A022998 -id:A022998 - cp's OEIS frontend

A083885 (4^n+2^n+0^n+(-2)^n)/4.

1, 1, 6, 16, 72, 256, 1056, 4096, 16512, 65536, 262656, 1048576, 4196352, 16777216, 67117056, 268435456, 1073774592, 4294967296, 17180000256, 68719476736, 274878431232, 1099511627776, 4398048608256, 17592186044416, 70368752566272

Offset: 0

Paul Barry, May 09 2003

Binomial transform of A083884.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (4,4,-16).

[(4^n+2^n+0^n+(-2)^n)/4: n in [0..20]]; // Vincenzo Librandi, Jun 16 2011

Join[{1},Table[(4^n+2^n+(-2)^n)/4,{n,30}]] (* or *) Join[{1}, LinearRecurrence[ {4,4,-16},{1,6,16},30]] (* Harvey P. Dale, Dec 12 2011 *)

a(n) = (4^n+2^n+0^n+(-2)^n)/4.
G.f.: (4*x^3-2*x^2-3*x+1)/((2*x+1)*(2*x-1)*(4*x-1)).
E.g.f.: exp(4*x)+exp(2*x)+exp(0)+exp(-2*x).
A007814(a(n)) = A022998(n-1). - Ralf Stephan, Feb 14 2004
a(0)=1, a(1)=1, a(2)=6, a(3)=16, a(n)=4*a(n-1)+4*a(n-2)-16*a(n-3) [From Harvey P. Dale, Dec 12 2011]

A118432 Denominator of sum of reciprocals of first n 5-simplex numbers A000389.

Original entry on oeis.org

1, 6, 14, 56, 504, 168, 264, 198, 286, 1001, 273, 1456, 1904, 2448, 15504, 969, 1197, 2926, 3542, 42504, 10120, 11960, 14040, 8190, 47502, 5481, 6293, 28768, 32736, 185504, 41888, 11781, 13209, 29526, 164502, 73112, 81016

Offset: 1

Jonathan Vos Post, Apr 28 2006

Numerators are A118431. Fractions are: 1/1, 7/6, 17/14, 69/56, 625/504, 209/168, 329/264, 247/198, 357/286, 1250/1001, 341/273, 1819/1456, 2379/1904, 3059/2448, 19375/15504, 1211/969, 1496/1197, 3657/2926, 4427/3542, 53125/42504, 12649/10120, 14949/11960, 17549/14040, 10237/8190, 59375/47502, 6851/5481, 7866/6293, 35959/28768, 40919/32736, 231875/185504, 52359/41888, 14726/11781, 16511/13209, 36907/29526, 205625/164502, 91389/73112, 101269/81016. The denominator of sum of reciprocals of first n triangular numbers is A026741. The denominator of sum of reciprocals of first n tetrahedral numbers is A118392. The denominator of sum of reciprocals of first n pentatope numbers is A118412.

			a(1) = 1 = denominator of 1/1.
a(2) = 6 = denominator of 7/6 = 1/1 + 1/6.
a(3) = 14 = denominator of 17/14 = 1/1 + 1/6 + 1/21.
a(4) = 56 = denominator of 69/56 = 1/1 + 1/6 + 1/21 + 1/56.
a(5) = 42 = denominator of 55/42 = 1/1 + 1/6 + 1/21 + 1/56 + 1/126.
a(10) = 1001 = denominator of 1250/1001 = 1/1+ 1/6 + 1/21 + 1/56 + 1/126 + 1/252 + 1/462 + 1/792 + 1/1287 + 1/2002.
a(20) = 42504 = denominator of 53125/42504 = 1/1 + 1/6 + 1/21 + 1/56 + 1/126 + 1/252 + 1/462 + 1/792 + 1/1287 + 1/2002 + 1/3003 + 1/4368 + 1/6188 + 1/8568 + 1/11628 + 1/15504 + 1/20349 + 1/26334 + 1/33649 + 1/42504.

G. C. Greubel, Table of n, a(n) for n = 1..5000

Cf. A000332, A000389, A022998, A026741, A118391, A118391, A118411, A118412, A118431.

