A127749 - OEIS

A127749

Inverse of number triangle A(n,k) = 1/(2n+1) if k <= n <= 2k, 0 otherwise.

%I #13 Apr 21 2021 11:33:18

%S 1,0,3,0,-3,5,0,3,-5,7,0,0,0,-7,9,0,-3,5,0,-9,11,0,0,0,0,0,-11,13,0,3,

%T -5,7,0,0,-13,15,0,0,0,0,0,0,0,-15,17,0,0,0,-7,9,0,0,0,-17,19,0,0,0,0,

%U 0,0,0,0,0,-19,21,0,-3,5

%N Inverse of number triangle A(n,k) = 1/(2n+1) if k <= n <= 2k, 0 otherwise.

%C Conjectures: row sums modulo 2 are the Fredholm-Rueppel sequence A036987; row sums of triangle modulo 2 are A111982. Row sums are A127750.

%C The first conjecture is equivalent to the row sums conjecture in A111967. - _R. J. Mathar_, Apr 21 2021

%F T(n,k) = (2*k+1)*A111967(n,k). - _R. J. Mathar_, Apr 21 2021

%e Triangle begins

%e 1;

%e 0, 3;

%e 0, -3, 5;

%e 0, 3, -5, 7;

%e 0, 0, 0, -7, 9;

%e 0, -3, 5, 0, -9, 11;

%e 0, 0, 0, 0, 0, -11, 13;

%e 0, 3, -5, 7, 0, 0, -13, 15;

%e 0, 0, 0, 0, 0, 0, 0, -15, 17;

%e 0, 0, 0, -7, 9, 0, 0, 0, -17, 19;

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, -19, 21;

%e 0, -3, 5, 0, -9, 11, 0, 0, 0, 0, -21, 23;

%e 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -23, 25;

%e Inverse of triangle

%e 1;

%e 0, 1/3;

%e 0, 1/5, 1/5;

%e 0, 0, 1/7, 1/7;

%e 0, 0, 1/9, 1/9, 1/9;

%e 0, 0, 0, 1/11, 1/11, 1/11;

%e 0, 0, 0, 1/13, 1/13, 1/13, 1/13;

%e 0, 0, 0, 0, 1/15, 1/15, 1/15, 1/15;

%e 0, 0, 0, 0, 1/17, 1/17, 1/17, 1/17, 1/17;

%e 0, 0, 0, 0, 0, 1/19, 1/19, 1/19, 1/19, 1/19;

%e 0, 0, 0, 0, 0, 1/21, 1/21, 1/21, 1/21, 1/21, 1/21;

%p A127749 := proc(n,k)

%p option remember ;

%p if k > n then

%p 0 ;

%p elif k = n then

%p 2*n+1 ;

%p else

%p -(2*k+1)*add( procname(n,i)/(2*i+1),i=k+1..min(n,2*k)) ;

%p end if;

%p end proc:

%p seq(seq( A127749(n,k),k=0..n),n=0..20) ; # _R. J. Mathar_, Feb 09 2021

%t nmax = 10;

%t A[n_, k_] := If[k <= n <= 2k, 1/(2n+1), 0];

%t invA = Inverse[Table[A[n, k], {n, 0, nmax}, {k, 0, nmax}]];

%t T[n_, k_] := invA[[n+1, k+1]];

%t Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* _Jean-François Alcover_, Oct 05 2020 *)

%Y Cf. A111967.

%K sign,tabl

%O 0,3

%A _Paul Barry_, Jan 28 2007