More details on the table A101020/A101021: Explicit form of the rational polynomials R(n,x):=hypergeom([-n,-n],[1/2],x/2): R(n,x) = 1+sum(((binomial(n,k)^2)/binomial(2*k,k))*(2*x)^k,k=1..n). The rational polynomials R(n,x):=hypergeom([-n,-n],[1/2],x/2) = sum(r(n,m)*x^m ,m=0..n), are, for n=0..10: n 0 1 1 1+x 2 1+4*x+2/3*x^2 3 1+9*x+6*x^2+2/5*x^3 4 1+16*x+24*x^2+32/5*x^3+8/35*x^4 5 1+25*x+200/3*x^2+40*x^3+40/7*x^4+8/63*x^5 6 1+36*x+150*x^2+160*x^3+360/7*x^4+32/7*x^5+16/231*x^6 7 1+49*x+294*x^2+490*x^3+280*x^4+56*x^5+112/33*x^6+16/429*x^7 8 1+64*x+1568/3*x^2+6272/5*x^3+1120*x^4+3584/9*x^5+1792/33*x^6+1024/429*x^7+128/6435*x^8 9 1+81*x+864*x^2+14112/5*x^3+18144/5*x^4+2016*x^5+5376/11*x^6+6912/143*x^7+1152/715*x^8+128/12155*x^9 10 1+100*x+1350*x^2+5760*x^3+10080*x^4+8064*x^5+33600/11*x^6+76800/143*x^7+5760/143*x^8+2560/2431*x^9+256/46189*x^10 The rational coefficients furnish the triangle r(n,m), m=0..n, for n=0..10: n 0 1 1 1 1 2 1 4 2/3 3 1 9 6 2/5 4 1 16 24 32/5 8/35 5 1 25 200/3 40 40/7 8/63 6 1 36 150 160 360/7 32/7 16/231 7 1 49 294 490 280 56 112/33 16/429 8 1 64 1568/3 6272/5 1120 3584/9 1792/33 1024/429 128/6435 9 1 81 864 14112/5 18144/5 2016 5376/11 6912/143 1152/715 128/12155 10 1 100 1350 5760 10080 8064 33600/11 76800/143 5760/143 2560/2431 256/46189 Written as rational sequence (n=0..10): [1, 1, 1, 1, 4, 2/3, 1, 9, 6, 2/5, 1, 16, 24, 32/5, 8/35, 1, 25, 200/3, 40, 40/7, 8/63, 1, 36, 150, 160, 360/7, 32/7, 16/231, 1, 49, 294, 490, 280, 56, 112/33, 16/429, 1, 64, 1568/3, 6272/5, 1120, 3584/9, 1792/33, 1024/429, 128/6435, 1, 81, 864, 14112/5, 18144/5, 2016, 5376/11, 6912/143, 1152/715, 128/12155, 1, 100, 1350, 5760, 10080, 8064, 33600/11, 76800/143, 5760/143, 2560/2431, 256/46189] The numerator sequence gives triangle A101020(n): [1, 1, 1, 1, 4, 2, 1, 9, 6, 2, 1, 16, 24, 32, 8, 1, 25, 200, 40, 40, 8, 1, 36, 150, 160, 360, 32, 16, 1, 49, 294, 490, 280, 56, 112, 16, 1, 64, 1568, 6272, 1120, 3584, 1792, 1024, 128, 1, 81, 864, 14112, 18144, 2016, 5376, 6912, 1152, 128, 1, 100, 1350, 5760, 10080, 8064, 33600, 76800, 5760, 2560, 256, ...] a(n,m) tabl head (triangle) for A101020 n\m 0 1 2 3 4 5 6 7 8 9 . . . 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 2 1 4 2 0 0 0 0 0 0 0 3 1 9 6 2 0 0 0 0 0 0 4 1 16 24 32 8 0 0 0 0 0 5 1 25 200 40 40 8 0 0 0 0 6 1 36 150 160 360 32 16 0 0 0 7 1 49 294 490 280 56 112 16 0 0 8 1 64 1568 6272 1120 3584 1792 1024 128 0 9 1 81 864 14112 18144 2016 5376 6912 1152 128 . . . The denominator sequence gives triangle A101021(n): [1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 1, 5, 35, 1, 1, 3, 1, 7, 63, 1, 1, 1, 1, 7, 7, 231, 1, 1, 1, 1, 1, 1, 33, 429, 1, 1, 3, 5, 1, 9, 33, 429, 6435, 1, 1, 1, 5, 5, 1, 11, 143, 715, 12155, 1, 1, 1, 1, 1, 1, 11, 143, 143, 2431, 46189, ...] a(n,m) tabl head (triangle) for A101021 n\m 0 1 2 3 4 5 6 7 8 9 . . . 0 1 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 2 1 1 3 0 0 0 0 0 0 0 3 1 1 1 5 0 0 0 0 0 0 4 1 1 1 5 35 0 0 0 0 0 5 1 1 3 1 7 63 0 0 0 0 6 1 1 1 1 7 7 231 0 0 0 7 1 1 1 1 1 1 33 429 0 0 8 1 1 3 5 1 9 33 429 6435 0 9 1 1 1 5 5 1 11 143 715 12155 . . . #######################################################################################################################################