A068601 -id:A068601 - cp's OEIS frontend

A158623 Denominator of the reduced fraction A158620(n)/A158621(n).

9, 18, 10, 45, 63, 28, 108, 135, 55, 198, 234, 91, 315, 360, 136, 459, 513, 190, 630, 693, 253, 828, 900, 325, 1053, 1134, 406, 1305, 1395, 496, 1584, 1683, 595, 1890, 1998, 703, 2223, 2340, 820, 2583, 2709, 946, 2970, 3105, 1081, 3384, 3528, 1225, 3825

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, ...

			a(2) = 9 = denominator of (2^3-1)/2^3+1 = 7/9. a(3) = 18 = denominator of ((2^3-1)*(3^3-1))/((2^3+1)*(3^3+1)) = (7 * 26)/ (9 * 28) = 182/252 = 13/18. a(4) = 10 = denominator 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) = 45 = denominator 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.

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

A158623 := proc(n) 2*(n^2+n+1)/3/n/(n+1) ; denom(%) ; end: seq(A158623(n),n=2..100) ; # R. J. Mathar, Mar 27 2009

Denominator of (Product_{k=2..n} (k^3-1)) / Product_{k=2..n} (k^3+1) = denominator of Product_{k=2..n} A068601(k)/A001093(k).
A158620(n)/A158621(n) = 2(n^2+n+1)/(3n(n+1)). Conjecture: a(n) = 3a(n-3) - 3a(n-6) + a(n-9), so trisections are A152996, A060544 and 3*A081266. - R. J. Mathar, Mar 27 2009
Empirical g.f.: -x^2*(x^8 - 2*x^5 + 9*x^4 + 18*x^3 + 10*x^2 + 18*x + 9) / ((x-1)^3*(x^2 + x + 1)^3). - Colin Barker, May 09 2013

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

A340242 Square array read by upward antidiagonals: T(n,k) is the number of n-ary strings of length k containing 000.

Original entry on oeis.org

1, 1, 3, 1, 5, 8, 1, 7, 21, 20, 1, 9, 40, 81, 47, 1, 11, 65, 208, 295, 107, 1, 13, 96, 425, 1021, 1037, 238, 1, 15, 133, 756, 2621, 4831, 3555, 520, 1, 17, 176, 1225, 5611, 15569, 22276, 11961, 1121, 1, 19, 225, 1856, 10627, 40091, 90085, 100768, 39667, 2391

Offset: 2

Robert P. P. McKone, Jan 01 2021

			For n = 4 and k = 5, there are 40 strings: {00000, 00001, 00002, 00003, 00010, 00011, 00012, 00013, 00020, 00021, 00022, 00023, 00030, 00031, 00032, 00033, 01000, 02000, 03000, 10000, 10001, 10002, 10003, 11000, 12000, 13000, 20000, 20001, 20002, 20003, 21000, 22000, 23000, 30000, 30001, 30002, 30003, 31000, 32000, 33000}.
Square table T(n,k):
      k=3: k=4:  k=5:   k=6:    k=7:     k=8:
n=2:    1    3     8     20      47      107
n=3:    1    5    21     81     295     1037
n=4:    1    7    40    208    1021     4831
n=5:    1    9    65    425    2621    15569
n=6:    1   11    96    756    5611    40091
n=7:    1   13   133   1225   10627    88717
n=8:    1   15   176   1856   18425   175967
n=9:    1   17   225   2673   29881   321281

Robert P. P. McKone, Antidiagonals n = 2..100, flattened

Rows: A050231 (n=2), A231430 (n=3).
Columns: A000567 (k=5), A103532 (k=6).
Cf. A340156 (containing 00).
Cf. A341050.
Cf. A000073, A068601.

