1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 3, 1, 0, 0, 0, 15, 60, 1, 0, 0, 0, 45, 360, 1080, 1, 0, 0, 0, 105, 1260, 7560, 20580, 1, 0, 0, 0, 210, 3360, 30240, 164640, 430080, 1, 0, 0, 0, 378, 7560, 90720, 740880, 3873240, 9920232, 1, 0, 0, 0, 630, 15120, 226800, 2469600, 19367460, 99406440, 252000000
Offset: 0
T(5,4) = 15 = 5*3, because there are 5 possibilities for a single node and T(4,4) = 3:
.1-2. .1-2. .1.2.
.|.|. ..X.. .|X|.
.3-4. .3-4. .3.4.
Triangle begins:
1;
1, 0;
1, 0, 0;
1, 0, 0, 0;
1, 0, 0, 0, 3;
1, 0, 0, 0, 15, 60;
A198148
a(n) = n*(n+2)*(9 - 7*(-1)^n)/16.
Original entry on oeis.org
0, 3, 1, 15, 3, 35, 6, 63, 10, 99, 15, 143, 21, 195, 28, 255, 36, 323, 45, 399, 55, 483, 66, 575, 78, 675, 91, 783, 105, 899, 120, 1023, 136, 1155, 153, 1295, 171, 1443, 190, 1599, 210, 1763, 231, 1935, 253, 2115, 276, 2303, 300, 2499, 325
Offset: 0
-
List([0..60], n -> n*(n+2)*(9-7*(-1)^n)/16); # G. C. Greubel, Feb 21 2019
-
[n*(n+2)*(9-7*(-1)^n)/16: n in [0..60]]; // Vincenzo Librandi, Nov 25 2011
-
A198148:=n->n*(n+2)*(9-7*(-1)^n)/16; seq(A198148(k), k=0..100); # Wesley Ivan Hurt, Oct 16 2013
-
LinearRecurrence[{0,3,0,-3,0,1},{0,3,1,15,3,35},60] (* Vincenzo Librandi, Nov 25 2011 *)
-
a(n)=n*(n+2)*(9-7*(-1)^n)/16 \\ Charles R Greathouse IV, Oct 16 2015
-
[n*(n+2)*(9-7*(-1)^n)/16 for n in (0..60)] # G. C. Greubel, Feb 21 2019
A234251
Triangle T(n, k) = Number of ways to choose k points from an n X n X n triangular grid so that no three of them form a 2 X 2 X 2 subtriangle. Triangle T read by rows.
Original entry on oeis.org
1, 1, 1, 3, 3, 1, 6, 15, 16, 6, 1, 10, 45, 111, 156, 120, 42, 2, 1, 15, 105, 439, 1191, 2154, 2583, 1977, 885, 189, 9, 1, 21, 210, 1305, 5565, 17052, 38337, 63576, 77208, 67285, 40512, 15750, 3480, 333, 9, 1, 28, 378, 3240, 19620, 88590, 307362, 833228, 1779219
Offset: 1
Triangle begins
1, 1;
1, 3, 3;
1, 6, 15, 16, 6;
1, 10, 45, 111, 156, 120, 42, 2;
1, 15, 105, 439, 1191, 2154, 2583, 1977, 885, 189, 9;
...
There are no more than T(4, 7) = 2 ways to choose 7 points (X) from a 4 X 4 X 4 grid so that no 3 of them form a 2 X 2 X 2 subtriangle:
X X
X . . X
. X X X X .
X X . X X . X X
A243214
Number of ways to place 5 points on a triangular grid of side n, so that no three of them are vertices of an equilateral triangle with sides parallel to the grid.
Original entry on oeis.org
0, 63, 1566, 13971, 77124, 319206, 1083723, 3181401, 8344854, 20006349, 44548227, 93248628, 185176866, 351410664, 640972980, 1129067352, 1928196432, 3203016813
Offset: 3
A053527
Number of bipartite graphs with 4 edges on nodes {1..n}.
