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
R. H. Hardin, Table of n, a(n) for n = 1..210
FORMULA
a(n) = n^4 + 4*n^3 + 2*n^2 + 9*n + 1 for n>1.
From Colin Barker, Nov 13 2018: (Start)
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
R. H. Hardin, Nov 19 2014
STATUS
approved