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%I #8 Nov 15 2024 23:33:31
%S 1,3,3,3,21,7,3,21,105,15,3,21,105,465,31,3,21,105,465,1953,63,3,21,
%T 105,465,1953,8001,127,3,21,105,465,1953,8001,32385,255,3,21,105,465,
%U 1953,8001,32385,130305,511,3,21,105,465,1953,8001,32385,130305,522753,1023
%N Denominators in a harmonic triangle; q-analog of A126615, here q = 2.
%C The harmonic triangle uses the terms of this sequence as denominators, numerators = 1. The inverse of the harmonic triangle has entries -2^(n-k-1) for 1<=k<n (subdiagonals) and 2^n - 1 (main diagonal).
%C Conjecture: Row sums of the harmonic triangle are A204243(n) / A005329(n).
%F T(n, k) = (2^k - 1) * (2^(k+1) - 1) for 1 <= k < n; T(n, n) = 2^n - 1.
%F Sum_{k=1..n} 2^(k-1) / T(n, k) = 1.
%F Product_{k=1..n} T(n, k)^((-1)^k) = 1.
%F Row sums are n + 4 * (2^n - 1) * (2^(n-1) - 1) / 3 = n + 4 * A006095(n).
%F G.f.: x*y*(1 + 2*x - 4*x*y + 4*x^2*y)/((1 - x)*(1 - x*y)(1 - 2*x*y)*(1 - 4*x*y)). - _Stefano Spezia_, Oct 23 2024
%e Triangle T(n, k) for 1 <= k <= n starts:
%e n\ k : 1 2 3 4 5 6 7 8 9 10
%e ================================================================
%e 1 : 1
%e 2 : 3 3
%e 3 : 3 21 7
%e 4 : 3 21 105 15
%e 5 : 3 21 105 465 31
%e 6 : 3 21 105 465 1953 63
%e 7 : 3 21 105 465 1953 8001 127
%e 8 : 3 21 105 465 1953 8001 32385 255
%e 9 : 3 21 105 465 1953 8001 32385 130305 511
%e 10 : 3 21 105 465 1953 8001 32385 130305 522753 1023
%e etc.
%e The harmonic triangle starts:
%e [1] 1/1
%e [2] 1/3 1/3
%e [3] 1/3 1/21 1/7
%e [4] 1/3 1/21 1/105 1/15
%e [5] 1/3 1/21 1/105 1/465 1/31
%e [6] 1/3 1/21 1/105 1/465 1/1953 1/63
%e etc.
%e The inverse of the harmonic triangle starts:
%e [1] 1
%e [2] -1 3
%e [3] -2 -1 7
%e [4] -4 -2 -1 15
%e [5] -8 -4 -2 -1 31
%e [6] -16 -8 -4 -2 -1 63
%e etc.
%o (PARI) T(n,k)=if(k<n,(2^k-1)*(2^(k+1)-1),2^n-1)
%Y Cf. A000225, A005329, A006095, A126615, A204243.
%K nonn,easy,tabl,frac
%O 1,2
%A _Werner Schulte_, Oct 22 2024