Original entry on oeis.org
0, 0, 0, 0, 3, 140, 1125, 5355, 19075, 56133, 143955, 332475, 706860, 1404975, 2640638, 4733820, 8149050, 13543390, 21825450, 34227018, 52388985, 78463350, 115233195, 166252625, 236008773, 330108075, 455489125, 620664525, 835994250
Offset: 0
- R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.5.
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).
-
List([0..30], n-> Binomial(n,4)*(n+2)*(n^3-5*n-36)/16 ) # G. C. Greubel, May 15 2019
-
[(n^5-4*n^4-n^3+16*n^2-12*n)*(n^3-5*n-36)/384: n in [0..30]]; // Vincenzo Librandi, May 08 2012
-
CoefficientList[Series[x^4*(3+113*x-27*x^2+18*x^3-2*x^4)/(1-x)^9, {x,0, 30}], x] (* Vincenzo Librandi, May 08 2012 *)
-
{a(n) = binomial(n,4)*(n+2)*(n^3-5*n-36)/16}; \\ G. C. Greubel, May 15 2019
-
[binomial(n,4)*(n+2)*(n^3-5*n-36)/16 for n in (0..30)] # G. C. Greubel, May 15 2019
A144163
Triangle T(n,k), n>=0, 0<=k<=n, read by rows: T(n,k) = number of simple graphs on n labeled nodes with k edges where each maximally connected subgraph is either a tree or a cycle.
Original entry on oeis.org
1, 1, 0, 1, 1, 0, 1, 3, 3, 1, 1, 6, 15, 20, 3, 1, 10, 45, 120, 150, 12, 1, 15, 105, 455, 1185, 1473, 70, 1, 21, 210, 1330, 5565, 14469, 18424, 465, 1, 28, 378, 3276, 19635, 81060, 213990, 280200, 3507, 1, 36, 630, 7140, 57393, 334656, 1385076, 3732300, 5029218, 30016
Offset: 0
T(4,3) = 20, because there are 20 simple graphs on 4 labeled nodes with 3 edges, where each maximally connected subgraph is either a tree or a cycle, 16 of these graphs consist of a single tree with 4 nodes and 4 consist of a cycle with 3 and a tree with 1 node:
.1-2. .1-2. .1.2. .1.2. .1-2. .1-2. .1-2. .1-2. .1-2. .1.2.
.|\.. ../|. ..\|. .|/.. .|... ...|. ../.. ..\.. .|.|. .|.|.
.4.3. .4.3. .4-3. .4-3. .4-3. .4-3. .4-3. .4-3. .4.3. .4-3.
.
.1.2. .1.2. .1-2. .1.2. .1.2. .1.2. .1.2. .1.2. .1-2. .1-2.
.|/|. .|\|. ..X.. ..X|. ..X.. .|X.. ../|. .|\.. .|/.. ..\|.
.4.3. .4.3. .4.3. .4.3. .4-3. .4.3. .4-3. .4-3. .4.3. .4.3.
Triangle begins:
1;
1, 0;
1, 1, 0;
1, 3, 3, 1;
1, 6, 15, 20, 3;
1, 10, 45, 120, 150, 12;
-
f:= proc(n,k) option remember; local j; if k=0 then 1 elif k<0 or n<=k then 0 elif k=n-1 then n^(n-2) else add(binomial(n-1,j) *f(j+1,j) *f(n-1-j,k-j), j=0..k) fi end:
c:= proc(n,k) option remember; local i,j; if k=0 then 1 elif k<0 or n
-
f[n_, k_] := f[n, k] = Which[k == 0, 1, k<0 || n <= k, 0, k == n-1, n^(n-2), True, Sum[Binomial[n-1, j]*f[j+1, j]*f[n-1-j, k-j], {j, 0, k}]]; c[n_, k_] := c[n, k] = Which[k == 0, 1 , k<0 || nJean-François Alcover, Jan 21 2014, translated from Alois P. Heinz's Maple code *)
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