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Triangle: row #n has n+1 terms. T(n,m) = 4^m (2n+1)! / ( (2n-2m)! (2m+1)! ).
2

%I #15 Mar 18 2018 14:30:31

%S 1,3,4,5,40,16,7,140,336,64,9,336,2016,2304,256,11,660,7392,21120,

%T 14080,1024,13,1144,20592,109824,183040,79872,4096,15,1820,48048,

%U 411840,1281280,1397760,430080,16384

%N Triangle: row #n has n+1 terms. T(n,m) = 4^m (2n+1)! / ( (2n-2m)! (2m+1)! ).

%C Row #n has the unsigned coefficients of a polynomial whose roots are 2 cot (Pi k / (2n+1)) for k=1..2n.

%C Polynomial of row #n = Sum_{m=0..n} (-1)^m*T(n,m)*x^(2n-2m).

%H Rui Duarte and António Guedes de Oliveira, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Duarte/duarte7.html">A Famous Identity of Hajós in Terms of Sets</a>, Journal of Integer Sequences, Vol. 17 (2014), #14.9.1.

%e 1

%e 3x^2 - 4

%e 5x^4 - 40x^2 + 16

%e 7x^6 - 140x^4 + 336x^2 - 64

%e 9x^8 - 336x^6 + 2016x^4 - 2304x^2 + 256

%e 11x^10 - 660x^8 + 7392x^6 - 21120x^4 + 14080x^2 - 1024

%e Polynomial #4 has eight roots: 2 cot (Pi k / 9) for k=1..8.

%o (PARI) T(n,m) = 4^m*(2*n+1)!/((2*n-2*m)!*(2*m+1)!);

%o tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ _Michel Marcus_, Mar 18 2018

%Y Cf. A085840.

%K nonn,tabl

%O 0,2

%A _Gary W. Adamson_, Jul 05 2003

%E Edited by _Don Reble_, Nov 13 2005

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Last modified September 24 03:45 EDT 2024. Contains 376185 sequences. (Running on oeis4.)