OFFSET
1,2
COMMENTS
From Clark Kimberling Aug 05 2022, Oct 24 2022: (Start)
R for "right side of expelled (number)",
L for "left side",
I for "inner", i.e., next to expelled, and
REFERENCES
D. Gale, Tracking the Automatic Ant: And Other Mathematical Explorations, ch. 5, p. 27. Springer, 1998.
R. K. Guy, Unsolved Problems in Number Theory, 3rd ed., Springer, 2004; Section E35, p. 359.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Enrique Pérez Herrero, Table of n, a(n) for n = 1..100000
Enrique Pérez Herrero, Kimberling's Expulsion Array
Clark Kimberling, Problem 1615, Crux Mathematicorum, Vol. 17 (2) 44 1991; Solution to Problem 1615, Crux Mathematicorum, Vol. 18, March 1992, pp. 82-83.
Eric Weisstein's World of Mathematics, Kimberling Sequence
FORMULA
a(theta(k)) = 3*theta(k)-(k+1), where theta(k) = Sum_{i=0..k-1} 2^floor(i/3). - Enrique Pérez Herrero, Feb 23 2010
From Connor Brown, May 05 2023 to Feb 01 2024: (Start)
14 sets of values which predictably appear within the sequence have been found, 1 by Richard Guy (1992) and 13 by Connor Brown (2023). Below, k is any positive integer unless otherwise specified.
a(3*2^k-3) = 9*2^k - 3*k - 10. (Guy, 1992)
a(5*2^k-3) = 15*2^k - 3*k - 12.
a(4*2^k-3) = 12*2^k - 3*k - 11.
a((20/3)*2^k-(4/3)) = 20*2^k - 3*k - 13 for odd k.
a((16/3)*2^k-(4/3)) = 16*2^k - 3*k - 12 for even k.
a((40/7)*2^k-(3/7)) = (120/7)*2^k - 3*k - (93/7) for k==1 (mod 3).
a((16/5)*2^k-(2/5)) = (48/5)*2^k - 3*k - (46/5) for k==1 (mod 4).
a((12/5)*2^k-(2/5)) = (36/5)*2^k - 3*k - (41/5) for k==0 (mod 4).
a((48/13)*2^k+(8/13)) = (144/13)*2^k - 3*k - (145/13) for k==1 (mod 12).
a((64/13)*2^k+(8/13)) = (192/13)*2^k - 3*k - (158/13) for k==3 (mod 12).
a((80/13)*2^k+(8/13)) = (240/13)*2^k - 3*k - (171/13) for k==8 (mod 12).
a((64/9)*2^k+(7/9)) = (64/3)*2^k - 3*k - (35/3) for k==1 (mod 6).
a((80/9)*2^k+(7/9)) = (80/3)*2^k - 3*k - (38/3) for k==4 (mod 6).
a((64/15)*2^k+(7/15)) = (64/5)*2^k - 3*k - (63/5) for k>4, k==1 (mod 4).
(End)
EXAMPLE
The eight diagonals described in Comments:
A007063 = RILI = (1, 3, 5, 4, 10, 7, 15, 8, 20, 9, 18, 24, 31, 14, ... )
A282348 = ROLO = (1, 3, 5, 2, 8, 9, 4, 10, 7, 20, 12, 24, 14, 23, ... )
A356376 = LORO = (1, 3, 5, 6, 4, 11, 12, 9, 13, 15, 23, 7, 27, 16, ... )
A356026 = LIRI = (1, 3, 5, 7, 4, 12, 10, 17, 6, 22, 15, 19, 24, 33, ... )
A356377 = ROLI = (1, 3, 5, 4, 8, 6, 10, 15, 2, 9, 13, 26, 11, 12, ... )
A356378 = RILO = (1, 3, 5, 2, 10, 9, 15, 8, 20, 19, 7, 21, 31, 6, ... )
A356379 = LORI = (1, 3, 5, 7, 4, 12, 11, 17, 10, 22, 21, 9, 23, 33, ... )
A356380 = LIRO = (1, 3, 5, 6, 4, 11, 13, 2, 7, 14, 24, 9, 10, 31, ... )
MATHEMATICA
