A068601 -id:A068601 - cp's OEIS frontend

A162615 Triangle read by rows in which row n lists n terms, starting with n, such that the difference between successive terms is equal to n^3 - 1 = A068601(n).

1, 2, 9, 3, 29, 55, 4, 67, 130, 193, 5, 129, 253, 377, 501, 6, 221, 436, 651, 866, 1081, 7, 349, 691, 1033, 1375, 1717, 2059, 8, 519, 1030, 1541, 2052, 2563, 3074, 3585, 9, 737, 1465, 2193, 2921, 3649, 4377, 5105, 5833, 10, 1009, 2008, 3007, 4006, 5005, 6004

Offset: 1

Omar E. Pol, Jul 12 2009

See also the triangles of A162614 and A162616.

			Triangle begins:
  1;
  2,   9;
  3,  29,  55;
  4,  67, 130, 193;
  5, 129, 253, 377, 501;
  6, 221, 436, 651, 866, 1081;
  ...

Harvey P. Dale, Table of n, a(n) for n = 1..10000

Cf. A000583, A068601, A159797, A162609, A162610, A162611, A162612, A162613, A162614, A162616.

A162615 := proc(n,k) n+(k-1)*(n^3-1) ; end proc: seq(seq(A162615(n,k),k=1..n),n=1..15) ; # R. J. Mathar, Feb 05 2010

Flatten[Table[c=n^3-1;NestList[#+c&,n,n-1],{n,10}]] (* Harvey P. Dale, Nov 13 2011 *)

Row sums: n*(n^4 - n^3 + n + 1)/2. - R. J. Mathar, Jul 20 2009

Terms beyond the 6th row from R. J. Mathar and Max Alekseyev, Feb 05 2010

A162616 Triangle read by rows in which row n lists n terms, starting with n^3 + n - 1, such that the difference between successive terms is equal to n^3 - 1 = A068601(n).

Original entry on oeis.org

1, 9, 16, 29, 55, 81, 67, 130, 193, 256, 129, 253, 377, 501, 625, 221, 436, 651, 866, 1081, 1296, 349, 691, 1033, 1375, 1717, 2059, 2401, 519, 1030, 1541, 2052, 2563, 3074, 3585, 4096, 737, 1465, 2193, 2921, 3649, 4377, 5105, 5833, 6561, 1009, 2008

Offset: 1

Omar E. Pol, Jul 12 2009

Note that the last term of the n-th row is the fourth power of n, A000583(n).

See also the triangles of A162614 and A162615.

			Triangle begins:
    1;
    9,  16;
   29,  55,  81;
   67, 130, 193, 256;
  129, 253, 377, 501,  625;
  221, 436, 651, 866, 1081, 1296;
  ...

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000583, A068601, A159797, A162609, A162610, A162611, A162612, A162613, A162614, A162615.

A162616 := proc(n,k) n^3+n-1+(k-1)*(n^3-1) ; end proc: seq(seq(A162616(n,k),k=1..n),n=1..15) ; # R. J. Mathar, Feb 05 2010

Table[NestList[#+n^3-1&,n^3+n-1,n-1],{n,10}]//Flatten (* Harvey P. Dale, Dec 17 2021 *)

Row sums: n*(n^2 + n - 1)*(n^2+1)/2. - R. J. Mathar, Jul 20 2009

More terms from R. J. Mathar, Feb 05 2010

A158620 Partial products of A068601.

Original entry on oeis.org

7, 182, 11466, 1421784, 305683560, 104543777520, 53421870312720, 38891121587660160, 38852230466072499840, 51673466519876424787200, 89240076679826585607494400, 195971208388899181994057702400

Offset: 2

Jonathan Vos Post, Mar 23 2009

A158621(n) = Product_{k=2..n} (k^3+1). A158622(n) is the numerator of the reduced fraction A158620(n)/A158621(n). A158623(n) is the denominator of the reduced fraction A158620(n)/A158621(n).

