A068601 -id:A068601 - cp's OEIS frontend

A081257 a(n) is the greatest prime factor of (n^3 - 1).

7, 13, 7, 31, 43, 19, 73, 13, 37, 19, 157, 61, 211, 241, 13, 307, 17, 127, 421, 463, 13, 79, 601, 31, 37, 757, 271, 67, 29, 331, 151, 1123, 397, 97, 43, 67, 1483, 223, 547, 1723, 139, 631, 283, 109, 103, 61, 181, 43, 2551, 379, 919, 409, 2971, 79, 103, 3307, 163

Offset: 2

Jan Fricke, Mar 14 2003

The record values here (as well as those for A081256) appear to match the terms of A002383 for n > 1. - Bill McEachen, Jun 19 2023

			a(7)=19 because 7^3 - 1 = 342 = 2*3*3*19.

Jon E. Schoenfield, Table of n, a(n) for n = 2..10000 (first 999 terms from R. J. Mathar).

Cf. A081256, A081258, A002383.

Cf. A096175 (n^3-1 is an odd semiprime), A096176 ((n^3-1)/(n-1) is prime).

A081257 := proc(n)
        A006530( n^3-1) ;
end proc: # R. J. Mathar, Jul 18 2015

FactorInteger[#][[-1,1]]&/@(Range[2,60]^3-1) (* Harvey P. Dale, Oct 09 2017 *)

a(n)=my(f=factor(n^3-1)); f[#f~,1] \\ Charles R Greathouse IV, Mar 08 2017

from sympy import primefactors
def A081257(n): return max(primefactors(n-1)+primefactors(n*(n+1)+1)) # Chai Wah Wu, Oct 15 2022

a(n) = A006530(A068601(n)). - Michel Marcus, Jun 19 2023

More terms from Hugo Pfoertner, Jun 21 2004

A258807 a(n) = n^5 - 1.

Original entry on oeis.org

0, 31, 242, 1023, 3124, 7775, 16806, 32767, 59048, 99999, 161050, 248831, 371292, 537823, 759374, 1048575, 1419856, 1889567, 2476098, 3199999, 4084100, 5153631, 6436342, 7962623, 9765624, 11881375, 14348906, 17210367, 20511148, 24299999, 28629150, 33554431

Offset: 1

Vincenzo Librandi, Jun 11 2015

Muniru A Asiru, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Subsequence of A181124.
Cf. A024003, A024049, A300656.
Sequences of the type n^k-1: A132411 (k=2), A068601 (k=3), A123865 (k=4), this sequence (k=5), A123866 (k=6), A258808 (k=7), A258809 (k=8), A258810 (k=9), A123867 (k=10), A258812 (k=11), A123868 (k=12).

List([1..35],n->n^5-1); # Muniru A Asiru, Oct 28 2018

Magma
```
[n^5-1: n in [1..50]];
```

I:=[0,31,242,1023, 3124,7775]; [n le 6 select I[n] else 6*Self(n-1)-15*Self(n-2)+20*Self(n-3)-15*Self(n-4)+ 6*Self(n-5)-Self(n-6): n in [1..50]];

seq(n^5-1,n=1..35); # Muniru A Asiru, Oct 28 2018

Table[n^5 - 1, {n, 1, 50}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 31, 242, 1023, 3124, 7775}, 50]

a(n)=n^5-1 \\ Charles R Greathouse IV, Jun 11 2015

for n in range(1, 50): print(n**5 - 1, end=', ') # Stefano Spezia, Oct 28 2018

[n^5-1 for n in (1..50)] # Bruno Berselli, Jun 11 2015

G.f.: x^2*(31 + 56*x + 36*x^2 - 4*x^3 + x^4)/(1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = -A024003(n). - Bruno Berselli, Jun 11 2015
Sum_{n>=2} 1/a(n) = Sum_{n>=1} (zeta(5*n) - 1) = 0.0379539032... - Amiram Eldar, Nov 06 2020

A100119 a(n) = n-th centered n-gonal number.

Original entry on oeis.org

1, 2, 7, 19, 41, 76, 127, 197, 289, 406, 551, 727, 937, 1184, 1471, 1801, 2177, 2602, 3079, 3611, 4201, 4852, 5567, 6349, 7201, 8126, 9127, 10207, 11369, 12616, 13951, 15377, 16897, 18514, 20231, 22051, 23977, 26012, 28159, 30421, 32801, 35302

Offset: 0

Jonathan Vos Post, Dec 26 2004

a(n) is n times the n-th triangular number plus 1. - Thomas M. Green, Nov 16 2009
From Gary W. Adamson, Jul 31 2010: (Start)
Equals (1, 2, 3, 4, ...) convolved with (1, 0, 4, 7, 10, 13, ...).
Example: a(5) = 76 = (6, 5, 4, 3, 2, 1) dot (1, 0, 4, 7, 10, 13) = (6 + 0 + 16 + 21 + 20 + 13). (End)

			a(2) = 2*3 + 1 = 7, a(3) = 3*6 + 1 = 19, a(4) = 4*10 + 1 = 41. - _Thomas M. Green_, Nov 16 2009

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

See also A101357 (Cumulative sums of the n-th n-gonal numbers).
A diagonal of A101321.
Cf. A060354, A098547, A101357.
Cf. A006003, A005920.

