A183207 -id:A183207 - cp's OEIS frontend

A093005 a(n) = n * ceiling(n/2).

1, 2, 6, 8, 15, 18, 28, 32, 45, 50, 66, 72, 91, 98, 120, 128, 153, 162, 190, 200, 231, 242, 276, 288, 325, 338, 378, 392, 435, 450, 496, 512, 561, 578, 630, 648, 703, 722, 780, 800, 861, 882, 946, 968, 1035, 1058, 1128, 1152, 1225, 1250, 1326, 1352, 1431, 1458

Offset: 1

Amarnath Murthy, Mar 29 2004

Old name was: The lone multiple of n among the next n numbers.
Another old name: a(n) = n*floor((n+1)/2).
Consider the triangle
   1
   2  3
   4  5  6
   7  8  9 10
  11 12 13 14 15
  16 17 18 19 20 21
  22 23 24 25 26 27 28
  ... Then sequence contains the multiple of n in the n-th row.
Interleaves A000384 and A001105. - Paul Barry, Jun 29 2006
Termwise products of the natural numbers and the natural numbers repeated.
Number of pairs (x,y) having the same parity, with x in {0,...,n} and y in {0,...,2n}. - Clark Kimberling, Jul 02 2012
Similar to generalized hexagonal numbers A000217. Other members of this family are A210977, A006578, A210978, A181995, A210981, A210982. - Omar E. Pol, Aug 09 2012
For even n, a(n) gives the sum of all the parts in the partitions of n into exactly two parts.  For odd n>1, a(n) gives n plus the sum of all the parts in the partitions of n into exactly two parts. - Wesley Ivan Hurt, Nov 14 2013
Number of regions of the plane that do not contain the origin, in the arrangement of lines with polar equations rho = 1/cos(theta-k*2*Pi/n), k=0..n-1; or, by inversion, number of bounded regions in the arrangement of circles with radius 1 and centers the n-th roots of unity. - Luc Rousseau, Feb 08 2019
Numbers k such that floor(sqrt(2k)+1/2) | k. - Wesley Ivan Hurt, Dec 01 2020

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Torleiv Kløve, Linear recurring sequences in boolean rings, Math. Scand., 33 (1973), 5-12.
Torleiv Kløve, Linear recurring sequences in boolean rings, Math. Scand., 33 (1973), 5-12. (Annotated scanned copy)
Luc Rousseau, Illustration, a(n) viewed as a number of regions in an arrangement of lines / of circles.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A000384, A001105, A006578, A008619 (a(n)/n), A181995.

Cf. A183207, A210977, A210978, A210981, A210982, A370980.

[n*(n+(n mod 2))/2: n in [1..70]]; // G. C. Greubel, Mar 14 2024

A093005:=n->n*ceil(n/2); seq(A093005(n), n=1..100); # Wesley Ivan Hurt, Nov 14 2013

a[n_Integer] := n*Floor[(n + 1)/2] (* Olivier Gérard, Jun 21 2007 *)
Table[n*Ceiling[n/2],{n,60}] (* or *) LinearRecurrence[{1,2,-2,-1,1},{1,2,6,8,15},60] (* Harvey P. Dale, May 08 2014 *)

a(n)=(n+1)\2*n \\ Charles R Greathouse IV, Jun 11 2015

for n in range(1,55):
    print(n*((n+1)//2), end=",")   # Alex Ratushnyak, Apr 26 2012

[n*(n +(n%2))/2 for n in range(1,71)] # G. C. Greubel, Mar 14 2024

a(n) = n*floor((n+1)/2).
a(n) = n*A008619(n).
a(2*n-1) = n*(2*n-1), a(2*n) = 2*n^2.
From Paul Barry, Jun 29 2006: (Start)
G.f.: x*(1+x+2*x^2)/((1+x)^2*(1-x)^3).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) = n*((2*n+1) - (-1)^n)/4. (End)
a(n) = n * ceiling(n/2). - Wesley Ivan Hurt, Nov 14 2013
E.g.f.: (1/2)*x*( (x+2)*cosh(x) + (x+1)*sinh(x) ). - G. C. Greubel, Mar 14 2024
Sum_{n>=1} 1/a(n) = Pi^2/12 + 2*log(2). - Amiram Eldar, Mar 15 2024
a(n) = A183207(n) - A370980(n) + 1, by Euler's formula. - Scott R. Shannon, Jul 07 2024

Corrected and extended by Joshua Zucker, May 08 2006
New name from Alex Ratushnyak, Apr 26 2012
New name from Wesley Ivan Hurt, Nov 14 2013

A183209 Tree generated by floor(3n/2): a(1) = 1, a(2n) = (3a(n))-1, a(2n+1) = floor((3a(n+1))/2).