Also the determinant of the n X n matrix given by m(i,j) = i^3 if i=j and 1 otherwise. For example, Det[{{1,1,1, 1},{1,8,1,1},{1,1,27,1},{1,1,1,64}}] = 11466 = a(4). - John M. Campbell, May 20 2011

			a(2) = 2^3-1 = 7.
a(3) = (2^3-1)*(3^3-1) = 7 * 26 = 182.
a(4) = (2^3-1)*(3^3-1)*(4^3-1) = 7 * 26 * 63 = 11466.

G. C. Greubel, Table of n, a(n) for n = 2..180

Cf. A000217, A005448, A016921, A068601, A158621-A158624, A010791.

Rest[FoldList[Times,1,Range[2,15]^3-1]] (* Harvey P. Dale, Apr 18 2015 *)

a(n) = prod(k = 2, n, k^3 - 1); \\ Michel Marcus, Sep 29 2013

Product_{k=2..n} (k^3-1) = Product_{k=2..n} A068601(k).

a(n) ~ 2^(3/2) * sqrt(Pi) * cosh(sqrt(3)*Pi/2) * n^(3*n+3/2) / (3 * exp(3*n)). - Vaclav Kotesovec, Jul 11 2015

A022003 Decimal expansion of 1/999.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1

Offset: 0

N. J. A. Sloane

Expansion in any base b of 1/(b^3-1). E.g., 1/7 in base 2, 1/26 in base 3, 1/63 in base 4, etc. - Franklin T. Adams-Watters, Nov 07 2006

a(n) = A130196(n) - A131534(n). - Reinhard Zumkeller, Nov 12 2009

			0.001001001001001001001...

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

Essentially the same as A079978.
Cf. A068601.
Partial sums are given by A002264(n+1).

Join[{0,0},RealDigits[1/999,10,120][[1]]] (* or *) PadRight[{},120,{0,0,1}] (* Harvey P. Dale, May 24 2012 *)

a(n)=n%3==2 \\ Jaume Oliver Lafont, Mar 24 2009

From Mario Catalani (mario.catalani(AT)unito.it), Jan 07 2003: (Start)
G.f.: x^2/(1-x^3).
a(n) = -(1/2)*((-1)^floor((2n-1)/3) + (-1)^floor((2n+1)/3)). (End)
From Hieronymus Fischer, May 29 2007: (Start)
a(n) = ((n+2) mod 3) mod 2.
a(n) = (1/2)*(1 - (-1)^(n + floor((n+2)/3))). (End)
a(n) = (1 + (-1)^Fibonacci(n+1))/2. - Hieronymus Fischer, Jun 14 2007
a(n) = (n^5 - n^2) mod 3. - Gary Detlefs, Mar 20 2010
a(n) = ((-1)^(a(n-1) + a(n-2)) + 1)/2 starting from n=3. - Adriano Caroli, Nov 21 2010
a(n) = 1 - Fibonacci(n+1) mod 2. - Gary Detlefs, Dec 26 2010
a(n) = floor((n+1)/3) - floor(n/3). - Tani Akinari, Oct 22 2012

A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 13, 13, 1, 1, 19, 25, 19, 1, 1, 25, 37, 37, 25, 1, 1, 31, 49, 55, 49, 31, 1, 1, 37, 61, 73, 73, 61, 37, 1, 1, 43, 73, 91, 97, 91, 73, 43, 1, 1, 49, 85, 109, 121, 121, 109, 85, 49, 1, 1, 55, 97, 127, 145, 151, 145, 127, 97, 55, 1, 1, 61, 109, 145, 169, 181, 181, 169, 145, 109, 61, 1