m[r_] := Normal[With[{p = 1/n}, SparseArray[{Band[{1, 2}] -> p, {i_, 1} /; i <= r -> 1 - p, {r + 1, r + 1} -> 1}]]];
T[n_, k_, r_] := MatrixPower[m[r], k][[1, r + 1]]*n^k;
Reverse[Table[T[n, k - n + 3, 3], {k, 2, 11}, {n, 2, k}], 2] // Flatten

my(x2='x^2+'x+1); T(n,k) = n^k - polcoeff(lift(x2*Mod('x, 'x^3-(n-1)*x2)^k), 2); \\ Kevin Ryde, Jan 02 2021

m(3) = [1 - 1/n, 1/n, 0, 0; 1 - 1/n, 0, 1/n, 0; 1 - 1/n, 0, 0, 1/n; 0, 0, 0, 1], is the probability/transition matrix for three consecutive "0" -> "containing 000".

A088036 Numbers k such that k^3 - 1 is divisible by a cube other than 1.

Original entry on oeis.org

9, 10, 17, 18, 19, 25, 28, 33, 37, 41, 46, 49, 55, 57, 64, 65, 73, 81, 82, 89, 91, 97, 100, 105, 109, 113, 118, 121, 126, 127, 129, 136, 137, 145, 153, 154, 161, 163, 169, 172, 177, 181, 185, 190, 193, 199, 201, 208, 209, 217, 225, 226, 233, 235, 241, 244, 249

Offset: 1

Amarnath Murthy, Sep 19 2003

nonn

R. J. Mathar, Table of n, a(n) for n = 1..1185

Cf. A068601, A088035.

isA046099 := proc(n)
    local p;
    for p in ifactors(n)[2] do
        if op(2,p) >= 3 then
            return true;
        end if;
    end do:
    false ;
end proc:
n := 1;
for i from 1 to 5000 do
    if isA046099(i^3-1) then
        printf("%d %d\n",n,i) ;
        n := n+1;
    end if;
end do: # R. J. Mathar, Dec 03 2015

n3dcQ[n_]:=Count[Divisors[n^3-1]],?(IntegerQ[Surd[#,3]]&)>1; Select[ Range[ 2,250],n3dcQ] (* _Harvey P. Dale, Oct 05 2017 *)

from sympy import factorint
def ok(n): return max(factorint(n**3 - 1).values()) >= 3
print(list(filter(ok, range(1, 250)))) # Michael S. Branicky, Sep 10 2021

Corrected and extended by Ray Chandler, Sep 23 2003

Definition clarified by Harvey P. Dale, Oct 05 2017

A128972 n^3 - 1 divided by its largest cube divisor.

Original entry on oeis.org

7, 26, 63, 124, 215, 342, 511, 91, 37, 1330, 1727, 2196, 2743, 3374, 4095, 614, 17, 254, 7999, 9260, 10647, 12166, 13823, 1953, 17575, 19682, 813, 24388, 26999, 29790, 32767, 4492, 39303, 42874, 46655, 1876, 54871, 59318, 63999, 8615, 74087, 79506

Offset: 2

Jonathan Vos Post, Apr 28 2007

In other words, cubefree part of n^3-1, or cubefree kernel of n^3-1. Cube analog of A068310.

			a(9) = (9^3-1)/8 = (2^3 * 7 * 13)/(2^3) = 728/8 = 91.
a(10) = (10^3-1)/27 = (3^3 * 37)/(3^3) = 999/27 = 37.
a(18) = (18^3-1)/343 = (7^3 * 17)/(7^3) = 5831/343 = 17.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A002350, A004709, A007948, A062378, A067872, A033314, A068310.

a:= n -> mul(f[1]^(f[2] mod 3), f = ifactors(n^3-1)[2]):
seq(a(n),n=2..100); # Robert Israel, Sep 24 2014

a(n) = A062378(A068601(n)) = A062378(n^3-1).

More terms from Carl R. White, Nov 09 2010

A135300 Positive X-values of solutions to the equation 1!X^4 - 2!(X + 1)^3 + 3!(X + 2)^2 - (4^2)(X + 3) + 5^2 = Y^3.

