OFFSET
0,3
COMMENTS
Sum of terms in row n = 3^n. The polynomials S(n,t) can be defined recursively by S(0,t)=1, S(n,t)=t^n - t^(n-1) + (2+t)S(n-1,t) for n>=1. S(n,t)=Sum(B(j,t), j=2^n .. 2^(n+1)-1), where B(n,t) are the Stern polynomials, defined by B(0,t)=0, B(1,t)=1, B(2n,t)=tB(n,t), B(2n+1,t)=B(n+1,t)+B(n,t) for n>=1 (see S. Klavzar et al. and A125184). For example, S(2,t)=B(4,t)+B(5,t)+B(6,t)+B(7,t).
Subtriangle of (0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Feb 26 2012
LINKS
S. Klavzar, U. Milutinovic and C. Petr, Stern polynomials, Adv. Appl. Math. 39 (2007), 86-95.
FORMULA
T(n,k)=2^(n-k-1)*(n+k+1)binomial(n,k)/(n-k+1) if k<n; T(n,n)=n+1. Rec. rel.: T(n,k)=2T(n-1,k)+T(n-1,k-1) if k<=n-2.
G.f.: (1-x)/((1-y*x)*(1-(y+2)*x)). - Philippe Deléham, Feb 26 2012
Recurrence : T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) - T(n-2,k-2) with T(0,0) = T(1,0) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Feb 26 2012
EXAMPLE
Triangle starts:
1;
1,2;
2,4,3;
4,10,9,4;
8,24,28,16,5;
16,56,80,60,25,6;
Triangle (0,1,1,0,0,0,...) DELTA (1,1,-1,1,0,0,0,0,...) begins:
1
0, 1
0, 1, 2
0, 2, 4, 3
0, 4, 10, 9, 4
0, 8, 24, 28, 16, 5
0, 16, 56, 80, 60, 25, 6
MAPLE
T:=proc(n, k) if k<n then 2^(n-k-1)*binomial(n, k)*(n+k+1)/(n-k+1) elif k=n then n+1 else 0 fi end: for n from 0 to 11 do seq(T(n, k), k=0..n) od; # yields sequence in triangular form
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Dec 04 2006
STATUS
approved