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a(n) = n + floor(n*t) + floor(n/t), where t is the tribonacci constant.
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%I #7 Mar 30 2012 18:37:25

%S 2,6,9,13,16,20,22,26,29,33,36,40,43,46,50,53,57,60,63,66,70,73,77,81,

%T 83,87,90,94,97,101,104,107,110,114,118,121,125,127,131,134,138,141,

%U 145,147,151,155,158,162,165,168,171,175,178,182,185,189,191,195,199,202,206,209,212,215,219,222,226,229,232,236,239,243,246,250,252,256,259,263,266,270,273,276,280,283,287,290,294,296,300,303,307,311,314,317,320,324,327,331,334,337,340,344,347,351,355,357,361,364,368,371,375,378,381,384,388,392,395,399,401,405,408,412,415,419,421,425,429,432

%N a(n) = n + floor(n*t) + floor(n/t), where t is the tribonacci constant.

%C This is one of three sequences that partition the positive integers.

%C Given t is the tribonacci constant, then the following sequences are disjoint:

%C . A184820(n) = n + [n/t] + [n/t^2],

%C . A184821(n) = n + [n*t] + [n/t],

%C . A184822(n) = n + [n*t] + [n*t^2], where []=floor.

%C This is a special case of Clark Kimberling's results given in A184812.

%F Limit a(n)/n = t^2 = 3.3829757679...

%F a(n) = n + floor(n*p/q) + floor(n*r/q), where p=t, q=t^2, r=t^3, and t is the tribonacci constant (see Clark Kimberling's formula in A184812).

%e Let t be the tribonacci constant, then t^2 = 1 + t + 1/t where:

%e t = 1.8392867552..., t^2 = 3.3829757679..., t^3 = 6.2222625231...

%o (PARI) {a(n)=local(t=real(polroots(1+x+x^2-x^3)[1]));n+floor(n*t)+floor(n/t)}

%Y Cf. A184820, A184822, A184812, A058265.

%K nonn

%O 1,1

%A _Paul D. Hanna_, Jan 22 2011