|
| |
|
|
A250353
|
|
Number of length 4 arrays x(i), i=1..4 with x(i) in i..i+n and no value appearing more than 2 times.
|
|
2
|
|
|
|
16, 75, 235, 581, 1221, 2287, 3935, 6345, 9721, 14291, 20307, 28045, 37805, 49911, 64711, 82577, 103905, 129115, 158651, 192981, 232597, 278015, 329775, 388441, 454601, 528867, 611875, 704285, 806781, 920071, 1044887, 1181985, 1332145
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
There are n+1 candidates for any of the 4 values in the 4-tuple. If there were no constraints, there were (n+1)^4 arrays. The constraint of not counting the quadruplets (4,4,4,4), (5,5,5,5), ..... (n+1,n+1,n+1,n+1) discards n-2 of the 4-tuples. [The case n=1 is special because there are not quadruplets]. Adding the constraint of not having triplets discards (3,3,3,*) and (*,n+2,n+2,n+2) where the star represents one of n+1 values; this is a total of 2*(n+1). The constraint of not having triplets also discards the (*,4,4,4), (4,*,4,4), (4,4,*,4), (4,4,4,*), (*,5,5,5),... (*,1+n,1+n,1+n),....(1+n,1+n,1+n,*) where the star represents one of n values (not n+1 here not to account for the quadruplets twice). There are binomial(4,1)*n*(n-2) of these triplets. The result is a(n) = (n+1)^4 -(n-2) -2*(n+1) -4*n*(n-2) = n^4+4*n^3+2*n^2+9*n+1. - R. J. Mathar, Oct 11 2020
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = n^4 + 4*n^3 + 2*n^2 + 9*n + 1 for n>1.
G.f.: x*(16 - 5*x + 20*x^2 - 4*x^3 - 4*x^4 + x^5) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>6.
(End)
|
|
|
EXAMPLE
|
Some solutions for n=6:
..2....0....2....3....3....3....2....4....4....4....1....0....2....5....5....5
..6....2....7....1....7....4....2....4....3....4....4....6....1....3....1....4
..2....4....3....5....6....7....7....6....8....2....7....2....6....6....2....5
..3....7....7....8....6....4....6....7....8....7....8....6....9....4....5....3
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