Denominator[Accumulate[1/Binomial[Range[5,50],5]]] (* Harvey P. Dale, Jul 17 2016 *)

A118411(n)/A118412(n) = Sum_{i=1..n} (1/A000389(n)).
A118411(n)/A118412(n) = Sum_{i=1..n} (1/C(n,5)).
A118411(n)/A118412(n) = Sum_{i=1..n} (1/(n*(n+1)*(n+2)*(n+3)*(n+4)/120)).

A159469 Maximum remainder when (k + 1)^n + (k - 1)^n is divided by k^2 for variable n and k > 2.

Original entry on oeis.org

6, 8, 20, 24, 42, 48, 72, 80, 110, 120, 156, 168, 210, 224, 272, 288, 342, 360, 420, 440, 506, 528, 600, 624, 702, 728, 812, 840, 930, 960, 1056, 1088, 1190, 1224, 1332, 1368, 1482, 1520, 1640, 1680, 1806, 1848, 1980, 2024, 2162, 2208, 2352, 2400, 2550, 2600

Offset: 3

Gaurav Kumar, Apr 13 2009

			For n = 3, maxr => 3*3 - 3 = 6 since 3 is odd.
For n = 4, maxr => 4*4 - 2*4 = 8 since 4 is even.

Iain Fox, Table of n, a(n) for n = 3..10000
Project Euler, Problem 120: Square remainders
Iain Fox, Proof for recursive definition and generating function
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A050187.

LinearRecurrence[{1,2,-2,-1,1},{6,8,20,24,42},50] (* Harvey P. Dale, Apr 18 2018 *)

a(n) = if (n % 2, n^2 - n, n^2 - 2*n); \\ Michel Marcus, Aug 26 2013

first(n) = Vec(x^3*(-6-2*x)/((x+1)^2*(x-1)^3) + O(x^(n+3))) \\ Iain Fox, Nov 26 2017

maxr(n) = n*n - 2*n if n is even, and n*n - n if n is odd.
G.f.: x^3*(-6-2*x)/((x+1)^2*(x-1)^3). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009 (proved by Iain Fox, Nov 26 2017)
a(n) = 2*A050187(n). - R. J. Mathar, Aug 08 2009 (proved by Iain Fox, Nov 27 2017)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 7. - Colin Barker, Oct 29 2017 (proved by Iain Fox, Nov 26 2017)
a(n) = n^2 - n*(3 + (-1)^n)/2. - Iain Fox, Nov 26 2017
From Iain Fox, Nov 27 2017: (Start)
a(n) = A000290(n) - A022998(n).
a(n) = 2*A093005(n-2) + A168273(n-1).
a(n) = (4*(A152749(n-2)) + A091574(n-1) - A010719(n-1))/3.
E.g.f.: x*(exp(x)*x - sinh(x)).
(End)

A168068 Array T(n,k) read by antidiagonals: T(n,2k+1) = 2k+1. T(n,2k) = 2^n*k.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 2, 3, 0, 1, 4, 3, 2, 0, 1, 8, 3, 4, 5, 0, 1, 16, 3, 8, 5, 3, 0, 1, 32, 3, 16, 5, 6, 7, 0, 1, 64, 3, 32, 5, 12, 7, 4, 0, 1, 128, 3, 64, 5, 24, 7, 8, 9, 0, 1, 256, 3, 128, 5, 48, 7, 16, 9, 5, 0, 1, 512, 3, 256, 5, 96, 7, 32, 9, 10, 11, 0, 1, 1024, 3, 512, 5, 192, 7, 64, 9, 20, 11, 6, 0, 1, 2048, 3, 1024, 5

Offset: 0

Paul Curtz, Nov 18 2009

The array is constructed multiplying the even-indexed A026741(k) by 2^n, and keeping the odd-indexed A026471(k) as they are.

Connections to the hydrogen spectrum: The squares of the second row are T(1,k)^2 = A001477(k)^2 = A000290(k) which are the denominators of the Lyman lines (see A171522). The squares of the row T(2,k) are in A154615, denominators of the Balmer series. Row T(3,k) is related to A106833 and A061038.

