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%I #28 Jan 21 2019 20:08:16
%S 12,16,4,24,28,32,36,8,44,48,52,56,12,64,68,72,76,16,84,88,92,96,20,
%T 104,108,112,116,24,124,128,132,136,28,144,148,152,156,32,164,168,172,
%U 176,36,184,188,192,196,40,204,208,212,216,44,224,228
%N Numerators of k/(10+k)+1 for k = 2*n-1.
%C For even k the expression k/(k+10)+1 yields A060791 as denominators, A096431 as numerators. For odd k it yields A096431 as denominators, the present sequence as numerators.
%C Note that A096431 is denominator of (9*(n^4 - 2n^3 + 2n^2 - n) + 2)/(2*(2*n-1)), equivalently denominator of (3*n^2 - 3*n + 1)*(3*n^2 - 3*n + 2)/(2*n-1), and that A060791 is n/gcd(n,5).
%F a(n) = numerator((2*n-1)/(2*n+9) + 1).
%F Conjecture: a(n) = 2*a(n-5) - a(n-10) = 4*A060791(n+2) with g.f. -4*x*(-3 - 4*x - x^2 - 6*x^3 - 7*x^4 - 2*x^5 - x^6 + x^8 + 2*x^9) / ( (x-1)^2*(x^4 + x^3 + x^2 + x + 1)^2 ). [_R. J. Mathar_, Dec 07 2010]
%e n=1: (2*1-1)/(2*1+9)+1 = 1/11+1 = 12/11, hence a(1) = 12;
%e n=2: (2*2-1)/(2*2+9)+1 = 3/13+1 = 16/13, hence a(2) = 16;
%e n=3: (2*3-1)/(2*3+9)+1 = 5/15+1 = 1/3+1 = 4/3, hence a(3) = 4;
%p A175784 := proc(n) local k ; k := 2*n-1 ; numer(k/(10+k)+1) ; end proc:
%p seq(A175784(n),n=1..30) ; # _R. J. Mathar_, Feb 05 2011
%t Numerator[Table[k/(10 + k) + 1, {k, 1, 100, 2}]]
%Y Cf. A096431, A060791.
%K nonn
%O 1,1
%A _Steven J. Forsberg_, Dec 04 2010
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