 247, 273, 273, 287, 315, 315, 330, 360, 360, 376, 408, 408, 425, 459, 459, 477, 513, 513, 532, 570, 570
Offset: 1
- Colin Barker, Table of n, a(n) for n = 1..1000
- Stan Wagon, Graph Theory Problems from Hexagonal and Traditional Chess, The College Mathematics Journal, Vol. 45, No. 4, September 2014, pp. 278-287.
- Eric Weisstein's World of Mathematics, Clique Covering Number.
- Eric Weisstein's World of Mathematics, Independence Number.
- Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).
-
LinearRecurrence[{1, 0, 2, -2, 0, -1, 1}, {1, 3, 3, 5, 9, 9, 12}, 50]
Table[1/18 ((n + 3) (3 n + 2) - 2 (n + 3) Cos[2 n Pi/3] - 2 Sqrt[3] (n + 1) Sin[2 n Pi/3]), {n, 50}]
Table[Piecewise[{{n (n + 3), Mod[n, 3] == 0}, {(n + 1) (n + 2), Mod[n, 3] == 1}, {(n + 1) (n + 4), Mod[n, 3] == 2}}]/6, {n, 50}]
-
Vec(x*(1 + 2*x) / ((1 - x)^3*(1 + x + x^2)^2) + O(x^60)) \\ Colin Barker, Jul 15 2017
A115265
Correlation triangle for floor((n+3)/3).
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 2, 3, 3, 3, 2, 2, 4, 4, 4, 4, 2, 3, 4, 5, 7, 5, 4, 3, 3, 5, 6, 8, 8, 6, 5, 3, 3, 6, 7, 9, 11, 9, 7, 6, 3, 4, 6, 8, 12, 12, 12, 12, 8, 6, 4, 4, 7, 9, 13, 15, 15, 15, 13, 9, 7, 4
Offset: 0
Triangle begins
1;
1,1;
1,2,1;
2,2,2,2;
2,3,3,3,2;
2,4,4,4,4,2;
3,4,5,7,5,4,3;
3,5,6,8,8,6,5,3;
3,6,7,9,11,9,7,6,3;
-
T[n_, k_] := Sum[Boole[j <= k] * Floor[(k - j + 3)/3] * Boole[j <= n-k] * Floor[(n - k - j + 3)/3], {j, 0, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 15 2017 *)
Showing 1-3 of 3 results.
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