A184822 - OEIS

%I #8 Mar 30 2012 18:37:25

%S 5,11,18,24,30,37,42,49,55,61,68,74,79,86,92,99,105,111,117,123,130,

%T 136,142,149,154,160,167,173,180,186,192,198,204,211,217,223,230,235,

%U 241,248,254,261,267,272,279,285,291,298,304,310,316,322,329,335,342,348,353,360,366,372,379,385,391,397,403,410,416,423,428,434,441,447,453,460,465,472,478,484,491,497,503,509,515,522,528,534,541,546,553,559,565,572,578,583,590,596,603,609,615,621,627,634,640,646,653,658,664,671,677,684,690,696,702,708,715,721,727,734,739,745,752,758,765,771

%N a(n) = n + floor(n*t) + floor(n*t^2), 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^3 = 6.2222625231...

%F a(n) = n + floor(n*q/p) + floor(n*r/p), 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^3 = 1 + t + t^2 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^2)}

%Y Cf. A184820, A184821, A184812, A058265.

%K nonn

%O 1,1

%A _Paul D. Hanna_, Jan 22 2011