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%I #18 Aug 27 2022 03:13:41
%S -1,-1,1,2,4,13,19,13,17,43,53,32,38,89,103,59,67,151,169,94,104,229,
%T 251,137,149,323,349,188,202,433,463,247,263,559,593,314,332,701,739,
%U 389,409,859,901,472,494,1033,1079,563,587,1223,1273,662,688,1429,1483,769,797,1651,1709,884,914,1889,1951,1007,1039,2143,2209,1138,1172,2413,2483,1277,1313,2699,2773,1424,1462,3001,3079,1579
%N Numerator of -A127276(n)/A001788(n).
%C The sequence of fractions starts -1/0, -1/1, 1/3, 2/3, 4/5, 13/15, 19/21, 13/14, 17/18, 43/45, 53/55, 32/33, 38/39, ...
%C The denominators are apparently A064038(n+1) = A061041(4+8*n) (i.e., specified as numerators in A061041).
%C The difference between denominator and numerator is A014695(n), n > 0.
%F Conjecture: a(n) = +3*a(n-1) -6*a(n-2) +10*a(n-3) -12*a(n-4) +12*a(n-5) -10*a(n-6) +6*a(n-7) -3*a(n-8) +a(n-9) with g.f. (x^4-x^3+3*x^2-x+1)*(x^4-x^3-2*x^2-x+1) / ( (x-1)^3*(x^2+1)^3 ). - _R. J. Mathar_, Dec 12 2010
%F a(n) = 3*a(n-4) -3*a(n-8) +a(n-12).
%p A001788 := proc(n) n*(n+1)*2^(n-2) ; end proc:
%p A127276 := proc(n) 2^n-A001788(n) ; end proc:
%p A176126 := proc(n) if n = 0 then -1 else 2^n/A001788(n)-1 ; numer(-%) ; end if; end proc:
%p seq(A176126(n),n=0..40) ;
%Y Cf. A001788, A127276.
%Y Cf. A061041, A064038, A014695.
%K sign,frac
%O 0,4
%A _Paul Curtz_, Dec 07 2010
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