Original entry on oeis.org

1, 2, 3, 5, 4, 8, 7, 14, 6, 11, 12, 23, 10, 20, 21, 41, 9, 17, 16, 32, 18, 35, 34, 68, 15, 29, 30, 59, 31, 62, 61, 122, 13, 26, 25, 50, 24, 47, 48, 95, 27, 53, 52, 104, 51, 101, 102, 203, 22, 44, 43, 86, 45, 89, 88, 176, 46, 92, 93, 185, 91, 182, 183, 365, 19, 38, 39, 77, 37

Offset: 1

Clark Kimberling, Dec 30 2010

A permutation of the positive integers. See the comment at A183079. Leftmost branch of tree is essentially A061418. Rightmost: A007051.

			First levels of the tree:
                      1
                      2
            3                   5
          4   8               7   14

Cf. A183079, A007051, A016789, A183207, A183208, A032766, A191450.
Similar permutations: A048673, A254103.
Inverse permutation: A259431.

import Data.List (transpose)
a183209 n k = a183209_tabf !! (n-1) !! (k-1)
a183209_row n = a183209_tabf !! (n-1)
a183209_tabf = [1] : iterate (\xs -> concat $
   transpose [map a032766 xs, map (a016789 . subtract 1) xs]) [2]
a183209_list = concat a183209_tabf
-- Reinhard Zumkeller, Jun 27 2015

f:= proc(n) option remember;
  if n::even then 3*procname(n/2)-1
  else floor(3*procname((n+1)/2)/2)
  fi
end proc:
f(1):= 1:
seq(f(n), n=1..100); # Robert Israel, Jan 26 2015

a[1]=1; a[n_] := a[n] = If[EvenQ[n], 3a[n/2]-1, Floor[3a[(n+1)/2]/2] ]; Array[a, 100] (* Jean-François Alcover, Feb 02 2018 *)

def a(n):
    if n==1: return 1
    if n%2==0: return 3*a(n//2) - 1
    else: return (3*a((n - 1)//2 + 1))//2
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 06 2017

Let L(n)=floor(3n/2).
Let U(n)=3n-1.  U is the complement of L.
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.
From Antti Karttunen, Jan 26 2015: (Start)
a(1) = 1, a(2n) = (3*a(n))-1, a(2n+1) = A032766(a(n+1)) = floor((3*a(n+1))/2).
Other identities:
a(2^n) = A007051(n) for all n >= 0. [A property shared with A048673 and A254103.]
(End)

Formula to the name-field added by Antti Karttunen, Jan 26 2015

A370980 If n is even, (n^2-2*n+2)/2, otherwise (n^2-n+2)/2.

Original entry on oeis.org

1, 1, 1, 4, 5, 11, 13, 22, 25, 37, 41, 56, 61, 79, 85, 106, 113, 137, 145, 172, 181, 211, 221, 254, 265, 301, 313, 352, 365, 407, 421, 466, 481, 529, 545, 596, 613, 667, 685, 742, 761, 821, 841, 904, 925, 991, 1013, 1082, 1105, 1177, 1201, 1276, 1301, 1379, 1405, 1486, 1513, 1597, 1625, 1712, 1741, 1831, 1861, 1954

Offset: 0

Scott R. Shannon and N. J. A. Sloane, Mar 23 2024

nonn

Total number of circles in A371373 and A371253, if in the later all the circular arcs are completed to form full circles.

The sequence also gives the number of vertices created from circle intersections when a circle of radius r is drawn around each of n equally spaced points on the circumference of a circle of radius r. The number of regions in these constructions is A093005(n) and the number of edges is A183207(n). See the attached images. - Scott R. Shannon, Jul 06 2024.

			a(n) = 1+n*floor((n-1)/2) = 1+n*A004526(n-1). - _Chai Wah Wu_, Mar 23 2024

Scott R. Shannon, Image for n = 3. In this and other images the center of each circle of shown as a white dot.
Scott R. Shannon, Image for n = 4.
Scott R. Shannon, Image for n = 5
Scott R. Shannon, Image for n = 10.
Scott R. Shannon, Image for n = 20.