I:=[1, 2, 7, 19]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 25 2012

Table[(n^3+n^2)/2+1,{n,0,6!}] (* Vladimir Joseph Stephan Orlovsky, Mar 06 2010 *)
LinearRecurrence[{4,-6,4,-1},{1,2,7,19},40] (* Vincenzo Librandi, Jun 25 2012 *)

a(n) = n^2*(n+1)/2+1; \\ Altug Alkan, Sep 21 2018

a(n) = 1 + n*(n + n^2)/2 = 1 + (1/2)*n^2 + (1/2) * n^3 = 1 + mean(n^2, n^3). - Joshua Zucker, May 03 2006
Equals A002411(n) + 1. - Olivier Gérard, Jun 20 2007
G.f.: (1 - 2*x + 5*x^2 - x^3) / (x-1)^4. - R. J. Mathar, Apr 04 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 25 2012
a(n) = (A098547(n)+1)/2. - Richard Turk, Jul 18 2017
a(n) = A060354(n+2) - A000290(n+1) = A006003(n+1) - A005563(n) and for n>0 A005920(n) - A068601(n+1). - Bruce J. Nicholson, Jun 23 2018

Corrected and extended by Joshua Zucker, May 03 2006

A158621 Partial products of A001093.

Original entry on oeis.org

9, 252, 16380, 2063880, 447861960, 154064514240, 79035095805120, 57695619937737600, 57753315557675337600, 76927416322823549683200, 133007502822161917402252800, 292350491203111894450151654400

Offset: 2

Jonathan Vos Post, Mar 23 2009

A158620(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).

			a(2) = 2^3+1 = 9. a(3) = (2^3+1)*(3^3+1) = 9 * 28 = 252. a(4) = (2^3+1)*(3^3+1)*(4^3+1) = 9 * 28 * 65 = 16380.

Cf. A001093, A016921, A068601, A158620, A158622-A158624, A255433.

Table[Product[(k^3+1),{k,2,n}],{n,2,20}] (* Vaclav Kotesovec, Jul 11 2015 *)
FoldList[Times,Range[2,20]^3+1] (* Harvey P. Dale, Oct 15 2017 *)

PRODUCT[k=2..n](k^3+1) = PRODUCT[k=2..n]A001093(k).

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

A185065 a(n) = n*(n^3 + 2).

Original entry on oeis.org

0, 3, 20, 87, 264, 635, 1308, 2415, 4112, 6579, 10020, 14663, 20760, 28587, 38444, 50655, 65568, 83555, 105012, 130359, 160040, 194523, 234300, 279887, 331824, 390675, 457028, 531495, 614712, 707339, 810060, 923583, 1048640

Offset: 0

Vincenzo Librandi, Mar 31 2011

Numbers a(n) such that a(n)^3 = x^3*(x-2). The values of x are in A084380.

A058895(n)^3 + A068601(n)^3 + A033562(n)^3 = a(n)^3, for n > 0. - Vincenzo Librandi, Mar 13 2012

			20^3 = 10^3*(10-2); 87^3 = 29^3*(29-2).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A084380, A058895, A068601, A033562.

Table[n(n^3+2),{n,0,40}]  (* Harvey P. Dale, Apr 11 2011 *)

a(n)=n*(n^3+2) \\ Charles R Greathouse IV, Oct 07 2015

G.f.: x*(3 + 5*x + 17*x^2 - x^3)/(1-x)^5. - Bruno Berselli, Mar 31 2011

A349508 a(n) is the numerator of binomial(n^3 + 6n^2 - 6n + 2, n^3 - 1)/n^3.

Original entry on oeis.org

1, 21318, 111399602430962720, 219754881677312748254868619396977023490, 91574665590547903212939476569574243557076290573519342040406738188187312

Offset: 1

Stefano Spezia, Nov 20 2021

a(n) is the numerator 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.).

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.
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, A349509 (denominators), A349510, A349511, A349512.

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

a(n)/A349509(n) <= A349510(n) < A349511(n) < A349512(n) (see Corollary 7 in Zhang et al., 2021).

a(n)/A349509(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).

