A186219 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f and g are the triangular numbers and squares. Complement of A186220.

1, 3, 5, 7, 8, 10, 12, 13, 15, 17, 19, 20, 22, 24, 25, 27, 29, 31, 32, 34, 36, 37, 39, 41, 43, 44, 46, 48, 49, 51, 53, 54, 56, 58, 60, 61, 63, 65, 66, 68, 70, 72, 73, 75, 77, 78, 80, 82, 83, 85, 87, 89, 90, 92, 94, 95, 97, 99, 101, 102, 104, 106, 107, 109, 111, 113, 114, 116, 118, 119, 121, 123, 124, 126, 128, 130, 131, 133, 135, 136, 138, 140, 142, 143, 145, 147, 148, 150, 152, 153, 155, 157, 159, 160, 162, 164, 165, 167, 169, 171

Offset: 1

Author

Clark Kimberling, Feb 15 2011

Keywords

• nonn

Comments

Suppose that f and g are strictly increasing functions for which (f(i)) and (g(j)) are integer sequences. If 0<|d|<1, the sets F={f(i): i>=1} and G={g(j)+d: j>=1} are clearly disjoint. Let f^=(inverse of f) and g^=(inverse of g). When the numbers in F and G are jointly ranked, the rank of f(n) is a(n):=n+floor(g^(f(n))-d), and the rank of g(n)+d is b(n):=n+floor(f^(g(n))+d). Therefore, the sequences a and b are a complementary pair.

Although the sequences (f(i)) and (g(j)) may not be disjoint, the sequences (f(i)) and (g(j)+d) are disjoint, and this observation enables two types of adjusted joint rankings:

(1) if 0

Using f(i)=ui^2+vi+w and g(j)=xj^2+yj+z, we can carry out adjusted joint rankings of any pair of polygonal sequences (triangular, square, pentagonal, etc.) In this case,

a(n)=n+floor((-y+sqrt(4x(un^2+vn+w-z-d)+y^2))/(2x)),

b(n)=n+floor((-v+sqrt(4u(xn^2+yn+z-w+d)+v^2)/(2u)),

where a(n) is the rank of un^2+vn+w and b(n) is the rank

of xn^2+yn+z+d, where d must be chosen small enough, in

absolute value, that the sets F and G are disjoint.

Example: f=A000217 (triangular numbers) and g=A000290 (squares) yield adjusted rank sequences a=A186219 and b=A186220 for d=1/4 and a=A186221 and b=A186222 for d=-1/4.

Examples

			First, write
1..3...6..10..15...21..28..36..45...  (triangular)
1....4.. 9......16...25....36....49.. (square)
Replace each number by its rank, where ties are settled by ranking the triangular number before the square:
a=(1,3,5,7,8,10,12,13,...)
b=(2,4,6,9,11,14,16,18,...).
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A186145 (joint ranks of squares and cubes),

A000217 (triangular numbers),

A000290 (squares),

A186220 (complement of A186119)

A186221 ("after" instead of "before")

A186222 (complement of A186221).

Programs

Magma

[n + Floor(Sqrt((n^2 + n)/2 - 1/4)): n in [1..100]]; // G. C. Greubel, Aug 26 2018

Mathematica

(* adjusted joint ranking of triangular numbers and squares, using general formula *)
d=1/4; u=1/2; v=1/2; w=0; x=1; y=0; z=0;
h[n_]:=-y+(4x(u*n^2+v*n+w-z-d)+y^2)^(1/2);
a[n_]:=n+Floor[h[n]/(2x)]; (* rank of triangular n(n+1)/2 *)
k[n_]:=-v+(4u(x*n^2+y*n+z-w+d)+v^2)^(1/2);
b[n_]:=n+Floor[k[n]/(2u)]; (* rank of square n^2 *)
Table[a[n],{n,1,100}] (* A186219 *)
Table[b[n],{n,1,100}] (* A186220 *)

PARI

vector(100, n, n + floor(sqrt((n^2 + n)/2 - 1/4))) \\ G. C. Greubel, Aug 26 2018

Formula

a(n) = n + floor(sqrt((n^2+n)/2 - 1/4)), (A186219).

b(n) = n + floor((-1 + sqrt(8*n^2+3))/2), (A186220).

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