A016755 -id:A016755 - cp's OEIS frontend

A057781 a(n) = n^4+4 = (n^2-2n+2)(n^2+2n+2) = ((n-1)^2+1)((n+1)^2+1).

4, 5, 20, 85, 260, 629, 1300, 2405, 4100, 6565, 10004, 14645, 20740, 28565, 38420, 50629, 65540, 83525, 104980, 130325, 160004, 194485, 234260, 279845, 331780, 390629, 456980, 531445, 614660, 707285, 810004, 923525, 1048580, 1185925

Offset: 0

Henry Bottomley, Nov 04 2000

Donald E. Knuth, The Art of Computer Programming, Addison-Wesley, Reading, MA, 1997, Vol. 1, exercise 1.2.1, Nr. 11, p. 19. [From Reinhard Zumkeller, Apr 11 2010]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A016755, A033999, A053755.

Cf. A000583, A002522, A002523, A272298.

[n^4+4: n in [0..40]]; // Vincenzo Librandi, Sep 07 2011

Table[n^4+4,{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)
LinearRecurrence[{5,-10,10,-5,1},{4,5,20,85,260},40] (* Harvey P. Dale, Aug 20 2020 *)

first(m)=vector(m,i,i--;i^4+4) \\ Anders Hellström, Sep 15 2015

G.f.: -(5*x^4-5*x^3+35*x^2-15*x+4) / (x-1)^5. - Colin Barker, Mar 29 2013
a(n) = A002523(n) + 3.
a(n) = A002522(n-1) * A002522(n+1).
Sum_{k=0..n} A033999(k)*A016755(k)/a(k) = A033999(n)*(n+1)/A053755(n+1), see Knuth reference. - Reinhard Zumkeller, Apr 11 2010
a(n) = (n^2)^2 + 2^2 = (n^2-2)^2 + (2*n)^2. - Thomas Ordowski, Sep 15 2015
a(n) = A272298(3*n)/3^4. - Bruno Berselli, Apr 29 2016
Sum_{n>=0} 1/a(n) = (Pi*coth(Pi) + 1)/8. - Amiram Eldar, Oct 04 2021

A287091 Expansion of Product_{k>=1} 1/(1 - x^((2*k-1)^3)).

Original entry on oeis.org

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, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5

Offset: 0

Ilya Gutkovskiy, May 19 2017

nonn

Number of partitions of n into odd cubes.

In general, if m > 0 and g.f. = Product_{k>=1} 1/(1 - x^((2*k-1)^m)), then a(n) ~ exp((m+1) * (Gamma(1/m) * Zeta(1+1/m) / (2*m^2))^(m/(m+1)) * n^(1/(m+1))) * (Gamma(1/m) * Zeta(1+1/m))^(m/(2*(m+1))) / (sqrt(Pi*(m+1)) * 2^((1+m*(m+3))/(2*(m+1))) * m^((m-1)/(2*(m+1))) * n^((2*m+1)/(2*(m+1)))). - Vaclav Kotesovec, Sep 19 2017

			a(27) = 2 because we have [27] and [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, 1].

Cf. A000009 (m=1), A167661 (m=2).
Cf. A003108, A016755, A259792, A279329, A280865.
Cf. A001156, A046042, A292547.

nmax = 110; CoefficientList[Series[Product[1/(1 - x^((2*k-1)^3)), {k, 1, Floor[nmax^(1/3)/2] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 18 2017 *)

G.f.: Product_{k>=1} 1/(1 - x^((2*k-1)^3)).

a(n) ~ exp(2^(5/4) * (Gamma(1/3) * Zeta(4/3))^(3/4) * n^(1/4) / 3^(3/2)) * (Gamma(1/3) * Zeta(4/3)/2)^(3/8) / (8 * 3^(1/4) * sqrt(Pi) * n^(7/8)). - Vaclav Kotesovec, Sep 18 2017

A290276 Numbers that are the sum of distinct odd positive cubes.

Original entry on oeis.org

1, 27, 28, 125, 126, 152, 153, 343, 344, 370, 371, 468, 469, 495, 496, 729, 730, 756, 757, 854, 855, 881, 882, 1072, 1073, 1099, 1100, 1197, 1198, 1224, 1225, 1331, 1332, 1358, 1359, 1456, 1457, 1483, 1484, 1674, 1675, 1701, 1702, 1799, 1800, 1826, 1827, 2060, 2061, 2087, 2088, 2185, 2186, 2197, 2198, 2212

Offset: 1

Ilya Gutkovskiy, Jul 25 2017

nonn

Complement of A292740.

