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%I #29 Jan 02 2023 19:13:13
%S 1,1,1,2,1,6,1,4,3,5,1,4,1,7,15,8,1,18,1,10,21,11,1,24,5,13,9,14,1,30,
%T 1,16,11,17,35,12,1,19,39,20,1,42,1,22,9,23,1,48,7,25,17,26,1,54,55,
%U 28,19,29,1,20,1
%N a(n) = denominator of B(0,n), where B(n,n) = 0, B(n-1,n) = 1/n and otherwise B(m,n) = B(m-1,n+1) - B(m-1,n).
%C The numerators are A189731(n).
%C B(0,n) = 0, 1, 1, 3/2, 2, 17/6, 4, 23/4, 25/3, 61/5, 18, 107/4, 40, 421/7, ...
%C is a super autosequence as defined in A242563.
%C The positive integers in B(0,n) give A064723(n). Corresponding rank: A006093(n+1). B(0,n) is linked to the primes A000040.
%C Divisor of B(0,n), n > 0: 1, 1, 1, 2, 2, 4, 5, ... = A172128(n+1).
%C Common (LCM) denominators for the antidiagonals: 1, 1, 1, 2, 2, 6, 6, 12, 12, ... = A139550(n+1)?.
%C 1 = 1
%C 1/2 + 3/2 = 2
%C 1/3 + 5/6 + 17/6 = 4
%C 1/4 + 7/12 + 7/4 + 23/4 = 25/3
%C etc.
%C The positive terms of the first bisection are the sum of the corresponding antidiagonal terms upon the 0's.
%C 0 followed by A001610(n), i.e., 0, 0, 2, 3, 6, 10, 17, ... is an autosequence of the second kind.
%F a(2n+1) = A175386(n).
%F a(n) = denominator(A001610(n)/(n+1)). [edited by _Michel Marcus_, Nov 14 2022]
%F a(n) = denominator((A000204(n+1) - 1)/(n+1)). - _Artur Jasinski_, Nov 06 2022
%t Table[Denominator[(LucasL[n+1]-1)/(n+1)], {n, 0, 100}] (* _Artur Jasinski_, Nov 06 2022 *)
%Y Cf. A000204, A175386, A001610, A189731, A139550, A242563.
%K nonn
%O 0,4
%A _Paul Curtz_, May 26 2014
%E a(24)-a(60) from _Jean-François Alcover_, May 26 2014
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