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%I #29 Feb 21 2019 06:39:26
%S 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,
%T 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,
%U 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
%N 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.
%C 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:
%C 0, 1/1, 1/1, 1/2, 0, -1/6, ...
%C 1/1, 0, -1/2, -1/2, -1/6, 1/6, ...
%C -1/1, -1/2, 0, 1/3, 1/3, 1/12, ...
%C 1/2, 1/2, 1/3, 0, -1/4, -1/4, ...
%C 0, -1/6, -1/3, -1/4, 0, 1/5, ...
%C -1/6, -1/6, 1/12, 1/4, 1/5, 0, ...
%C 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.
%C 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, ...
%C It appears that the sequence of numerators in the first row is 6-periodic: 0, 1, 1, 1, 0, -1.
%F 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).
%F 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
%p 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:
%p A189733 := proc(n) denom(B(0,n)) ; end proc:
%p seq(A189733(n),n=0..80) ; # _R. J. Mathar_, Jun 04 2011
%t 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 *)
%K nonn,frac
%O 0,4
%A _Paul Curtz_, May 23 2011
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