Original entry on oeis.org

1, 7, 26, 63, 124, 215, 342, 511, 728, 999, 1330, 1727, 2196, 2743, 3374, 4095, 4912, 5831, 6858, 7999, 9260, 10647, 12166, 13823, 15624, 17575, 19682, 21951, 24388, 26999, 29790, 32767, 35936, 39303, 42874

Offset: 1

Mohamed Bouhamida, Dec 04 2007

To prove that X = 1 or X = n^3 - 1: Y^3 = 1!*X^4 - 2!*(X + 1)^3 + 3!*(X + 2)^2 - (4^2)*(X + 3) + 5^2 = X^4 - 2*(X + 1)^3 + 6*(X + 2)^2 - 16(X + 3) + 25 = X^4 - 2*X^3 + 2X - 1 = (X + 1)(X^3 - 3*X^2 + 3X - 1) = (X + 1)*(X - 1)^3, which means that X = 1 or (X + 1) must be a cube, so (X, Y) = (1, 0) or (X, Y) = (n^3 - 1, n(n^3 - 2)) with n >= 2.

Apart from the first term, the same as A068601. - R. J. Mathar, Apr 29 2008

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001093, A068601.

Join[{1},LinearRecurrence[{4,-6,4,-1},{7,26,63,124},40]] (* Harvey P. Dale, Jul 12 2015 *)

a(n)=if(n>1, n^3-1, 1) \\ Charles R Greathouse IV, Oct 09 2016

a(1) = 1 and a(n) = n^3 - 1 with n >= 2.
G.f.: x*(1 + 3*x + 4*x^2 - 3*x^3 + x^4)/(1-x)^4. - Colin Barker, Oct 25 2012
E.g.f.: (1 + x)*(1 - exp(x)*(1 - 2*x - x^2)). - Stefano Spezia, Apr 22 2023

A365373 a(n) = n(3n^4 + 15n^3 + 25n^2 - 15*n - 28)/60.

Original entry on oeis.org

0, 0, 7, 40, 136, 356, 791, 1568, 2856, 4872, 7887, 12232, 18304, 26572, 37583, 51968, 70448, 93840, 123063, 159144, 203224, 256564, 320551, 396704, 486680, 592280, 715455, 858312, 1023120, 1212316, 1428511, 1674496, 1953248, 2267936, 2621927, 3018792, 3462312

Offset: 0

Stefano Spezia, Sep 02 2023

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Antidiagonal sums of A365372.

Cf. A068601 (2nd differences).

a[n_]:= n*(3*n^4 + 15*n^3 + 25*n^2 - 15*n - 28)/60; Array[a,37,0]

def A365373(n): return n*(n*(n*(n*(3*n + 15) + 25) - 15) - 28)//60 # Chai Wah Wu, Sep 04 2023

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) for n > 5.
O.g.f.: x^2*(7 - 2*x + x^2)/(1 - x)^6.
E.g.f.: exp(x)*x^2*(210 + 190*x + 45*x^2 + 3*x^3)/60.

A385591 Numbers k such that both k^3 - 1 and k^3 + 1 are triprimes.

Original entry on oeis.org

66, 132, 180, 228, 240, 288, 294, 336, 378, 420, 462, 600, 612, 660, 678, 702, 882, 918, 960, 1116, 1164, 1278, 1302, 1320, 1800, 2550, 2562, 3270, 3300, 3372, 3408, 3438, 3822, 3882, 3990, 4050, 4422, 4536, 4812, 5040, 5088, 5208, 5250, 5418, 5748, 5754, 5778, 5838, 6882, 6960, 7128, 7182, 7254

Offset: 1

Robert Israel, Aug 09 2025

nonn

Numbers k such that k^3 - 1 and k^3 + 1 each have 3 prime factors, counted with multiplicity.
All terms are divisible by 6.
The Generalized Bunyakovsky Conjecture implies there are infinitely many j such that 6+7*j, 32 + 35*j, 1225 * j^2 + 2205 * j + 993 and 175 * j^2 + 305 * j + 133 are all prime.  For such j, 31 + 35*j is a term of the sequence.  Thus the conjecture implies the sequence is infinite. The first two such j are 1 and 31, corresponding to a(1) = 66 and a(20) = 1116.

			a(3) = 180 is a term because 180^3 - 1 = 5831999 = 31 * 179 * 1051 and 5832001 = 7 * 181 * 4603 are each products of 3 primes.

