A178528 -id:A178528 - cp's OEIS frontend

A183170 First of two trees generated by the Beatty sequence of sqrt(2).

1, 3, 4, 10, 5, 13, 14, 34, 7, 17, 18, 44, 19, 47, 48, 116, 9, 23, 24, 58, 25, 61, 62, 150, 26, 64, 66, 160, 67, 163, 164, 396, 12, 30, 32, 78, 33, 81, 82, 198, 35, 85, 86, 208, 87, 211, 212, 512, 36, 88, 90, 218, 93, 225, 226, 546, 94, 228

Offset: 1

Clark Kimberling, Dec 28 2010

This tree grows from (L(1),U(1))=(1,3). The other tree, A183171, grows from (L(2),U(2))=(2,6). Here, L is the Beatty sequence A001951 of r=sqrt(2); U is the Beatty sequence A001952 of s=r/(r-1). The two trees are complementary; that is, every positive integer is in exactly one tree. (L and U are complementary, too.) The sequence formed by taking the terms of this tree in increasing order is A183172.

			First levels of the tree:
.......................1
.......................3
..............4...................10
.........5..........13........14........34
.......7..17......18..44....19..47....48..116

Ivan Neretin, Table of n, a(n) for n = 1..8192

Cf. A183171, A183172, A001951, A001952, A178528, A074049.

a = {1, 3}; row = {a[[-1]]}; r = Sqrt[2]; s = r/(r - 1); Do[a = Join[a, row = Flatten[{Floor[#*{r, s}]} & /@ row]], {n, 5}]; a (* Ivan Neretin, May 25 2015 *)

See the formula at A178528, but use r=sqrt(2) instead of r=sqrt(3).

A026254 a(n) = j if n = [ jsqrt(3) ], else a(n) = k if n = [ (k/2)(3 + sqrt(3)) ].

Original entry on oeis.org

1, 1, 2, 2, 3, 4, 3, 5, 4, 6, 5, 7, 8, 6, 9, 7, 10, 8, 11, 12, 9, 13, 10, 14, 15, 11, 16, 12, 17, 13, 18, 19, 14, 20, 15, 21, 16, 22, 23, 17, 24, 18, 25, 19, 26, 27, 20, 28, 21, 29, 30, 22, 31, 23, 32, 24, 33, 34, 25, 35, 26, 36, 27, 37, 38, 28

Offset: 1

Clark Kimberling

nonn

Every positive integer occurs exactly twice. a(n) is the parent of n in the tree A178528 generated by the Beatty sequence of sqrt(3).

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

Cf. A178528, A022838, A054406.

N:= 100: # to get a(1)..a(N)
for j from 1 do m:= floor(j*sqrt(3)); if m > N then break fi; A[m]:= j od:
for k from 1 do m:= floor(k/2*(3+sqrt(3))); if m > N then break fi; A[m]:= k od:
seq(A[i],i=1..N); # Robert Israel, Jan 08 2018

Comment by Clark Kimberling, Dec 24 2010

A183080 Tree generated by the Beatty sequence of 3-sqrt(2).

Original entry on oeis.org

1, 2, 3, 5, 4, 8, 7, 13, 6, 10, 12, 21, 11, 18, 20, 35, 9, 16, 15, 27, 19, 32, 33, 56, 17, 29, 28, 48, 31, 54, 55, 94, 14, 24, 25, 43, 23, 40, 42, 73, 30, 51, 50, 86, 52, 89, 88, 151, 26, 46, 45, 78, 44, 75, 76, 129, 49, 83, 85, 146, 87, 148, 149, 254

Offset: 1

Clark Kimberling, Dec 23 2010

A permutation of the positive integers. See the note at A183079.

			First five rows:
1
2
3 5
4 8 7 13
6 10 12 21 11 18 20 35

Cf. A183079, A178528, A074049.

a = {1, 2}; row = {a[[-1]]}; r = 3 - Sqrt[2]; s = r/(r - 1); Do[a = Join[a, row = Flatten[{Floor[#*{r, s}]} & /@ row]], {n, 5}]; a (* Ivan Neretin, Nov 09 2015 *)

Let L(n)=floor(n*r), U(n)=floor(n*s), where r=3-sqrt(2) and s=r/(r-1).

The tree-array T(n,k) is then given by rows: T(0,0) = 1; T(1,0) = 2; T(n,2j) = L(T(n-1),j); T(n,2j+1) = U(T(n-1),j); for j=0,1,...,2^(n-1)-1, n>=2.

A183081 Tree generated by the Beatty sequence of 4-sqrt(5).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 8, 11, 10, 13, 12, 16, 15, 20, 14, 18, 19, 25, 17, 23, 22, 30, 21, 27, 28, 36, 26, 34, 35, 46, 24, 32, 31, 41, 33, 43, 44, 57, 29, 39, 40, 53, 38, 50, 52, 69, 37, 48, 47, 62, 49, 64, 63, 83, 45, 60, 59, 78, 61, 80, 81

Offset: 1

Clark Kimberling, Dec 23 2010

A permutation of the positive integers. See the note at A183079.

			Top 5 rows:
1
2
3 4
5 6 7 9
8 11 10 13 12 16 15 20

Cf. A074049, A178528, A183079.

a = {1, 2}; row = {a[[-1]]}; r = 4 - Sqrt[5]; s = r/(r - 1); Do[a = Join[a, row = Flatten[{Floor[#*{r, s}]} & /@ row]], {n, 5}]; a (* Ivan Neretin, Nov 09 2015 *)

Let L(n)=floor(n*r), U(n)=floor(n*s), where r=4-sqrt(5) and s=r/(r-1).
The tree-array T(n,k) is then given by rows:
T(0,0) = 1; T(1,0) = 2; T(n,2j) = L(T(n-1),j); T(n,2j+1) = U(T(n-1),j);
for j=0,1,...,2^(n-1)-1, n>=2.

cp's OEIS Frontend

A183170 First of two trees generated by the Beatty sequence of sqrt(2).

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A026254 a(n) = j if n = [ j*sqrt(3) ], else a(n) = k if n = [ (k/2)*(3 + sqrt(3)) ].

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A183080 Tree generated by the Beatty sequence of 3-sqrt(2).

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A183081 Tree generated by the Beatty sequence of 4-sqrt(5).

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Mathematica

Formula

A026254 a(n) = j if n = [ jsqrt(3) ], else a(n) = k if n = [ (k/2)(3 + sqrt(3)) ].