A129294 Number of divisors of n^3 - 1 that are not greater than n.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 5, 3, 5, 2, 7, 2, 4, 7, 5, 3, 6, 2, 6, 6, 6, 2, 13, 4, 4, 4, 8, 4, 10, 3, 6, 5, 6, 5, 14, 2, 6, 5, 8, 3, 10, 3, 8, 10, 5, 3, 16, 3, 9, 5, 9, 2, 11, 5, 8, 7, 4, 3, 20, 2, 5, 9, 11, 4, 18, 4, 6, 5, 8, 3, 14, 5, 4, 8, 6, 4, 17, 2, 21, 5, 6, 3, 16, 6, 10, 8, 8, 2, 14, 5, 9, 7, 6, 5, 16

Offset: 2

Reinhard Zumkeller, Apr 09 2007

nonn

a(n) = #{d: d<=n and A068601(n) mod d = 0};

a(n)>1 for n>2, see A129295 for m such that a(m)=2: a(A129295(n))=2.

			a(100) = #{1,3,7,9,11,13,21,27,33,37,39,63,77,91,99} = 15.

R. Zumkeller, Table of n, a(n) for n = 2..1000

Cf. A129292, A129296.

Table[Count[Divisors[n^3-1],?(#<n+1&)],{n,2,100}] (* _Harvey P. Dale, Sep 27 2018 *)

a(n) = sumdiv(n^3-1, d, d <= n); \\ Michel Marcus, Aug 01 2018

a(1)=1 removed by Michel Marcus, Aug 01 2018

A257238 Triangle T(n, k) = n^3 - k^3, 0 <= k < = n.

Original entry on oeis.org

0, 1, 0, 8, 7, 0, 27, 26, 19, 0, 64, 63, 56, 37, 0, 125, 124, 117, 98, 61, 0, 216, 215, 208, 189, 152, 91, 0, 343, 342, 335, 316, 279, 218, 127, 0, 512, 511, 504, 485, 448, 387, 296, 169, 0, 729, 728, 721, 702, 665, 604, 513, 386, 217, 0, 1000, 999, 992, 973, 936, 875, 784, 657, 488, 271, 0

Offset: 0

Wolfdieter Lang, May 12 2015

See the comments in A025581 and A079904 on a problem by François Viète (Vieta)(1593). Regarding that problem, note the simple identity: n^3 - k^3 = (n - k)^3 + 3*n*k*(n - k), for n > = k >= 0.
Row sums give A126274(n-1) for n >= 1, and 0 for n=0.
Alternating row sums are ars(2*n) = ars(2*n-1) = (4*n-3)*n^2 = A103532(n-1), for n >= 1, and ars(0) = 0.

			The triangle T(n, k) begins:
  n\k    0   1   2   3   4   5   6   7   8   9  10
   0:    0
   1:    1   0
   2:    8   7   0
   3:   27  26  19   0
   4:   64  63  56  37   0
   5:  125 124 117  98  61   0
   6:  216 215 208 189 152  91   0
   7:  343 342 335 316 279 218 127   0
   8:  512 511 504 485 448 387 296 169   0
   9:  729 728 721 702 665 604 513 386 217   0
  10: 1000 999 992 973 936 875 784 657 488 271   0
  ...

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)

Cf. A126274, A103532, A025581, A079904, A335821.

for n from 0 to 10 do seq(n^3-k^3,k=0..n) od; # Robert Israel, May 10 2018

Table[n^3-k^3,{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Jan 02 2021 *)

T(n, k) =  A025581(n, k)*(A025581(n, k)^2 + 3* A079904(n, k)) (see the identity mentioned in a comment).
Columns (with one leading zero and offset 0): k=0: l^3 = A000578(l), k=1: (l+1)^3 - 1 = A068601(l+1), k=2: l*(l^2 + 6*l + 12), k=3: l*(l^2 + 9*l + 27), k=4: l*(l^2 + 12*l + 48), k=5: l*(l^2 + 15*l + 75), ...
G.f. for T(n,k): (1+4*x+4*x*y+x^2-14*x^2*y+x^2*y^2-2*x^3*y-2*x^3*y^2+7*x^4*y^2)*x/((1-x*y)^3*(1-x)^4). - Robert Israel, May 10 2018

A319213 a(n) = phi(n^3 - 1)/3 where phi is A000010.