Offset: 0

Kolosov Petro, Aug 31 2017

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(1, n, k), i.e T(n, k) is partial case of L(m, n, k) for m = 1.
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
  ----------------------------------------
  k=    0   1   2   3   4   5   6   7   8
  ----------------------------------------
  n=0:  1;
  n=1:  1,  1;
  n=2:  1,  7,  1;
  n=3:  1, 13, 13,  1;
  n=4:  1, 19, 25, 19,  1;
  n=5:  1, 25, 37, 37, 25,  1;
  n=6:  1, 31, 49, 55, 49, 31,  1;
  n=7:  1, 37, 61, 73, 73, 61, 37,  1;
  n=8:  1, 43, 73, 91, 97, 91, 73, 43,  1;

Georg Fischer, Table of n, a(n) for n = 0..495 [rows 0..10 and 12..30 from Kolosov Petro]
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Columns k=0..6 give A000012, A016921, A017533, A161705, A103214, A128470, A158065.
Column sums k=0..4 give A000027, A000567, A051866, A051872, A255185.
Row sums give A001093.
Various cases of L(m, n, k): This sequence (m=1), A300656(m=2), A300785(m=3). See comments for L(m, n, k).
Differences of cubes n^3 are T(A000124(n), 1).
Cf. A000124, A000578, A003154, A003215, A007318, A008458, A016969.
Cf. A038593, A055012, A068601, A077028, A094053, A166873, A227776.
Cf. A294317, A302971, A304042.

Flat(List([0..11],n->List([0..n],k->6*k*(n-k)+1))); # Muniru A Asiru, Oct 09 2018

/* As triangle */ [[6*k*(n-k) + 1: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 26 2018

T := (n, k) -> 6*k*(n-k) + 1:
seq(seq(T(n, k), k=0..n), n=0..11); # Muniru A Asiru, Oct 09 2018

T[n_, k_] := 6 k (n - k) + 1; Column[Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Jun 02 2019 *)

t(n, k) = 6*k*(n-k)+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */
trianglerows(9) \\ Felix Fröhlich, Jan 09 2018

def A287326(n,k): return 6*k*(n-k) + 1
flatten([[A287326(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 25 2024

T(n, k) = 6*k*(n-k) + 1.
G.f. of column k: n^k*(1+(6*k-1)*n)/(1-n)^2.
G.f.: (1 - x - x*y + 7*x^2*y)/((1 - x)^2*(1 - x*y)^2). - Stefano Spezia, Oct 09 2018 [Adapted by Stefano Spezia, Sep 25 2024]
From Kolosov Petro, Jun 05 2019: (Start)
T(n, k) = 1/2 * T(A294317(n, k), k) + 1/2.
T(n+1, k) = 2*T(n, k) - T(n-1, k), for n >= k.
T(n, k) = 6*A077028(n, k) - 5.
T(2n, n) = A227776(n).
T(2n+1, n) = A003154(n+1).
T(2n+3, n) = A166873(n+1).
Sum_{k=0..n-1} T(n, k) = Sum_{k=1..n} T(n, k) = A000578(n).
Sum_{k=1..n-1} T(n, k) = A068601(n).
(n+1)^3 - n^3 = T(A000124(n), 1). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = (-1/2)*(1 + (-1)^n)*A016969(floor(n/2) - 1). - G. C. Greubel, Sep 25 2024

A058895 a(n) = n^4 - n.

Original entry on oeis.org

0, 0, 14, 78, 252, 620, 1290, 2394, 4088, 6552, 9990, 14630, 20724, 28548, 38402, 50610, 65520, 83504, 104958, 130302, 159980, 194460, 234234, 279818, 331752, 390600, 456950, 531414, 614628, 707252, 809970, 923490, 1048544, 1185888, 1336302, 1500590, 1679580

Offset: 0

Henry Bottomley, Jan 08 2001

a(n) is the number of ways to assign 4 different students to n different dorm rooms, each of which can hold at most 3 students. In other words, a(n) is the number of functions f:[4]->[n] with the size of the pre-image set of each element of the codomain at most 3. - Dennis P. Walsh, Mar 21 2013

a(n) are the values of m that yield integer solutions to this family of equations: x = sqrt(m + sqrt(x)), which may also be viewed as an infinitely recursive radical. The real solutions for x at each m = a(n) is n^2, except at n = 1 (m = 0) where x = 0 or 1 is a solution. - Richard R. Forberg, Oct 15 2014

Harry J. Smith, Table of n, a(n) for n = 0..2000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A002061, A002378, A024001, A027444, A027482, A068601, A033562, A185065, A339605.