K[i_, j_] := i + j - 1 /; (j >= 2 i - 3);
K[i_, j_] := K[i - 1, i - j/2 - 1] /; (EvenQ[j] && (j < 2 i - 3));
K[i_, j_] := K[i - 1, i + (j - 1)/2] /; (OddQ[j] && (j < 2 i - 3));
A007063[i_] := A007063[i] = K[i, i]; SetAttributes[A007063, Listable] (* Enrique Pérez Herrero, Feb 09 2010 *)
(* Next program generates the 8 arrays with highlighted diagonal sequences. *)
len = 1000;
roli = Join[{{1}},
NestList[
Join[#[[Riffle[Range[Length[#], (Length[#] + 3)/2, -1],
Range[(Length[#] - 1)/2, 1, -1]]]],
Range[#, # + 2] &[(3 Length[#] + 1)/2]] &, {2, 3, 4}, len]];
rili = Join[{{1}},
NestList[Join[#[[Riffle[Range[(Length[#] + 3)/2, Length[#]],
Range[(Length[#] - 1)/2, 1, -1]]]],
Range[#, # + 2] &[(3 Length[#] + 1)/2]] &, {2, 3, 4},
len]]; (*A007063*)
rolo = Join[{{1}},
NestList[Join[#[[Riffle[Range[Length[#], (Length[#] + 3)/2, -1],
Range[1, (Length[#] - 1)/2]]]],
Range[#, # + 2] &[(3 Length[#] + 1)/2]] &, {2, 3, 4},
len]]; (*A282348*)
rilo = Join[{{1}},
NestList[Join[#[[Riffle[Range[(Length[#] + 3)/2, Length[#]],
Range[1, (Length[#] - 1)/2, 1]]]],
Range[#, # + 2] &[(3 Length[#] + 1)/2]] &, {2, 3, 4}, len]];
lori = Join[{{1}},
NestList[
Join[#[[Riffle[Range[1, (Length[#] - 1)/2],
Range[(Length[#] + 3)/2, Length[#]]]]],
Range[#, # + 2] &[(3 Length[#] + 1)/2]] &, {2, 3, 4}, len]];
liri = Join[{{1}},
NestList[Join[#[[Riffle[Range[(Length[#] - 1)/2, 1, -1],
Range[(Length[#] + 3)/2, Length[#]]]]],
Range[#, # + 2] &[(3 Length[#] + 1)/2]] &, {2, 3, 4},
len]]; (*A356026*)
loro = Join[{{1}},
NestList[Join[#[[Riffle[Range[1, (Length[#] - 1)/2],
Range[Length[#], (Length[#] + 3)/2, -1]]]],
Range[#, # + 2] &[(3 Length[#] + 1)/2]] &, {2, 3, 4}, len]];
liro = Join[{{1}},
NestList[
Join[#[[Riffle[Range[(Length[#] - 1)/2, 1, -1],
Range[Length[#], (Length[#] + 3)/2, -1]]]],
Range[#, # + 2] &[(3 Length[#] + 1)/2]] &, {2, 3, 4}, len]];
(Map[{#, Take[Flatten[Map[Take[#, {(Length[#] + 1)/2}] &, #]], 200] &[
ToExpression[#]]} &, {"rolo", "rilo", "roli", "rili", "loro",
"liro", "lori", "liri"}]) // ColumnForm
rows = 10; Map[{#,
Grid[Map[Map[StringPadLeft[ToString[#], 2] &, #] &,
Take[ToExpression[#], rows]],
Frame -> {None, None, Map[{#, #} -> True &, Range[rows]]},
FrameStyle -> Directive[Red]]} &, {"rolo", "rilo", "roli", "rili",
"loro", "liro", "lori", "liri"}]
(* Peter J. C. Moses, Oct 24 2022; Clark Kimberling, Oct 24 2022 *)
PROG
(PARI) K(i, j) = { my(i1, j1); i1=i; j1=j;
while(j1<(2*i1-3), if(j1%2, j1=i1+((j1-1)/2), j1=i1-((j1+2)/2)); i1--; );
return(i1+j1-1); }
A007063(i)=K(i, i); \\ Enrique Pérez Herrero, Feb 21 2010
CROSSREFS
KEYWORD
nonn,nice,easy,changed
AUTHOR
EXTENSIONS
More terms from James A. Sellers, Dec 23 1999
STATUS
approved