			881 is in the sequence because 881 = 27 + 125 + 729 = 3^3 + 5^3 + 9^3.

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for sequences related to sums of cubes

Cf. A003997, A016755, A287091, A290275.

N:= 10000: # to get all terms <= N
M:= floor(N^(1/3)):
G:= mul(1+x^(j^3),j=1..M,2):
S:= series(G,x,N+1):
select(t -> coeff(S,x,t)>0, [$1..N]); # Robert Israel, Jul 26 2017

max = 2212; f[x_] := Product[1 + x^(2 k + 1)^3, {k, 0, 8}]; Exponent[#, x] & /@ List @@ Normal[Series[f[x], {x, 0, max}]] // Rest

A309335 a(n) = n^3 if n odd, 7*n^3/8 if n even.

Original entry on oeis.org

0, 1, 7, 27, 56, 125, 189, 343, 448, 729, 875, 1331, 1512, 2197, 2401, 3375, 3584, 4913, 5103, 6859, 7000, 9261, 9317, 12167, 12096, 15625, 15379, 19683, 19208, 24389, 23625, 29791, 28672, 35937, 34391, 42875, 40824, 50653, 48013, 59319, 56000, 68921, 64827, 79507, 74536, 91125

Offset: 0

Ilya Gutkovskiy, Jul 24 2019

Moebius transform of A007331.

Amiram Eldar, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Cf. A000578, A007331, A016755, A026741, A059376, A248882, A303383 (partial sums), A308422, A309336.

a[n_] := If[OddQ[n], n^3, 7 n^3/8]; Table[a[n], {n, 0, 45}]
nmax = 45; CoefficientList[Series[x (1 + 7 x + 23 x^2 + 28 x^3 + 23 x^4 + 7 x^5 + x^6)/(1 - x^2)^4, {x, 0, nmax}], x]
LinearRecurrence[{0, 4, 0, -6, 0, 4, 0, -1}, {0, 1, 7, 27, 56, 125, 189, 343}, 46]
Table[n^3 (15 - (-1)^n)/16, {n, 0, 45}]

G.f.: x * (1 + 7*x + 23*x^2 + 28*x^3 + 23*x^4 + 7*x^5 + x^6)/(1 - x^2)^4.
G.f.: Sum_{k>=1} J_3(k) * x^k/(1 - x^(2*k)), where J_3() is the Jordan function (A059376).
Dirichlet g.f.: zeta(s-3) * (1 - 1/2^s).
a(n) = n^3 * (15 - (-1)^n)/16.
a(n) = Sum_{d|n, n/d odd} J_3(d).
Sum_{n>=1} 1/a(n) = 57*zeta(3)/56 = 1.223522205001729897639...
Multiplicative with a(2^e) = 7*2^(3*e-3), and a(p^e) = p^(3*e) for odd primes p. - Amiram Eldar, Oct 26 2020
Euler transform is A248882. - Georg Fischer, Nov 10 2020

A383368 Number of intercalates in pine Latin squares of order 2n.

Original entry on oeis.org

1, 12, 27, 80, 125, 252, 343, 576, 729, 1100, 1331, 1872, 2197, 2940, 3375, 4352, 4913, 6156, 6859, 8400, 9261, 11132, 12167, 14400, 15625

Offset: 1

Eduard I. Vatutin, Apr 24 2025

Pine Latin square is a none canonical composite Latin square of order N=2*K formed from specially arranged cyclic Latin squares of order K.
By construction, pine Latin square is determined one-to-one by the cyclic square used, so number of pine Latin squares of order N is equal to number of cyclic Latin squares of order N/2.
All pine Latin squares are horizontally symmetric column-inverse Latin squares.
All pine Latin squares for selected order N are isomorphic one to another as Latin squares, so they have same properties (number of transversals, intercalates, etc.).
Pine Latin squares have interesting properties, for example, maximum known number of intercalates for some orders N (at least N in {2, 4, 6, 10, 18}).
Pine Latin squares do not exist for odd orders due to they are horizontally symmetric.
Pine Latin squares of order N=2n exists for all even orders due to existing of corresponding cyclic Latin squares of order n. According to this, maximum number of intercalates in a Latin square A092237(N) >= (2k)^2 * (2k + 1) for N=4k and A092237(N) >= (2k+1)^3 for N=4k+2. Therefore, asimptotically maximum number of intercalates in Latin squares of even orders N greater or equal than o(k1*N^3), where k1 = 1/8.