Original entry on oeis.org

2, 4, 12, 20, 56, 36, 144, 96, 216, 144, 520, 240, 840, 480, 576, 816, 1568, 756, 2520, 1232, 1872, 1560, 4400, 1440, 4320, 3024, 4860, 3168, 7056, 2640, 9000, 5984, 7920, 6144, 10080, 4752, 17784, 7992, 13104, 9184, 22080, 7560, 23688, 12960, 14688, 15840, 33120

Offset: 2

Seiichi Manyama, Sep 13 2018

Seiichi Manyama, Table of n, a(n) for n = 2..1000
Eric Weisstein's World of Mathematics, Totient Function.
Wikipedia, Euler's totient function.

Row 3 of A369291.
Cf. A000010, A068601 (n^3-1).
phi(n^b - 1)/b: A319210 (b=2), this sequence (b=3), A319214 (b=5).

EulerPhi[Range[2, 50]^3 - 1]/3 (* Paolo Xausa, Jun 18 2024 *)

PARI
```
{a(n) = eulerphi(n^3-1)/3}
```

Sum_{k=1..n} a(k) = c * n^4 + O((n*log(n))^3), where c = (2/27) * Product_{p prime == 1 (mod 3)} (1 - 3/p^2) * Product_{p prime == 2 (mod 3)} (1 - 1/p^2) = 0.047313356295... . - Amiram Eldar, Dec 09 2024

A158622 Numerator of the reduced fraction A158620(n)/A158621(n).

Original entry on oeis.org

7, 13, 7, 31, 43, 19, 73, 91, 37, 133, 157, 61, 211, 241, 91, 307, 343, 127, 421, 463, 169, 553, 601, 217, 703, 757, 271, 871, 931, 331, 1057, 1123, 397, 1261, 1333, 469, 1483, 1561, 547, 1723, 1807, 631, 1981, 2071, 721, 2257, 2353, 817, 2551, 2653, 919, 2863

Offset: 2

Jonathan Vos Post, Mar 23 2009

A158620(n) = Product_{k=2..n} (k^3-1). 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). The reduced fractions are 7/9, 13/18, 7/10, 31/45, 43/63, 19/28, 73/108, 91/135, 37/55, 133/198, ...
Is this the same as A046163? - R. J. Mathar, Mar 27 2009
Apparently a(n) = A130770(n) for 2 <= n <= 53. - Georg Fischer, Oct 24 2018

			a(2) = 7 = numerator of (2^3-1)/2^3+1 = 7/9.
a(3) = 13 = numerator of ((2^3-1)*(3^3-1))/((2^3+1)*(3^3+1)) = (7 * 26)/ (9 * 28) = 182/252 = 13/18.
a(4) = 7 = = numerator of ((2^3-1)*(3^3-1)*(4^3-1))/((2^3+1)*(3^3+1)*(4^3+1)) = (7 * 26 * 63)/(9 * 28 * 65) = 11466/16380 = 7/10.
a(5) = 31 = numerator of ((2^3-1)(3^3-1)(4^3-1)(5^3-1))/((2^3+1)(3^3+1)(4^3+1)(5^3+1)) = 1421784/2063880 = 31/45.

Harvey P. Dale, Table of n, a(n) for n = 2..1000

Cf. A001093, A016921, A068601, A130770, A158620-A158621, A158623.

A158622 := proc(n) 2*(n^2+n+1)/3/n/(n+1) ; numer(%) ; end: seq(A158622(n),n=2..100) ; # R. J. Mathar, Mar 27 2009

Table[Product[k^3-1,{k,2,n}]/Product[k^3+1,{k,2,n}],{n,2,60}]//Numerator (* Harvey P. Dale, Feb 26 2020 *)

Numerator of (Product_{k=2..n} (k^3-1))/Product_{k=2..n} (k^3+1) = numerator of Product_{k=2..n} A068601(k)/A001093(k).
A158620(n)/A158621(n) = 2(n^2+n+1)/(3n(n+1)). - R. J. Mathar, Mar 27 2009
Empirical g.f.: -x^2*(x^8 + x^7 + x^6 - 2*x^5 + 4*x^4 + 10*x^3 + 7*x^2 + 13*x + 7) / ((x-1)^3*(x^2 + x + 1)^3). - Colin Barker, May 09 2013

More terms from R. J. Mathar, Mar 27 2009

cp's OEIS Frontend

A081257 a(n) is the greatest prime factor of (n^3 - 1).

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A258807 a(n) = n^5 - 1.

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A100119 a(n) = n-th centered n-gonal number.

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A158621 Partial products of A001093.

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A185065 a(n) = n*(n^3 + 2).

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A349508 a(n) is the numerator of binomial(n^3 + 6*n^2 - 6*n + 2, n^3 - 1)/n^3.

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A129294 Number of divisors of n^3 - 1 that are not greater than n.

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A349508 a(n) is the numerator of binomial(n^3 + 6n^2 - 6n + 2, n^3 - 1)/n^3.