|
| |
|
|
A132892
|
|
Square array T(m,n) read by antidiagonals; T(m,n) is the number of equivalence classes in the set of sequences of n nonnegative integers that sum to m, generated by the equivalence relation defined in the following manner: we write a sequence in the form a[1]0a[2]0...0a[p], where each a[i] is a (possibly empty) sequence of positive integers; two sequences in this form, a[1]0a[2]0...0a[p] and b[1]0b[2]0...0b[q] are said to be equivalent if p=q and b[1],b[2],...,b[q] is a cyclic permutation of a[1],a[2],...a[p].
|
|
0
|
|
|
|
1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 3, 1, 1, 5, 9, 7, 4, 1, 1, 6, 13, 14, 10, 4, 1, 1, 7, 19, 25, 22, 12, 5, 1, 1, 8, 25, 41, 42, 30, 15, 5, 1, 1, 9, 33, 63, 79, 66, 43, 19, 6, 1, 1, 10, 41, 92, 131, 132, 99, 55, 22, 6, 1, 1, 11, 51, 129, 213, 245, 217, 143, 73, 26, 7, 1, 1, 12, 61, 175
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,5
|
|
|
COMMENTS
|
T(n,n) = A000108(n) (the Catalan numbers; see R. P. Stanley, Catalan addendum, problem starting "Equivalence classes of the equivalence relation ..."). T(m,m+1) = A007595(m+1); T(m,m+2) = A003441(m+1); T(m,m+3) = A003444(m+3); T(n+2,n) = A001453(n+1) (Catalan numbers - 1); T(m,1)=1; T(m,2)=m; T(m,3) = A080827(m) = A099392(m+1); T(m,4) = A004006(m).
|
|
|
LINKS
|
R. P. Stanley, Catalan addendum. See the interpretation (www, "Vertices of height n-1 of the tree T ...").
|
|
|
FORMULA
|
T(m,n) = Sum_{d | gcd(m,n+1)} (phi(n)*(C((m+n+1)/d-1, (n+1)/d-1) - C(m/d-1, (n+1)/d-1))/(n+1).
|
|
|
EXAMPLE
|
T(2,4) = 3 because we have {2000, 0200, 0020, 0002}, {1100, 0110, 0011} and {1010, 0101, 1001}.
T(4,2) = 4 because we have {40, 04}, {31}, {13} and {22}.
The square array starts:
1....1.....1.....1......1.....1.....1...
1....2.....3.....3......4.....4.....4...
1....3.....5.....7.....10....12....15...
1....4.....9....14.....22....30....43...
1....5....13....25.....42....66....99...
|
|
|
MAPLE
|
with(numtheory): T:=proc(m, n) local r, div, N: r:=igcd(m, n+1): div:=divisors(r): N:=nops(div): (sum(phi(div[j])*(binomial((m+n+1)/div[j]-1, (n+1)/div[j]-1) -binomial(m/div[j]-1, (n+1)/div[j]-1)), j=1..N))/(n+1) end proc: for m to 12 do seq(T(m, n), n=1..12) end do; # yields the upper left 12 by 12 block of the infinite matrix T(m, n)
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
