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Triangle read by rows: T(n,k) is the coefficient of t^k in the polynomial S(n,t)=[(1+t)(2+t)^n+(1-t)t^n]/2 (0<=k<=n).
4

%I #19 Sep 17 2024 21:04:32

%S 1,1,2,2,4,3,4,10,9,4,8,24,28,16,5,16,56,80,60,25,6,32,128,216,200,

%T 110,36,7,64,288,560,616,420,182,49,8,128,640,1408,1792,1456,784,280,

%U 64,9,256,1408,3456,4992,4704,3024,1344,408,81,10,512,3072,8320,13440,14400

%N Triangle read by rows: T(n,k) is the coefficient of t^k in the polynomial S(n,t)=[(1+t)(2+t)^n+(1-t)t^n]/2 (0<=k<=n).

%C 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).

%C 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

%H S. Klavzar, U. Milutinovic and C. Petr, <a href="http://dx.doi.org/10.1016/j.aam.2006.01.003">Stern polynomials</a>, Adv. Appl. Math. 39 (2007), 86-95.

%F 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.

%F G.f.: (1-x)/((1-y*x)*(1-(y+2)*x)). - _Philippe Deléham_, Feb 26 2012

%F 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

%F Sum_{k=0..n} T(n,k)*x^k = A033999(n), A011782(n), A000244(n), A010036(n), A081625(n) for x = -1, 0, 1, 2, 3 respectively. - _Philippe Deléham_, Feb 26 2012

%e Triangle starts:

%e 1;

%e 1,2;

%e 2,4,3;

%e 4,10,9,4;

%e 8,24,28,16,5;

%e 16,56,80,60,25,6;

%e Triangle (0,1,1,0,0,0,...) DELTA (1,1,-1,1,0,0,0,0,...) begins:

%e 1

%e 0, 1

%e 0, 1, 2

%e 0, 2, 4, 3

%e 0, 4, 10, 9, 4

%e 0, 8, 24, 28, 16, 5

%e 0, 16, 56, 80, 60, 25, 6

%p 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

%Y Cf. A125184.

%K nonn,tabl

%O 0,3

%A _Emeric Deutsch_, Dec 04 2006