Bisections are A001844 and A084849.
Cf. A371373, A371374, A371375.
Cf. A004526, A371253, A371254, A371253, A183297, A093005, A183207.

def A370980(n): return n*(n-1>>1)+1 # Chai Wah Wu, Mar 23 2024

a(n) = A183207(n) - A093005(n) + 1, by Euler's formula. - Scott R. Shannon, Jul 07 2024

A374827 Place n equally spaced points on the circumference of a circle of radius r and then connect each pair of points with straight lines whose intersections create A007569(n) - n additional points. Draw a circle of radius r around each of the A007569(n) points. The sequence gives the total number of curved edges formed from all circle intersections.

Original entry on oeis.org

1, 2, 9, 28, 150, 636, 3290, 6192, 35145, 57380, 230494, 192588, 1055535, 1177148

Offset: 1

Scott R. Shannon, Jul 21 2024

Scott R. Shannon, Image for n = 2. In this and other images the starting circles' centers are shown as white dots.
Scott R. Shannon, Image for n = 3.
Scott R. Shannon, Image for n = 4.

Cf. A374825 (vertices), A374826 (regions), A374828 (k-gons), A007569 (total circles), A183207, A374339.

a(n) = A374825(n) + A374826(n) - 1, by Euler's formula.

A183211 First of two trees generated by floor[(3n-1)/2].

Original entry on oeis.org

1, 3, 4, 9, 5, 12, 13, 27, 7, 15, 17, 36, 19, 39, 40, 81, 10, 21, 22, 45, 25, 51, 53, 108, 28, 57, 58, 117, 59, 120, 121, 243, 14, 30, 31, 63, 32, 66, 67, 135, 37, 75, 76, 153, 79, 159, 161, 324, 41, 84, 85, 171, 86, 174, 175, 351, 88, 177

Offset: 1

Clark Kimberling, Dec 30 2010

nonn

This tree grows from (L(1),U(1))=(1,3). The second tree, A183212, grows from (L(2),U(2))=(2,6). Here, L(n)=floor[(3n-1)/2] and U(n)=3n. The two trees are complementary in the sense that every positive integer is in exactly one tree. The sequence formed by taking the terms of this tree in increasing order is A183213. Leftmost branch of this tree: A183207. Rightmost: A000244. See A183170 and A183171 for the two trees generated by the Beatty sequence of sqrt(2).

			First four levels of the tree:
.......................1
.......................3
..............4..................9
............5...12............13....27

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

Cf. A183170, A183212, A001651, A008585, A183209, A183213.

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

See the formula at A183209, but use L(n)=floor[(3n-1)/2] and U(n)=3n instead of L(n)=floor(3n/2) and U(n)=3n-1.

A183208 Iterates of f(x)=floor((3x-1)/2) from x=6.

Original entry on oeis.org

6, 8, 11, 16, 23, 34, 50, 74, 110, 164, 245, 367, 550, 824, 1235, 1852, 2777, 4165, 6247, 9370, 14054, 21080, 31619, 47428, 71141, 106711, 160066, 240098, 360146, 540218, 810326, 1215488, 1823231, 2734846, 4102268, 6153401, 9230101, 13845151

Offset: 1

Clark Kimberling, Dec 30 2010

nonn

			a(1)=6; a(2)=f(a(1))=8; a(3)=f(a(2))=11.

Cf. A183207, A178931.

x=6; Join[{x}, Table[x=Floor[(3*x-1)/2], {n,0,36}]]
f[x_]:=Floor[(3x-1)/2]; NestList[f,6,37]

v=vector(100); v[1]=6; for(n=2,#v,v[n]=floor((3*v[n-1]-1)/2)); v

Iterates of f(x)=floor((3x-1)/2) from x=6.

cp's OEIS Frontend

A093005 a(n) = n * ceiling(n/2).

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A183209 Tree generated by floor(3n/2): a(1) = 1, a(2n) = (3*a(n))-1, a(2n+1) = floor((3*a(n+1))/2).

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A370980 If n is even, (n^2-2*n+2)/2, otherwise (n^2-n+2)/2.

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A183211 First of two trees generated by floor[(3n-1)/2].

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A183208 Iterates of f(x)=floor((3x-1)/2) from x=6.

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A183209 Tree generated by floor(3n/2): a(1) = 1, a(2n) = (3a(n))-1, a(2n+1) = floor((3a(n+1))/2).