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%I #29 Aug 13 2022 06:22:44
%S 1,5,14,55,91,68,95,253,325,203,248,595,703,410,473,1081,1225,689,770,
%T 1711,1891,1040,1139,2485,2701,1463,1580,3403,3655,1958,2093,4465,
%U 4753,2525,2678,5671,5995,3164,3335,7021,7381,3875,4064,8515,8911
%N Numerator of (3n+1)*(3n+2)/4.
%C A trisection of A064038.
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (3,-6,10,-12,12,-10,6,-3,1).
%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^2+4*x+1)*(x^6-2*x^5+12*x^4-13*x^3+12*x^2-2*x+1) ) / ( (x-1)^3*(x^2+1)^3 ). - _R. J. Mathar_, Dec 12 2010
%F The conjecture is correct. - _Charles R Greathouse IV_, Feb 08 2012
%F a(n) ~ -27/8*n^2 - 27/8*n. - _Ralf Stephan_, Jun 16 2014
%F Sum_{n>=0} 1/a(n) = (4/(3*sqrt(3)) - 1/3)*Pi. - _Amiram Eldar_, Aug 13 2022
%t Table[Numerator[(3 n + 1) (3 n + 2)/4], {n, 0, 50}] (* _Wesley Ivan Hurt_, Jun 14 2014 *)
%t LinearRecurrence[{3,-6,10,-12,12,-10,6,-3,1},{1,5,14,55,91,68,95,253,325},50] (* _Harvey P. Dale_, Jan 18 2020 *)
%Y Cf. A064038, A127922.
%K nonn,frac,less,easy
%O 0,2
%A _Paul Curtz_, Dec 09 2010
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