Search: a097513 -id:a097513
|
| |
|
|
A275416
|
|
Triangle read by rows: T(n,k) is the number of multisets of k odd numbers with a cap of the total sum set to n.
|
|
+10
3
|
|
|
|
1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 3, 4, 3, 1, 1, 3, 8, 5, 3, 1, 1, 4, 10, 10, 5, 3, 1, 1, 4, 16, 15, 11, 5, 3, 1, 1, 5, 20, 27, 17, 11, 5, 3, 1, 1, 5, 29, 38, 32, 18, 11, 5, 3, 1, 1, 6, 35, 60, 49, 34, 18, 11, 5, 3, 1, 1, 6, 47, 84, 83, 54, 35, 18, 11, 5, 3, 1, 1, 7, 56, 122, 123
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,4
|
|
|
COMMENTS
|
By considering the partitions of n into k parts we set a cap on the odd numbers of each part and count the multisets (ordered k-tuples) of odd numbers where each number is not larger than the cap of its part.
|
|
|
LINKS
|
|
|
|
FORMULA
|
T(n,k) = Sum_{c_i*N_i=n,i=1..k} binomial(T(N_i,1)+c_i-1,c_i) for 1 < k <= n.
G.f.: Product_{j>=1} (1-y*x^j)^(-ceiling(j/2)). - Alois P. Heinz, Apr 13 2017
|
|
|
EXAMPLE
|
T(6,2) = 3+2+3 = 8 counts {1,1} {1,3}, and {3,3} from taking two odd numbers <= 3; it counts {1,1} and {1,3} from taking an odd number <= 2 and an odd number <= 4; and it counts {1,1}, {1,3} and {1,5} from taking an odd number <= 1 and an odd number <= 5.
T(6,3) = 1+2+2 = 5 counts {1,1,1} from taking three odd numbers <= 2; it counts {1,1,1} and {1,1,3} from taking an odd number <= 1 and an odd number <= 2 and an odd number <= 3; and it counts {1,1,1} and {1,1,3} from taking two odd numbers <= 1 and an odd number <= 4.
1
1 1
2 1 1
2 3 1 1
3 4 3 1 1
3 8 5 3 1 1
4 10 10 5 3 1 1
4 16 15 11 5 3 1 1
5 20 27 17 11 5 3 1 1
5 29 38 32 18 11 5 3 1 1
6 35 60 49 34 18 11 5 3 1 1
6 47 84 83 54 35 18 11 5 3 1 1
7 56 122 123 94 56 35 18 11 5 3 1 1
7 72 164 192 146 99 57 35 18 11 5 3 1 1
|
|
|
MAPLE
|
b:= proc(n, i, p) option remember; `if`(p>n, 0, `if`(n=0, 1,
`if`(min(i, p)<1, 0, add(b(n-i*j, i-1, p-j)*
binomial(ceil(i/2)+j-1, j), j=0..min(n/i, p)))))
end:
T:= (n, k)-> b(n$2, k):
|
|
|
MATHEMATICA
|
b[n_, i_, p_] := b[n, i, p] = If[p > n, 0, If[n == 0, 1, If[Min[i, p] < 1, 0, Sum[b[n - i*j, i - 1, p - j]*Binomial[Ceiling[i/2] + j - 1, j], {j, 0, Min[n/i, p]}]]]];
T[n_, k_] := b[n, n, k];
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
| |
|
|
A054473
|
|
Number of ways of numbering the faces of a cube with nonnegative integers so that the sum of the 6 numbers is n.
|
|
+10
0
|
|
|
|
1, 1, 3, 5, 10, 15, 29, 41, 68, 98, 147, 202, 291, 386, 528, 688, 906, 1151, 1480, 1841, 2310, 2833, 3484, 4207, 5099, 6076, 7259, 8562, 10104, 11796, 13785, 15948, 18462, 21201, 24339, 27747, 31633, 35827, 40572, 45695, 51436, 57618, 64520, 71918
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
Here we consider the symmetries of the cube in 3D space (mirror reflections are not allowed), cf. A097513. - Geoffrey Critzer, Sep 28 2013
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f.: (3*x^6+x^5+x^4+1)/((1-x^4)*(1-x^3)^2*(1-x^2)^2*(1-x)).
|
|
|
MATHEMATICA
|
nn=43; f[x_]=1/(1-x); CoefficientList[Series[1/24 (f[x]^6+6f[x]^2f[x^4]+3f[x]^2f[x^2]^2+8f[x^3]^2+6f[x^2]^3), {x, 0, nn}], x] (* Geoffrey Critzer, Sep 28 2013 *)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
Search completed in 0.008 seconds
|