[n^4-n: n in [0..40]]; // Vincenzo Librandi, Feb 23 2012

seq(n*(n^3-1),n=0..25) ; # R. J. Mathar, Dec 10 2015

Table[n^4 - n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2012 *)

a(n) = { n^4-n } \\ Harry J. Smith, Jun 23 2009

(2*x^2*(7+4*x+x^2)/(1-x)^5).series(x, 37).coefficients(x, sparse=False) # Stefano Spezia, Jul 09 2021

a(n) = n*(n-1)*(n^2+n+1) = A000583(n) - n = A002061(n+1) * A002378(n-1) = (n-1) * A027444(n) = -n * A024001(n).
a(n) = 2*A027482(n). - Zerinvary Lajos, Jan 28 2008
a(n) = floor(n^7/(n^3+1)). - Gary Detlefs, Feb 11 2010
a(n)^3 = (a(n)/n)^4 + (a(n)/n)^3. - Vincenzo Librandi, Feb 23 2012
a(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^3, for n > 0. - Vincenzo Librandi, Mar 13 2012
G.f.: 2*x^2*(7 + 4*x + x^2)/(1 - x)^5. - Colin Barker, Apr 23 2012
a(n) = 14*C(n,2) + 36*C(n,3) + 24*C(n,4). - Dennis P. Walsh, Mar 21 2013
Sum_{n>=2} (-1)^n/a(n) = (Pi/3)*sech(Pi*sqrt(3)/2) + 4*log(2)/3 - 1 = 0.06147271494... . - Amiram Eldar, Jul 04 2020
Sum_{n>=2} 1/a(n) = A339605. - R. J. Mathar, Jan 08 2021
E.g.f.: exp(x)*x^2*(7 + 6*x + x^2). - Stefano Spezia, Jul 09 2021
a(n) = 12*A000332(n+2) + 2*A000537(n-1). - Yasser Arath Chavez Reyes, Apr 05 2024

A162614 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^3 - 1.

Original entry on oeis.org

0, 1, 1, 2, 9, 16, 3, 29, 55, 81, 4, 67, 130, 193, 256, 5, 129, 253, 377, 501, 625, 6, 221, 436, 651, 866, 1081, 1296, 7, 349, 691, 1033, 1375, 1717, 2059, 2401, 8, 519, 1030, 1541, 2052, 2563, 3074, 3585, 4096, 9, 737, 1465, 2193, 2921, 3649, 4377, 5105, 5833

Offset: 0

Omar E. Pol, Jul 15 2009

Note that the last term of the n-th row is the fourth power of n, A000583(n).

See also the triangles of A162615 and A162616.

			Triangle begins:
  0;
  1,   1;
  2,   9,  16;
  3,  29,  55,  81;
  4,  67, 130, 193, 256;
  5, 129, 253, 377, 501,  625;
  6, 221, 436, 651, 866, 1081, 1296;
  ...

Cf. A000583, A068601, A159797, A162609, A162610, A162611, A162612, A162613, A162615, A162616, A162622, A162623, A162624.

def A162614(n,k):
    return n+k*(n**3-1)
print([A162614(n,k) for n in range(20) for k in range(n+1)])
# R. J. Mathar, Oct 20 2009

Sum_{k=0..n} T(n,k) = n*(n^2-n+1)*(n+1)^2/2 (row sums). - R. J. Mathar, Jul 20 2009

T(n,k) = n + k*(n^3-1). - R. J. Mathar, Oct 20 2009

More terms from R. J. Mathar, Oct 20 2009

A033562 a(n) = 2*n^3 + 1.