			The array starts in row n=0 with columns k>=0 as:
0,1,1,3,2,5,3,7,4, A026741
0,1,2,3,4,5,6,7,8, A001477
0,1,4,3,8,5,12,7,16, A022998
0,1,8,3,16,5,24,7,32, A144433
0,1,16,3,32,5,48,7,64,
0,1,32,3,64,5,96,7,128,

A168068 := proc(n,k) if type(k,'odd') then k; else 2^(n-1)*k ; end if; end proc: # R. J. Mathar, Jan 22 2011

A173598 Period 6: repeat [1, 8, 7, 2, 4, 5].

Original entry on oeis.org

1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8, 7, 2, 4, 5, 1, 8

Offset: 0

Paul Curtz, Nov 23 2010

For A141425 = 1,2,4,5,7,8 permutations, see A153130. a(n) is based on A022998. Successive differences are linked to A070366.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Cf. A022998, A070366, A141425, A153130, A166138.

&cat [[1, 8, 7, 2, 4, 5]^^20]; // Wesley Ivan Hurt, Jun 23 2016

A173598:=n->[1, 8, 7, 2, 4, 5][(n mod 6)+1]: seq(A173598(n), n=0..100); # Wesley Ivan Hurt, Jun 23 2016

PadRight[{}, 100, {1, 8, 7, 2, 4, 5}] (* Wesley Ivan Hurt, Jun 23 2016 *)

a(n)=[1,8,7,2,4,5][n%6+1] \\ Charles R Greathouse IV, Jun 02 2011

a(n) = A166138(n) mod 9.
a(2n+1) + a(2n+2) = 9.
G.f.: (1+8*x+7*x^2+2*x^3+4*x^4+5*x^5) / ((1-x)*(1+x)*(1+x+x^2)*(x^2-x+1)). - R. J. Mathar, Mar 08 2011
From Wesley Ivan Hurt, Jun 23 2016: (Start)
a(n) = a(n-6) for n>5.
a(n) = (9 - cos(n*Pi) - 6*cos(2*n*Pi/3) + 2*sqrt(3)*sin(n*Pi/3))/2. (End)

A177883 Period 6: repeat [4, 5, 7, 2, 1, 8].

Original entry on oeis.org

4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5, 7, 2, 1, 8, 4, 5

Offset: 0

Paul Curtz, Dec 14 2010

Represents also the decimal expansion of 16934/37037 and the continued fractions of 0.23839... = (sqrt(496555)-667)/158 or of 4.194699... = (667+sqrt(496555))/327. - R. J. Mathar, Dec 20 2010

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Cf. A173598, A141425, A153130 (permutations).

&cat[[4, 5, 7, 2, 1, 8]: n in [0..20]]; // Wesley Ivan Hurt, Jun 18 2016

A177883:=n->[4, 5, 7, 2, 1, 8][(n mod 6)+1]: seq(A177883(n), n=0..100); # Wesley Ivan Hurt, Jun 18 2016

PadRight[{}, 120, {4,5,7,2,1,8}] (* Harvey P. Dale, Feb 11 2016 *)

a(n)=[4, 5, 7, 2, 1, 8][n%6+1] \\ Charles R Greathouse IV, Jul 17 2016

a(n) = A166304(n) mod 9 = A022998(3n+2) mod 9.
a(2n) + a(2n+1) = 9.
G.f.: (4+5*x+7*x^2+2*x^3+x^4+8*x^5) / ( (1-x)*(1+x)*(1+x+x^2)*(x^2-x+1) ). - R. J. Mathar, Dec 20 2010
From Wesley Ivan Hurt, Jun 18 2016: (Start)
a(n) = a(n-6) for n>5.
a(n) = (9 -cos(n*Pi) + 3*cos(n*Pi/3) - 3*cos(2*n*Pi/3) + sqrt(3)*sin(n*Pi/3) - 3*sqrt(3)*sin(2*n*Pi/3))/2. (End)

A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 15, 6, 14, 10, 45, 15, 33, 21, 91, 28, 60, 36, 153, 45, 95, 55, 231, 66, 138, 78, 325, 91, 189, 105, 435, 120, 248, 136, 561, 153, 315, 171, 703, 190, 390, 210, 861, 231, 473, 253, 1035, 276, 564, 300, 1225, 325

