W.Lang, Mar 07 2007 a(n,m) tabl head (triangle) for A053125 n\m 0 1 2 3 4 5 6 7 8 0 1 0 0 0 0 0 0 0 0 1 4 -2 0 0 0 0 0 0 0 2 16 -16 3 0 0 0 0 0 0 3 64 -96 40 -4 0 0 0 0 0 4 256 -512 336 -80 5 0 0 0 0 5 1024 -2560 2304 -896 140 -6 0 0 0 6 4096 -12288 14080 -7680 2016 -224 7 0 0 7 16384 -57344 79872 -56320 21120 -4032 336 -8 0 8 65536 -262144 430080 -372736 183040 -50688 7392 -480 9 ###################################################################################################### This triangle A053125, with offset n=1 and doubled entries, is used for Chebyshev U(2*n-1,x) coefficients (without zeros)and decreasing odd powers of x U(2*n-1,x) = sum(U(n,m)*x^(2*(n-m)-1),m=0..n-1), n>=1. U(n,m)=((-1)^m)*binomial(2*n-m-1,m)*2^(2*(n-m)-1), n>=m+1>=1, else 0. U(n,m)=((-1)^m)*sum(binomial(m+p,p)*binomial(2*n,2*(m+p)+1),p=0..n-1-m) U(n,m) tabl head (triangle) n\m 0 1 2 3 4 5 6 7 8 1 2 0 0 0 0 0 0 0 0 2 8 -4 0 0 0 0 0 0 0 3 32 -32 6 0 0 0 0 0 0 4 128 -192 80 -8 0 0 0 0 0 5 512 -1024 672 -160 10 0 0 0 0 6 2048 -5120 4608 -1792 280 -12 0 0 0 7 8192 -24576 28160 -15360 4032 -448 14 0 0 8 32768 -114688 159744 -112640 42240 -8064 672 -16 0 9 131072 -524288 860160 -745472 366080 -101376 14784 -960 18 Tis U(n,m) triangle gives therefore the formulae for sin((2*n)*phi)/sin(phi) in terms of odd powers of cos: n=1: sin(2*phi)/sin(phi) = 2*cos(phi) n=2: sin(4*phi)/sin(phi) = 8*cos(phi)^3 -4*cos(phi) n=3: sin(6*phi)/sin(phi) = 32*cos(phi)^5 -32*cos(phi)^3 +6*cos(phi) . . . The scaled triangle u(n,m):= U(n,m)/2^(2*(n-m)-1) is then u(n,m) tabl head (triangle) n\m 0 1 2 3 4 5 6 7 8 9 1 1 0 0 0 0 0 0 0 0 0 2 1 -2 0 0 0 0 0 0 0 0 3 1 -4 3 0 0 0 0 0 0 0 4 1 -6 10 -4 0 0 0 0 0 0 5 1 -8 21 -20 5 0 0 0 0 0 6 1 -10 36 -56 35 -6 0 0 0 0 7 1 -12 55 -120 126 -56 7 0 0 0 8 1 -14 78 -220 330 -252 84 -8 0 0 9 1 -16 105 -364 715 -792 462 -120 9 0 10 1 -18 136 -560 1365 -2002 1716 -792 165 -10 . . . This coincides with A053123 which has, however, offset n=0. This produces the formulae for sin(2*n*phi)/(2*sin(phi)) in terms of powers of (2^(2*(n-m)-2))*cos(phi)^(2*(n-m)-1) n=1: (1/2)*sin(2*phi)/sin(phi) = 1*(2^0)*cos(phi)^1 = cos(phi) n=2: (1/2)*sin(4*phi)/sin(phi) = 1*(2^2)*cos(phi)^3 -2*(2^0)*cos(phi)= 4*cos(phi)^3 - 2*cos(phi) n=3: (1/2)*sin(6*phi)/sin(phi) = 1*(2^4)*cos(phi)^5 -4*((2^2)*cos(phi)^3) + 3*cos(phi) . . . ############################################ e.o.f. #############################################################