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a(n) = n + [ns/r] + [nt/r] + [nu/r] + [nv/r] + [nw/r], where r=sin(x), s=cos(x), t=tan(x), u=csc(x), v=sec(x), w=cot(x), x=Pi/5.
8

%I #8 Apr 09 2021 22:14:37

%S 9,19,31,41,52,64,74,85,97,107,118,129,140,151,162,173,184,195,205,

%T 216,227,239,249,261,271,282,294,304,313,327,337,348,359,370,381,392,

%U 402,412,425,436,447,457,469,480,490,501,513,523,534,545,557,567,578,589,599,612,621,631,644,655,665,677,687,699

%N a(n) = n + [ns/r] + [nt/r] + [nu/r] + [nv/r] + [nw/r], where r=sin(x), s=cos(x), t=tan(x), u=csc(x), v=sec(x), w=cot(x), x=Pi/5.

%C This is one of six sequences that partition the positive integers. In general, suppose that r, s, t, u, v, w are positive real numbers for which the sets {i/r : i>=1}, {j/s : j>=1}, {k/t : k>=1, {h/u : h>=1}, {p/v : p>=1}, {q/w : q>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the six sets are jointly ranked. Define b(n), c(n), d(n), e(n), f(n) as the ranks of n/s, n/t, n/u, n/v, n/w respectively. It is easy to prove that

%C a(n) = n + [ns/r] + [nt/r] + [nu/r] + [nv/r] + [nw/r],

%C b(n) = [nr/s] + [nt/s] + [nu/s] + [nv/s] + [nw/s],

%C c(n) = [nr/t] + [ns/t] + [nu/t] + [nv/t] + [nw/t],

%C d(n) = n + [nr/u] + [ns/u] + [nt/u] + [nv/u] + [nw/u],

%C e(n) = n + [nr/v] + [ns/v] + [nt/v] + [nu/v] + [nw/v],

%C f(n) = n + [nr/w] + [ns/w] + [nt/w] + [nu/w] + [nv/w], where []=floor. Choosing r=sin(x), s=cos(x), t=tan(x), u=csc(x), v=sec(x), w=cot(x), x=Pi/5, gives a=A190513, b=A190514, c=A190515, d=A190516, e=A190517, f=A190518.

%t x = Pi/5;

%t r=Sin[x]; s=Cos[x]; t=Tan[x]; u=1/r; v=1/s; w=1/t;

%t p[n_, h_, k_] := Floor[n*h/k]

%t a[n_]:=n+p[n,s,r]+p[n,t,r]+p[n,u,r]+p[n,v,r]+p[n,w,r]

%t b[n_]:=n+p[n,r,s]+p[n,t,s]+p[n,u,s]+p[n,v,s]+p[n,w,s]

%t c[n_]:=n+p[n,r,t]+p[n,s,t]+p[n,u,t]+p[n,v,t]+p[n,w,t]

%t d[n_]:=n+p[n,r,u]+p[n,s,u]+p[n,t,u]+p[n,v,u]+p[n,w,u]

%t e[n_]:=n+p[n,r,v]+p[n,s,v]+p[n,t,v]+p[n,u,v]+p[n,w,v]

%t f[n_]:=n+p[n,r,w]+p[n,s,w]+p[n,t,w]+p[n,u,w]+p[n,v,w]

%t Table[a[n], {n,1,120}] (*A190513*)

%t Table[b[n], {n,1,120}] (*A190514*)

%t Table[c[n], {n,1,120}] (*A190515*)

%t Table[d[n], {n,1,120}] (*A190516*)

%t Table[e[n], {n,1,120}] (*A190517*)

%t Table[f[n], {n,1,120}] (*A190518*)

%Y Cf. A190514, A190515, A190516, A190517, A190518 (the other members of the partition).

%K nonn

%O 1,1

%A _Clark Kimberling_, May 11 2011