A174528 - OEIS

%I #17 May 11 2019 02:12:25

%S 1,1,1,1,8,1,1,39,39,1,1,166,708,166,1,1,677,11584,11584,677,1,1,2724,

%T 186171,753590,186171,2724,1,1,10915,2981685,48417191,48417191,

%U 2981685,10915,1,1,43682,47718190,3101684114,12443227012,3101684114,47718190

%N Triangle T(n,m) = 2*A022168(n,m) - binomial(n, m), 0 <= m <= n, read by rows.

%C Row sums are 1, 2, 10, 80, 1042, 24524, 1131382, 102819584, 18742118986, 6775774063892, 4926666912583390, ... = 2*A006118(n) - 2^n.

%C This triangle essentially compares a Gaussian binomial equivalent to Pascal's triangle and Pascal's triangle itself. - _Alonso del Arte_, Nov 14 2011

%e Triangle begins

%e   1;

%e   1,     1;

%e   1,     8,        1;

%e   1,    39,       39,          1;

%e   1,   166,      708,        166,           1;

%e   1,   677,    11584,      11584,         677,          1;

%e   1,  2724,   186171,     753590,      186171,       2724,        1;

%e   1, 10915,  2981685,   48417191,    48417191,    2981685,    10915,     1;

%e   1, 43682, 47718190, 3101684114, 12443227012, 3101684114, 47718190, 43682, 1;

%p A174528 := proc(n,k)

%p         2*A022168(n,k)-binomial(n,k) ;

%p end proc:

%p seq(seq(A174528(n,m),m=0..n),n=0..10) ; # _R. J. Mathar_, Nov 14 2011

%t c[n_, q_] = Product[1 - q^i, {i, 1, n}];

%t t[n_, m_, q_] = 2*c[n, q]/(c[m, q]*c[n - m, q]) - Binomial[n, m];

%t Table[Flatten[Table[Table[t[n, m, q], {m, 0, n}], {n, 0, 10}]], {q, 2, 12}]

%t (* alternate program *)

%t (* First run the program for A022168 to define gaussianBinom *)

%t Column[Table[2gaussianBinom[n, k, 4] - Binomial[n, k], {n, 0, 8}, {k, 0, n}], Center] (* _Alonso del Arte_, Nov 14 2011 *)

%K nonn,tabl

%O 0,5

%A _Roger L. Bagula_ and _Gary W. Adamson_, Mar 21 2010