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

Cf. A001093, A068601, A014612. Intersection of A115403 and A386915.

filter:= k -> numtheory:-bigomega(k-1) + numtheory:-bigomega(k^2 + k + 1) = 3 and
numtheory:-bigomega(k+1) + numtheory:-bigomega(k^2 - k + 1) = 3:
select(filter, [seq(i,i=6 .. 10000, 6);

Select[Range[7500], PrimeOmega[#^3 - 1] == PrimeOmega[#^3 + 1] == 3 &] (* Amiram Eldar, Aug 10 2025 *)

A386915 Numbers k such that k^3 - 1 is a triprime.

Original entry on oeis.org

4, 5, 15, 27, 32, 42, 44, 48, 50, 59, 60, 66, 72, 75, 78, 84, 98, 104, 108, 114, 119, 132, 140, 143, 147, 152, 162, 167, 174, 180, 182, 188, 200, 203, 206, 212, 215, 218, 224, 228, 234, 236, 240, 252, 258, 264, 266, 270, 279, 288, 290, 294, 308, 318, 336, 338, 342, 350, 374, 378, 383, 384, 390

Offset: 1

Robert Israel, Aug 07 2025

nonn

Numbers k such that either k-1 is prime and k^2 + k + 1 is a semiprime, or k-1 is a semiprime and k^2 + k + 1 is prime.
If k is odd, k-1 = 2*p for a prime p such that 4*p^2 + 6*p + 3 is prime.  The Generalized Bunyakovsky conjecture implies that there are infinitely many of these.
The Generalized Bunyakovsky conjecture also implies that there are infinitely many j such that 14*j + 3, 28*j^2 + 18*j + 3, 7*j + 2 and 196*j^2 + 154*j + 31 are all prime.  This implies that both k = 14*j + 4 and k + 1 are terms of the sequence.
There are no k where k, k + 1 and k + 2 are all terms of the sequence, since there are no terms == 1 (mod 3) except 4 (if k == 1 (mod 3), then k^3 == 1 (mod 9)).

			a(3) = 15 is a term because 15^3 - 1 = 3374 = 2 * 7 * 241 is the product of three primes.

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

Cf. A001358, A014612, A068601, A115403.

select(t -> numtheory:-bigomega(t^3-1)=3, [$1..1000]);

Select[Range[400], PrimeOmega[#^3 - 1] == 3 &] (* Amiram Eldar, Aug 08 2025 *)

isok(k) = bigomega(k^3-1) == 3; \\ Michel Marcus, Aug 08 2025

A204767 Quadruples (a,b,c,d) of the form ( n(n^3-1), n^3-1, 2n^3+1, n*(n^3+2) ).

Original entry on oeis.org

0, 0, 3, 3, 14, 7, 17, 20, 78, 26, 55, 87, 252, 63, 129, 264, 620, 124, 251, 635, 1290, 215, 433, 1308, 2394, 342, 687, 2415, 4088, 511, 1025, 4112, 6552, 728, 1459, 6579, 9990, 999, 2001, 10020, 14630, 1330, 2663, 14663, 20724, 1727, 3457, 20760, 28548, 2196, 4395, 28587