Original entry on oeis.org

1, 3, 17, 55, 129, 251, 433, 687, 1025, 1459, 2001, 2663, 3457, 4395, 5489, 6751, 8193, 9827, 11665, 13719, 16001, 18523, 21297, 24335, 27649, 31251, 35153, 39367, 43905, 48779, 54001, 59583, 65537, 71875, 78609, 85751, 93313, 101307, 109745, 118639, 128001

Offset: 0

N. J. A. Sloane

A058895(n)^3 + A068601(n)^3 + a(n)^3 = A185065(n)^3, for n>0. - Vincenzo Librandi, Mar 13 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A058895, A068601, A185065.

List([0..50], n-> 2*n^3+1); # G. C. Greubel, Oct 12 2019

[2*n^3+1: n in [0..50]]; // G. C. Greubel, Oct 12 2019

seq(2*n^3+1, n=0..50); # G. C. Greubel, Oct 12 2019

2*Range[0,50]^3+1 (* Vladimir Joseph Stephan Orlovsky, Feb 14 2011*)
CoefficientList[Series[1+x*(3+5x+5x^2-x^3)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Mar 13 2012 *)
LinearRecurrence[{4,-6,4,-1},{1,3,17,55},50] (* Harvey P. Dale, Aug 14 2023 *)

a(n)=2*n^3+1 \\ Charles R Greathouse IV, Mar 11 2012

[2*n^3+1 for n in range(50)] # G. C. Greubel, Oct 12 2019

G.f.: 1 + x*(3 + 5*x + 5*x^2 - x^3)/(1-x)^4. - Vincenzo Librandi, Mar 13 2012

E.g.f.: (1 + 2*x + 6*x^2 + 2*x^3)*exp(x). - G. C. Greubel, Oct 12 2019

Terms a(34) onward added by G. C. Greubel, Oct 12 2019

A123866 a(n) = n^6 - 1.

Original entry on oeis.org

0, 63, 728, 4095, 15624, 46655, 117648, 262143, 531440, 999999, 1771560, 2985983, 4826808, 7529535, 11390624, 16777215, 24137568, 34012223, 47045880, 63999999, 85766120, 113379903, 148035888, 191102975, 244140624, 308915775, 387420488

Offset: 1

Reinhard Zumkeller, Oct 16 2006

a(n) mod 7 = 0 iff n mod 7 > 0: a(A008589(n))=6; a(A047304(n)) = 0; a(n) mod 7 = 6*(1-A082784(n)).

a(n) = A005563(n-1)*A059826(n) = A068601(n)*A001093(n). - Reinhard Zumkeller, Feb 02 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A024004, A001014, A005563, A123865, A123867, A123868.

List([1..30], n-> n^6 -1); # G. C. Greubel, Aug 08 2019

a123866 = (subtract 1) . (^ 6)  -- Reinhard Zumkeller, Mar 11 2014

[n^6 - 1: n in [1..30]]; // Vincenzo Librandi, May 01 2011

A123866:=n->n^6 - 1; seq(A123866(n), n=1..40); # Wesley Ivan Hurt, Feb 26 2014

Table[n^6-1,{n,40}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)
LinearRecurrence[{7,-21,35,-35,21,-7,1}, {0,63,728,4095, 15624,46655, 117648}, 30] (* Harvey P. Dale, Nov 18 2012 *)

