A265644 Triangle read by rows: T(n,m) is the number of quaternary words of length n with m strictly increasing runs (0 <= m <= n).
1, 0, 4, 0, 6, 10, 0, 4, 40, 20, 0, 1, 65, 155, 35, 0, 0, 56, 456, 456, 56, 0, 0, 28, 728, 2128, 1128, 84, 0, 0, 8, 728, 5328, 7728, 2472, 120, 0, 0, 1, 486, 8451, 27876, 23607, 4950, 165
Offset: 0
Examples
Triangle starts: 1; 0, 4; 0, 6, 10; 0, 4, 40, 20; 0, 1, 65, 155, 35; 0, 0, 56, 456, 456, 56; . T(3,2) = 40, which accounts for the following words: [0 <= a <= 0, 1 | 0 <= b <= 1] = 2 [0 <= a <= 1, 2 | 0 <= b <= 2] = 6 [0 <= a <= 2, 3 | 0 <= b <= 3] = 12 [0 <= a <= 3 | 0, 1 <= b <= 3] = 12 [1 <= a <= 3 | 1, 2 <= b <= 3] = 6 [2 <= a <= 3 | 2, 3 <= b <= 3] = 2
References
- R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 24, p. 154.
Links
- MathPages, Balls In Bins With Limited Capacity
Crossrefs
Programs
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MuPAD
T:=(n,m)->_plus((-1)^(m-j)*binomial(n+1, m-j)*binomial(4*j, n)$j=0..m):
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PARI
T(n, k) = sum(j=0, k, (-1)^(k-j)*binomial(n+1, k-j)*binomial(4*j, n)); tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print()); \\ Michel Marcus, Feb 09 2016
Formula
Refer to comment to A120987 concerning formulas for general values of r and considerations.
Therefrom we get
T(n, m) = Qsc(3, n, m) =
Nb(4*m-n, 3, n+1) = Nb(4*(n-m)+3, 3, n+1) =
Sum_{j=0..n+1} (-1)^j*Cb(n+1, j)*Cb(4*(m-j), 4*(m-j)-n) =
Sum_{j=0..m} (-1)^(m-j)*Cb(n+1, m-j)*Cb(4*j, n) =
(in this last version Cb(n,m) can be replaced by binomial(n,m))
Sum_{j=0..m} (-1)^(m-j)*binomial(n+1, m-j)*binomial(4*j, n) = [z^n, t^m](1-t)/(1-t(1+(1-t)z)^4) where [x^n]F(x) denotes the coefficient of x^n in the formal power series expansion of F(x),
Nb(s,r,n) denotes the (r+1)-nomial coefficient [x^s](1+x+..+x^r)^n,(Nb(s,3,n) = A008287(n,s)).
Cb(x,m) denotes the binomial coefficient in its extended falling factorial notation (Cb(x,m)= x^_m/m! iff m is a nonnegative integer, 0 otherwise), as defined in the Graham et al. reference.
The diagonal T(n, n) = Nb(3, 3, n+1) = Sum_{j=0..n} (-1)^(n-j)*Cb(n+1, n-j)*Cb(4*j, n) = Cb(n+3, 3) = binomial(n+3, 3) = A000292(n+1).
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