A189733 Denominator of B(0,n) where B(n,n)=0, B(n-1,n) = (-1)^(n+1)/n, and B(m,n) = B(m+1,n-1) + B(m,n-1), n >= 0, m >= 0, is an array of fractions.

1, 1, 1, 2, 1, 6, 1, 4, 3, 5, 1, 12, 1, 7, 5, 8, 1, 18, 1, 10, 7, 11, 1, 24, 1, 13, 9, 14, 1, 30, 1, 16, 11, 17, 1, 36, 1, 19, 13, 20, 1, 42, 1, 22, 15, 23, 1, 48, 1, 25, 17, 26, 1, 54, 1, 28, 19, 29, 1, 60, 1, 31, 21, 32, 1, 66, 1, 34, 23, 35, 1, 72, 1, 37, 25, 38, 1, 78, 1, 40, 27

Offset: 0

Author

Paul Curtz, May 23 2011

Keywords

• nonn
• frac

Comments

The square array B(m,n) is defined by values on the diagonal and first subdiagonal and the recurrence of building first differences. It begins in row m=0 as:

0, 1/1, 1/1, 1/2, 0, -1/6, ...

1/1, 0, -1/2, -1/2, -1/6, 1/6, ...

-1/1, -1/2, 0, 1/3, 1/3, 1/12, ...

1/2, 1/2, 1/3, 0, -1/4, -1/4, ...

0, -1/6, -1/3, -1/4, 0, 1/5, ...

-1/6, -1/6, 1/12, 1/4, 1/5, 0, ...

The inverse binomial transform of the first row B(0,n) is the first column up to a sign: B(n,0) = (-1)^(n+1)*B(0,n). In this sense, B(0,n) is an eigensequence of the binomial transform.

B(0,n) = 0, 1/1, 1/1, 1/2, 0, -1/6, 0, 1/4, 1/3, 1/5, 0, -1/12, 0, 1/7, 1/5, 1/8, 0, -1/18, 0, 1/10, 1/7, 1/11, ...

It appears that the sequence of numerators in the first row is 6-periodic: 0, 1, 1, 1, 0, -1.

Programs

Maple

B := proc(m,n) option remember; if m=n then 0; elif n = m+1 then (-1)^(n+1)/n ; elif n > m then procname(m,n-1)+procname(m+1,n-1) ; elif n < m then procname(m-1,n+1)-procname(m-1,n) ; end if; end proc:
A189733 := proc(n) denom(B(0,n)) ; end proc:
seq(A189733(n),n=0..80) ; # R. J. Mathar, Jun 04 2011

Mathematica

b[m_, n_] := b[m, n] = Which[m == n, 0, n == m+1, (-1)^(n+1)/n, n > m, b[m, n-1] + b[m+1, n-1], n < m, b[m-1, n+1] - b[m-1, n]]; a[n_] := b[0, n] // Denominator; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Jan 07 2013 *)

Formula

a(n) = denominator(B(0,n)). Conjecture: a(6*n)=1, a(1+6*n)=1+3*n, a(2+6*n)=1+2*n, a(3+6*n)=2+3*n, a(4+6*n)=1, a(5+6*n)=6+6*n. a(n) = 2*a(n-6) - a(n-12).

Empirical g.f.: (1 + x + x^2 + 2*x^3 + x^4 + 6*x^5 - x^6 + 2*x^7 + x^8 + x^9 - x^10) / ((1 - x)^2 * (1 + x)^2 * (1 - x + x^2)^2 * (1 + x + x^2)^2). - Colin Barker, Nov 11 2016