A104706 First terms in the rearrangements of integer numbers (see comments).

1, 2, 3, 1, 4, 5, 1, 2, 6, 1, 7, 3, 1, 2, 8, 1, 9, 4, 1, 2, 10, 1, 3, 11, 1, 2, 5, 1, 12, 13, 1, 2, 3, 1, 4, 6, 1, 2, 14, 1, 15, 3, 1, 2, 7, 1, 5, 4, 1, 2, 16, 1, 3, 17, 1, 2, 8, 1, 18, 6, 1, 2, 3, 1, 4, 19, 1, 2, 5, 1, 9, 3, 1, 2, 20, 1, 21, 4, 1, 2, 7, 1, 3, 10, 1, 2, 22, 1, 5, 6, 1, 2, 3, 1, 4, 23, 1

Offset: 1

Author

Zak Seidov, Mar 19 2005

Keywords

• easy
• nonn

Comments

Take the sequence of natural numbers:

s0=1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,

Move the first term s(1)=1 to 2*1 places to the right:

s1=2,3,1,4,5,6,7,8,9,10,11,12,13,14,15,16,

Move the first term s(1)=2 to 2*2 places to the right:

s2=3,1,4,5,2,6,7,8,9,10,11,12,13,14,15,16,

Repeating the procedure we get successively:

s3=1,4,5,2,6,7,3,8,9,10,11,12,13,14,15,16,

s4=4,5,1,2,6,7,3,8,9,10,11,12,13,14,15,16,

s5=5,1,2,6,7,3,8,9,4,10,11,12,13,14,15,16,

s6=1,2,6,7,3,8,9,4,10,11,5,12,13,14,15,16,

.......................................................................

s100=8,3,1,2,24,25,4,5,12,7,6,26,9,27,13,28,29,10,14,30,31,15,11,

32,33,16,34,35,17,36,37,18,38,39,19,40,41,20,42,43,21,44,45,22,

46,47,23,48,49,50,51,52,53,54,55,56,57,58,59,60,

The sequence A104706 gives the first terms in the rearrangements s0,s1,s2,...,s100. Cf. A104705

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A104705, A028920.

Programs

Maple

A104706:= proc(N) # to produce a(1) .. a(N)
    local A, R, n,M;
    M:= N;
    R:= $1..M;
    A[1]:= 1;
    for n from 2 to N do
       if 2*R[1]+1 > M then
         R:= R, [$M+1..M+N]
       fi;
       R:= R[2..2*R[1]+1],R[1],R[2*R[1]+2..N];
       A[n]:= R[1];
    od:
    seq(A[n], n=1..N);
end proc:
A104706(100); # Robert Israel, Dec 04 2015

Mathematica

s=Range[100];bb={1};Do[s=Drop[Insert[s, s[[1]], 2+2s[[1]]], 1];bb=Append[bb, s[[1]]], {i, 100}];bb
NestList[Rest[Insert[#, #[[1]], 2 + 2 #[[1]]]] &, Range[30], 30][[All, 1]] (* Birkas Gyorgy, Mar 03 2011 *)

Sage

def A104706(n):
    m, N = 2, 2*n-1
    while true:
        if m.divides(N): return m-2
        N = N*(m-1)//m
        m += 1
print([A104706(n) for n in (1..97)]) # Peter Luschny, Dec 04 2015

Formula

a(n) = A028920(2n-1)-1. - Benoit Cloitre, Mar 09 2007