A187224 Rank transform of the sequence floor(3*n/2).

1, 3, 5, 7, 8, 11, 12, 14, 16, 18, 19, 21, 23, 25, 27, 29, 30, 32, 34, 36, 38, 40, 41, 43, 45, 47, 48, 51, 52, 54, 56, 58, 60, 61, 63, 65, 67, 69, 70, 72, 74, 76, 78, 80, 81, 83, 85, 87, 89, 91, 92, 94, 96, 98, 100, 102, 103, 105, 107, 109, 110, 113, 114, 116, 118, 120, 121, 123, 125, 127, 129, 131, 132, 135, 136, 138, 140, 142, 143, 145, 147, 149, 151, 153, 154, 156, 158, 160, 162, 163, 165, 167, 169, 171, 172, 175, 176, 178, 180, 182, 183, 185, 187, 189, 191, 193, 194, 196, 198, 200

Offset: 1

Author

Clark Kimberling, Mar 07 2011

Keywords

• nonn

Comments

Complement of A187225.

The notion of the rank transform of a sequence is introduced as follows. Suppose that a=(a(n)), for n>=1, is a nondecreasing sequence of nonnegative integers, where a(1)<=1, and suppose that b=(b(n)), for n>=1, is an increasing sequence of positive integers.

Define h(1)=a(1), and for n>1, define h(n)=the number of numbers b(i) satisfying a(n-1)<=b(i)

Define r(1)=1, and for n>1, define r(n)=b(n-1)+h(n)+1.

The sequence r is the adjusted rank sequence when a and b are jointly ranked, with a(i) before b(j) when a(i)=b(j). (For a discussion of adjusted joint rank sequences, see A186219 and A186350.)

If r(n)=b(n) for all n>=1, we call r the rank transform of a and denote it by R(a). To summarize,

(1) initial values: r(1)=1, h(1)=a(1);

(2) counting function: h(n)= # r(i) in the half-open

interval [a(n-1),a(n));

(3) recurrence: r(n)=r(n-1)+h(n)+1.

Assuming a unbounded, let c be the number of a(i)<=1, let c(1)=c+1, and for n>1, let c(n) be the rank of r(n) when all the numbers a(i)<=r(n) and r(1),...,r(n-1), r(n) are jointly ranked. Then, clearly, a(n)<=r(n)<=c(n) for n>=1, and the sequences r and c are a complementary pair.

What conditions on the sequence a will ensure that R(a) exists? That is, what conditions will ensure that the counting function in (2) can be determined inductively, so that the recurrence (3) can be used to self-generate the sequence r? The answer is this: a(n)<=c(n-1)+1; viz., if a(n)>c(n-1)+1, then c(n-1)+1=r(n), but then a(n)>r(n), a contradiction, but if a(n)<=c(n-1)+1, there is no such obstacle.

Examples:

R(A000012)=A000027

R(A000027)=A000201, the lower Wythoff sequence

R(A004526)=A026367

R(A005408)=A005408

Returning now to a and b as above, let (r(1,k)) be the adjusted joint rank sequence (AJRS) of a and b, with a(i) before b(j) when a(i)=b(j). Let (r(2,k)) be the AJRS of a and (r(1,k)); and inductively, let (r(n,k)) be the AJRS of a and (r(n-1,k)). If R(a) exists, then the limit of (r(n,k)) is R(a).

Thus, any choice of initial sequence b can be used to determine the first thousand terms of R(a). In the Mathematica program below, b=(1,2,3,4,...)=A000027.

Examples

			a... 1..3..4..6..7...9...10..12..13..15..16..18..19..
r... 1..3..5..7..8...11..12..14..16..18..19..21..23..
c... 2..4..6..9..10..13..15..17..20..22..24..26..28..
h... 1..1..1..1..0...2...0...1...1...1...0...1...1...
The sequences which converge to R(a), starting with
a=A187224 and b=A000027:
a(k)....1..3..4..6..7...9...10..12..13..15...
b(k)....1..2..3..4..5...6...7...8...9...10...
r(1,k)..1..4..6..9..11..14..16..19..21..24...
r(2,k)..1..3..4..6..8...9...11..13..14..16...
r(3,k)..1..3..5..7..9...11..13..15..16..19...
r(4,k)..1..3..5..7..8...10..12..14..15..17...
r(5,k)..1..3..5..7..8...11..12..14..16..18...
		

Crossrefs

Cf. A186219, A186350, A187225.

Programs

Mathematica

seqA=Table[Floor[3*n/2], {n,1,220}]     (* A032766 *)
seqB=Table[n, {n,1,120}];               (* A000027 *)
jointRank[{seqA_,seqB_}]:={Flatten@Position[#1,{,1}],Flatten@Position[#1,{,2}]}&[Sort@Flatten[{{#1,1}&/@seqA,{#1,2}&/@seqB},1]];
limseqU=FixedPoint[jointRank[{seqA,#1[[1]]}]&,jointRank[{seqA,seqB}]][[1]]                     (* A187224 *)
Complement[Range[Length[seqA]],limseqU] (* A187225 *)
(* by Peter J. C. Moses, Mar 07 2011 *)

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