Offset: 0

Paul Curtz, Mar 12 2012

a(n) is divisible by the n-th term of the sequence 3, 3, 1, 1, 3, 3 (periodically repeated with period 6).
a(n) is divisible by b(floor((n-1)/3)), where b(n) = 1, 3, 2, 3, 7, 3, 5, 3, 13, 3, 8, 3, 19, 3,... , n>=0, is defined by inserting a 3 after each entry of A165355.
(n+1)*(n+2)*(n+3)/2=3*A000292(n+1) is divisible by a(n+2), so there is an integer sequence c(n)= 3*A000292(n+1)/a(n+2) = 3, 12, 10, 20, 7, 28, 18,... with c(2*n)=A123167(n+1) and c(n)/A109613(n+2)=A176895(n).
The sequence of denominators of a(n+2)/n has period length 8: 1, 2, 1, 4, 1, 1, 1, 4.
A table T(k,c) = a(1+c*(1+2k)) of (2*k+1)-sections starts as follows:
0  1  1   3   3   15...
0  3  6  45  21   60...
0 15 15  60  55  325...
0 14 28 231 105  315...
0 45 45 189 171 1035...
The table of T'(k,c) = T(k,c)/(2k+1), columns c>=0, looks as follows, construction similar to A165943:
0 1 1  3  3  15  6  14   k=0
0 1 2 15  7  20 15  77   k=1
0 3 3 12 11  65 24  63   k=2
0 2 4 33 15  45 33 175   k=3
0 5 5 21 19 115 42 112   k=4
0 3 6 51 23  70 51 273   k=5
The entries T'(k,c) are divisible by A060819(c).
Differences are T'(2,c)-T'(0,c) = T'(4,c)-T'(2,c) = 0, 2, 2, 9, 8, 50, 18, 49, 32, ... which is A168077(c) multiplied by the c-th term of the period-4 sequence 2, 2, 2, 1.
Differences are T'(3,c)- T'(1,c) = T'(5,c)-T'(3,c) = 0, 1, 2, 18, 8, 25, 18, 98, 32,... which is  A168077(c) multiplied by the period-4 sequence 2, 1, 2, 2.
The reduced fractions T'(0,c)/T'(1,c) = 1, 1/2, 1/5, 3/7, 3/4, 2/5, 2/11, 5/13, 5/7, 3/8, 3/17, 7/19, .., c>=1, have a numerator sequence A026741(floor(c/2)+1). The denominator sequence is f(c) = 1, 2, 5, 7, 4, 5,.. = A001651(c+1)/A130658(c+1), with f(2*c+1) +f(2*c+2) = 3, 12, 9, 24 .. =3*A022998(c).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000217, A014634, A026741, A033991, A064038, A060819, A165355, A208950.

A185138 := proc(n)
        if n mod 4 = 0 then
                return n/4*(n-1) ;
        elif n mod 2 = 1 then
                return (n-1)*(n+1)/8 ;
        else
                return (n-1)*n/2 ;
        end if;
end proc: # R. J. Mathar, Apr 05 2012

Clear[b];b[1] = 0; b[2] = 0; b[3] = 1; b[4] = 1; b[5] = 3; b[6] = 3; b[7] = 15;b[8] = 6;b[n_Integer] := b[n] = ((-2 + n) (-4 (-4 + n) (-3 + n) (-2 + n) (8 + n (-9 + 2 n)) b[-3 + n] + (-5 + n) ((-3 +n) ((-4 + n) (211 + 2 n (-215 + n (147 + n (-41 + 4 n)))) - 4 (-1 + n) (19 + n (-13 + 2 n)) b[-2 + n]) - 4 (-4 + n)^2 (8 + n (-9 + 2 n)) b[-1 + n])))/(4 (-5 + n) (-4 + n) (-3 + n)^2 (19 + n (-13 + 2 n)))
a = Table[b[n], {n, 1, 52}] (* Roger L. Bagula, Mar 14 2012 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,0,1,1,3,3,15,6,14,10,45,15},60] (* Harvey P. Dale, Nov 23 2015 *)

x='x+O('x^50); concat([0,0], Vec(-x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3))) \\ G. C. Greubel, Jun 23 2017