A123866(n):=n^6 -1 $ makelist(A123866(n),n,1,30); /* Martin Ettl, Nov 05 2012 */

a(n)=n^6-1 \\ Charles R Greathouse IV, Oct 07 2015

[n^6 -1 for n in (1..30)] # G. C. Greubel, Aug 08 2019

G.f.: x^2*(63 + 287*x + 322*x^2 + 42*x^3 + 7*x^4 - x^5)/(1-x)^7. - Colin Barker, May 08 2012
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7); a(1)=0, a(2)=63, a(3)=728, a(4)=4095, a(5)=15624, a(6)=46655, a(7)=117648. - Harvey P. Dale, Nov 18 2012
Sum_{n>=2} 1/a(n) = 11/12 - Pi*sqrt(3)*tanh(Pi*sqrt(3)/2)/6. - Vaclav Kotesovec, Feb 14 2015
E.g.f.: 1 + (-1 + x + 31*x^2 + 90*x^3 + 65*x^4 + 15*x^5 + x^6)*exp(x). - G. C. Greubel, Aug 08 2019
Product_{n>=2} (1 + 1/a(n)) = 6*Pi^2*sech(sqrt(3)*Pi/2)^2. - Amiram Eldar, Jan 20 2021

A349509 a(n) is the denominator of binomial(n^3 + 6n^2 - 6n + 2, n^3 - 1)/n^3.

Original entry on oeis.org

1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1

Offset: 1

Stefano Spezia, Nov 20 2021

a(n) is the denominator of an upper bound of the number of vertices of the polytope of the n X n X n stochastic tensors, or equivalently, of the number of Latin squares of order n, or equivalently, of the number of n X n X n line-stochastic (0,1)-tensors (see Chang et al. and Zhang et al.).
Conjecture: 1 and 3 are the only terms that appear in this sequence.
This conjecture is correct, see formula. - Kevin Ryde, Jul 01 2023

Kevin Ryde, Table of n, a(n) for n = 1..10000
Haixia Chang, Vehbi Emrah Paksoy, and Fuzhen Zhang, Polytopes of Stochastic Tensors, Ann. Funct. Anal. 7(3): 386-393 (August 2016). arXiv:1608.03203 [math.CO], 2016. See p. 6.
Kevin Ryde, PARI/GP Code and Notes
Fuzhen Zhang and Xiao-Dong Zhang, Comparison of the upper bounds for the extreme points of the polytopes of line-stochastic tensors, arXiv:2110.12337 [math.CO], 2021. See p. 3.

Cf. A000578, A068601.
Cf. A349506, A349507, A349508 (numerators), A349510, A349511, A349512.
Cf. A363739 (run lengths), A349929 (indices of 3's).

a[n_]:=Denominator[Binomial[n^3+6n^2-6n+2,n^3-1]/n^3]; Array[a,90]

PARI
```
\\ See links.
```

from math import gcd, comb
def A349509(n): return n**3//gcd(comb(n*(n*(n + 6) - 6) + 2,6*n*(n-1)+3),n**3) # Chai Wah Wu, Dec 06 2021

A349508(n)/a(n) <= A349510(n) < A349511(n) < A349512(n) (see Corollary 7 in Zhang et al., 2021).
A349508(n)/a(n) ~ 2^(-4 + 6*n - 6*n^2)*3^(-7/2 + 6*n - 6*n^2)*e^(-75 + 233/n + 18*n + 6*n^2)*n^(-1 - 6*n + 6*n^2)/sqrt(Pi).
a(n) = 1 if n=1 or any x[i] + y[i] >= 3 where x and y are the ternary digits of n^3 = Sum x[i]*3^i and 6*n^2 - 6*n + 3 = Sum y[i]*3^i; and a(n) = 3 otherwise. - Kevin Ryde, Jul 01 2023

cp's OEIS Frontend

A162615 Triangle read by rows in which row n lists n terms, starting with n, such that the difference between successive terms is equal to n^3 - 1 = A068601(n).

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A162616 Triangle read by rows in which row n lists n terms, starting with n^3 + n - 1, such that the difference between successive terms is equal to n^3 - 1 = A068601(n).

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A158620 Partial products of A068601.

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A022003 Decimal expansion of 1/999.

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A287326 Triangle read by rows: T(n, k) = 6*k*(n-k) + 1; n >= 0, 0 <= k <= n.

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A058895 a(n) = n^4 - n.

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A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

A349509 a(n) is the denominator of binomial(n^3 + 6n^2 - 6n + 2, n^3 - 1)/n^3.