Offset: 1

Vincenzo Librandi, Mar 04 2012

Four consecutive (a,b,c,d) in the sequence are solutions to a^3+b^3+c^3 = d^3, that is a(4k+1)^3+a(4k+2)^3+a(4k+3)^3 = a(4k+4)^3.
Also, A058895(n)^3 + A068601(n)^3 + A033562(n)^3 = A185065(n)^3.
The sequence corresponds to the case m=1 in the identity (n*(n^3-m^3))^3+(m*(n^3-m^3))^3+(m*(2*n^3+m^3))^3 = (n*(n^3+2*m^3))^3.
G. H. Hardy and E. M. Wright gave this identity in their "An Introduction to the Theory of Numbers" together with (n*(n^3-2*m^3))^3+(m*(n^3+m^3))^3+(m*(2*n^3-m^3))^3 = (n*(n^3+m^3))^3 (see References). - Bruno Berselli, Mar 13 2012

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, 2008 (Sixth edition), Par. 13.7.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A058895, A068601, A033562, A185065.

&cat[[n*(n^3-1), n^3-1, 2*n^3+1, n*(n^3+2)]: n in [1..40]];

Flatten[Table[{n^4 - n, n^3 - 1, 2 n^3 + 1, n^4 + 2 n}, {n, 1, 40}]] (* Vincenzo Librandi, Jan 02 2014 *)

A234860 Sum of the divisors of n^3 - 1.

Original entry on oeis.org

8, 42, 104, 224, 264, 780, 592, 1680, 1520, 2880, 1896, 5642, 2968, 5808, 8736, 9548, 7200, 15360, 8440, 19488, 19032, 23040, 14448, 49920, 23560, 31836, 32912, 53312, 34200, 77688, 38912, 70812, 62088, 74088, 67584, 152320, 56392, 107520, 99736

Offset: 2

Vincenzo Librandi, Jan 01 2014

Vincenzo Librandi, Table of n, a(n) for n = 2..1001

Cf. A000203, A062835, A065764, A068601, A175926, A193433, A333240.

Magma
```
[SumOfDivisors(n^3-1): n in [2..50]];
```

Table[Total[Divisors[n^3 - 1]], {n, 2, 50}]

a(n) = sigma(n^3-1); \\ Amiram Eldar, Dec 09 2024

From Amiram Eldar, Dec 09 2024: (Start)
a(n) = A000203(A068601(n)).
Sum_{k=2..n} a(k) = c * n^4 + O((n*log(n))^3), where c = (83/288) * Product_{primes p == 1 (mod 3)} ((p^2+2)/(p^2-1)) * Product_{primes p == 2 (mod 3)} (p^2/(p^2-1)) = 0.4499262799... . (End)

Offset corrected by Amiram Eldar, Dec 09 2024

cp's OEIS Frontend

A158623 Denominator of the reduced fraction A158620(n)/A158621(n).

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A340242 Square array read by upward antidiagonals: T(n,k) is the number of n-ary strings of length k containing 000.

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A088036 Numbers k such that k^3 - 1 is divisible by a cube other than 1.

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A128972 n^3 - 1 divided by its largest cube divisor.

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A135300 Positive X-values of solutions to the equation 1!*X^4 - 2!*(X + 1)^3 + 3!*(X + 2)^2 - (4^2)*(X + 3) + 5^2 = Y^3.

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A365373 a(n) = n*(3*n^4 + 15*n^3 + 25*n^2 - 15*n - 28)/60.

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A385591 Numbers k such that both k^3 - 1 and k^3 + 1 are triprimes.

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A386915 Numbers k such that k^3 - 1 is a triprime.

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A135300 Positive X-values of solutions to the equation 1!X^4 - 2!(X + 1)^3 + 3!(X + 2)^2 - (4^2)(X + 3) + 5^2 = Y^3.

A365373 a(n) = n(3n^4 + 15n^3 + 25n^2 - 15*n - 28)/60.

A204767 Quadruples (a,b,c,d) of the form ( n(n^3-1), n^3-1, 2n^3+1, n*(n^3+2) ).