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*n) = A064038(2*n), a(2*n+1) = A000217(n).
a(n) = 3*A208950(n)/A109613(n).
a(n+1) = A060819(n) * A026741(n+2)(floor(n/2)).
G.f.: -x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3). - R. J. Mathar, Mar 22 2012
a(n) = (4*n^2-3*n-1+(2*n^2-3*n+1)*(-1)^n + n*(n-1)*(1+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4))/16. - Luce ETIENNE, May 13 2016
Sum_{n>=2} 1/a(n) = 2 - Pi/4 + 7*log(2)/2. - Amiram Eldar, Aug 12 2022

A201208 One 1, two 2's, three 1's, four 2's, five 1's, ...

Original entry on oeis.org

1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 1

Paul Curtz, Nov 28 2011

			May be written as a triangle:
  1
  2 2
  1 1 1
  2 2 2 2
  1 1 1 1 1
  2 2 2 2 2 2
  1 1 1 1 1 1 1
Row sums are A022998(n+1).

Cf. A057212, A022998.

Cf. A000034.

a201208 n = a201208_list !! (n-1)
a201208_list = concat $ zipWith ($) (map replicate [1..]) a000034_list
-- Reinhard Zumkeller, Dec 02 2011

ReplaceAll[ColumnForm[Table[Mod[k, 2], {k, 12}, {n, k}], Center], 0 -> 2] (* Alonso del Arte, Nov 28 2011 *)

a(n) = A057212(n) + 1. - T. D. Noe, Nov 28 2011

Edited by N. J. A. Sloane, Dec 02 2011

A246943 a(4n) = 4n , a(2n+1) = 8n+4 , a(4n+2) = 2*n+1.

Original entry on oeis.org

0, 4, 1, 12, 4, 20, 3, 28, 8, 36, 5, 44, 12, 52, 7, 60, 16, 68, 9, 76, 20, 84, 11, 92, 24, 100, 13, 108, 28, 116, 15, 124, 32, 132, 17, 140, 36, 148, 19, 156, 40, 164, 21, 172, 44, 180, 23, 188, 48, 196, 25, 204, 52, 212, 27, 220, 56, 228

Offset: 0

Paul Curtz, Sep 08 2014

Consider the denominators of the Balmer series A061038(n) = 0, 4, 1, 36, 16, 100,... (a permutation of the squares of the nonnegative numbers i.e. A000290(n)) divided by A028310(n)=1,1,2,... . The numerators are a(n). The denominators are A138191(n).
Note that A061038(3n)=9*A061038(n), n>=1.
a(3n) is divisible by the period 3 sequence: repeat 9, 3, 3.

			Numerators of a(0)=0/1=0, a(1)=4/1=4, a(2)=1/2, a(3)=36/3=12,... .

J. J. Balmer, Notiz über die Spectrallinien des Wasserstoffs, Annalen der Physik, vol. 261, 5 (1885) 80-87. First published June 25 1884 (Basel).
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A145979, A138191, A061038, A028310, A022998.

A246943:=n->n*(19-(-1)^n*13+2*cos(n*Pi/2))/8: seq(A246943(n), n=0..100); # Wesley Ivan Hurt, Apr 18 2017

LinearRecurrence[{0,0,0,2,0,0,0,-1},{0,4,1,12,4,20,3,28},60] (* Harvey P. Dale, Jun 22 2022 *)

concat(0, Vec(x*(4*x^6+x^5+12*x^4+4*x^3+12*x^2+x+4)/((x-1)^2*(x+1)^2*(x^2+1)^2) + O(x^100))) \\ Colin Barker, Sep 08 2014

Numerators of A061038(n)/A028310(n).
a(2n) = A022998(n).
G.f.: x*(4*x^6+x^5+12*x^4+4*x^3+12*x^2+x+4) / ((x-1)^2*(x+1)^2*(x^2+1)^2). - Colin Barker, Sep 08 2014
a(n) = n*(19-13*(-1)^n+(1+(-1)^n)*(-1)^((2*n-1+(-1)^n)/4))/8. - Luce ETIENNE, May 26 2015
a(n) = n*(19-(-1)^n*13+2*cos(n*Pi/2))/8. - Giovanni Resta, May 26 2015

A300153 Square array T(n, k) read by antidiagonals upwards, n > 0 and k > 0: T(n, k) is the number of parts inscribed in a rose or rhodonea curve with polar coordinates r = cos(t * (k/n)).

