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a(n) = n + floor(n*t) + floor(n*t^2) + floor(n*t^3) + floor(n*t^4), where t is the pentanacci constant.
5

%I #13 Dec 12 2019 15:11:44

%S 26,56,85,115,144,174,204,232,262,291,321,351,380,410,438,468,497,526,

%T 556,585,615,645,673,703,732,762,792,821,851,878,908,938,966,996,1025,

%U 1055,1085,1113,1143,1172,1202,1232,1261,1291,1319,1349,1379,1408,1437,1466,1496,1525,1554,1584,1613,1643,1673,1702,1731,1759,1789,1819,1848,1878,1906,1936,1965,1994,2024,2053,2083,2113,2142,2172,2200,2230,2260,2289,2319,2348,2377,2406,2435,2465,2494,2524,2554,2583,2611,2640,2670

%N a(n) = n + floor(n*t) + floor(n*t^2) + floor(n*t^3) + floor(n*t^4), where t is the pentanacci constant.

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

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

%C . A184835(n) = n + [n/t] + [n/t^2] + [n/t^3] + [n/t^4],

%C . A184836(n) = n + [n*t] + [n/t] + [n/t^2] + [n/t^3],

%C . A184837(n) = n + [n*t] + [n*t^2] + [n/t] + [n/t^2],

%C . A184838(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n/t],

%C . A184839(n) = n + [n*t] + [n*t^2] + [n*t^3] + [n*t^4], where []=floor.

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

%H Harvey P. Dale, <a href="/A184839/b184839.txt">Table of n, a(n) for n = 1..1000</a>

%F Limit a(n)/n = t^5 = 29.367054786236720687050865...

%F a(n) = n + floor(n*q/p) + floor(n*r/p) + floor(n*s/p) + floor(n*u/p), where p=t, q=t^2, r=t^3, s=t^4, u=t^5, and t is the pentanacci constant.

%e Given t = pentanacci constant, then t^5 = 1 + t + t^2 + t^3 + t^4,

%e t = 1.965948236645..., t^2 = 3.864952469169..., t^3 = 7.598296491482..., t^4 = 14.93785758893..., t^5 = 29.36705478623...

%t With[{t=Root[x^5-x^4-x^3-x^2-x-1,1]},Table[n+Total@@Through[ Floor[ n*t^Range[4]]],{n,100}]] (* _Harvey P. Dale_, Dec 12 2019 *)

%o (PARI) {a(n)=local(t=real(polroots(1+x+x^2+x^3+x^4-x^5)[1])); n+floor(n*t)+floor(n*t^2)+floor(n*t^3)+floor(n*t^4)}

%Y Cf. A184835, A184836, A184837, A184838; A184812, A103814.

%K nonn

%O 1,1

%A _Paul D. Hanna_, Jan 23 2011