Original entry on oeis.org

1, 4, 4, 2, 1, 3, 8, 12, 12, 8, 3, 4, 1, 4, 5, 12, 20, 24, 24, 20, 12, 4, 2, 9, 1, 10, 3, 7, 16, 28, 4, 40, 40, 4, 28, 16, 5, 8, 12, 12, 1, 12, 14, 8, 9, 20, 36, 48, 56, 60, 60, 56, 48, 36, 20, 6, 3, 2, 4, 20, 1, 21, 4, 3, 5, 11, 24, 44, 60, 72, 80, 84, 84, 80

Offset: 1

Rémy Sigrist, Feb 26 2018

For any real p > 0, the rose or rhodonea curve with polar coordinates r = cos(t * p):
- is dense in the unit disk when p is irrational,
- is closed when p is rational, say p = u/v in reduced form; in that case, the number of parts inscribed in the curve is T(v, u),
- see also the illustration in Links section.

			Array T(n, k) begins:
  n\k|    1    2    3    4    5    6    7    8    9
  ---+---------------------------------------------
    1|    1    4    3    8    5   12    7   16    9
    2|    4    1   12    4   20    3   28    8   36
    3|    2   12    1   24   10    4   14   48    3
    4|    8    4   24    1   40   12   56    4   72
    5|    3   20    9   40    1   60   21   80   27
    6|   12    2    4   12   60    1   84   24   12
    7|    4   28   12   56   20   84    1  112   36
    8|   16    8   48    4   80   24  112    1  144
    9|    5   36    2   72   25   12   35  144    1
   10|   20    3   60   20    4    9  140   40  180
   11|    6   44   18   88   30  132   42  176   54
...
The following diagram shows the curve for T(2, 1) and the corresponding 4 parts:
                         |
               ########     ########
           #####      #######      #####
        ###          ###   ###          ###
      ###           ##   |   ##           ###
     ##            ##         ##            ##
    ##             #  Part #2  #             ##
   ##              ##         ##              ##
   #                ###  |  ###                #
  -#- - - Part #3  - -#######- -  Part #1 - - -#-
   #                ###  |  ###                #
   ##              ##         ##              ##
    ##             #  Part #4  #             ##
     ##            ##         ##            ##
      ###           ##   |   ##           ###
        ###          ###   ###          ###
           #####      #######      #####
               ########     ########
                         |

Rémy Sigrist, Illustration of the first terms
Eric Weisstein's World of Mathematics, Rose
Wikipedia, Rose (mathematics)

Cf. A022998, A029578.

T(1, k) = A022998(k).
T(n, k) = T(n/gcd(n, k), k/gcd(n, k)).
Empirically, when gcd(n, k) = 1, we have the following formulas depending on the parity of n and of k:
             | k is odd                       | k is even
   ----------+--------------------------------+--------------------
   n is odd  | T(n, k) =     k * A029578(n+1) | T(n, k) = 2 * k * n
   n is even | T(n, k) = 2 * k * A029578(n+1) | N/A

cp's OEIS Frontend

A083885 (4^n+2^n+0^n+(-2)^n)/4.

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A118432 Denominator of sum of reciprocals of first n 5-simplex numbers A000389.

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A159469 Maximum remainder when (k + 1)^n + (k - 1)^n is divided by k^2 for variable n and k > 2.

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A168068 Array T(n,k) read by antidiagonals: T(n,2k+1) = 2k+1. T(n,2k) = 2^n*k.

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Maple

A173598 Period 6: repeat [1, 8, 7, 2, 4, 5].

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A177883 Period 6: repeat [4, 5, 7, 2, 1, 8].

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A185138 a(4*n) = n*(4*n-1); a(2*n+1) = n*(n+1)/2; a(4*n+2) = (2*n+1)*(4*n+1).

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A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).

A246943 a(4n) = 4n , a(2n+1) = 8n+4 , a(4n+2) = 2*n+1.