			For order N=8 pine Latin square
  0 1 2 3 4 5 6 7
  1 2 3 0 7 4 5 6
  2 3 0 1 6 7 4 5
  3 0 1 2 5 6 7 4
  4 5 6 7 0 1 2 3
  5 6 7 4 3 0 1 2
  6 7 4 5 2 3 0 1
  7 4 5 6 1 2 3 0
have 80 intercalates.
.
For order N=10 pine Latin square
  0 1 2 3 4 5 6 7 8 9
  1 2 3 4 0 9 5 6 7 8
  2 3 4 0 1 8 9 5 6 7
  3 4 0 1 2 7 8 9 5 6
  4 0 1 2 3 6 7 8 9 5
  5 6 7 8 9 0 1 2 3 4
  6 7 8 9 5 4 0 1 2 3
  7 8 9 5 6 3 4 0 1 2
  8 9 5 6 7 2 3 4 0 1
  9 5 6 7 8 1 2 3 4 0
have 125 intercalates.
.
For order N=12 pine Latin square
  0 1 2 3 4 5 6 7 8 9 10 11
  1 2 3 4 5 0 11 6 7 8 9 10
  2 3 4 5 0 1 10 11 6 7 8 9
  3 4 5 0 1 2 9 10 11 6 7 8
  4 5 0 1 2 3 8 9 10 11 6 7
  5 0 1 2 3 4 7 8 9 10 11 6
  6 7 8 9 10 11 0 1 2 3 4 5
  7 8 9 10 11 6 5 0 1 2 3 4
  8 9 10 11 6 7 4 5 0 1 2 3
  9 10 11 6 7 8 3 4 5 0 1 2
  10 11 6 7 8 9 2 3 4 5 0 1
  11 6 7 8 9 10 1 2 3 4 5 0
have 252 intercalates.

R. Bean, Critical sets in Latin squares and associated structures, Ph.D. Thesis, The University of Queensland, 2001.
Eduard I. Vatutin, About the properties of pine Latin squares (in Russian).
Eduard I. Vatutin, Proving list (examples).
Index entries for sequences related to Latin squares and rectangles.

Cf. A002860, A016755, A089207, A092237, A099721, A338522, A383570.

Hypothesis: For all known pine Latin squares of orders N=4k+2 number of intercalates a(n) = a(N/2)= a(2k+1) = (N/2)^3 = (2k+1)^3 = A016755((n-1)/2) (verified for N<29).

Hypothesis: For all known pine Latin squares of orders N=4k number of intercalates a(n) = a(N/2) = a(2k) = (N/2)^2 + (N/2)^3 = 4*k^2 + 8*k^3 = (2k)^2 * (2k+1) = 2*A089207(n/2) = 4*A099721(n/2) (verified for N<29).

A254473 24-hedral numbers: a(n) = (2n + 1)(8n^2 + 14n + 7).

Original entry on oeis.org

7, 87, 335, 847, 1719, 3047, 4927, 7455, 10727, 14839, 19887, 25967, 33175, 41607, 51359, 62527, 75207, 89495, 105487, 123279, 142967, 164647, 188415, 214367, 242599, 273207, 306287, 341935, 380247, 421319, 465247, 512127, 562055, 615127, 671439, 731087

Offset: 0

Luciano Ancora, Mar 26 2015

This sequence is very close to the A046142 sequence: a(n) is asymptotic to A046142(n) as n tends to infinity.

The formula for A046142, the Haüy rhombic dodecahedral number, is remarkably similar, (2*n-1)*(8*n^2-14*n+7), where the first factor of the dodecahedral formula has "+1" versus "-1" in the 24-hedral formula, and the second factor of the former has "-14n" versus the latter of "+14n". Note that the rhombic dodecahedron has 24 faces, further explaining the relationship. The difference of these sequences is diff(n)=72*n^2 + 14. - Peter M. Chema, Jan 09 2016

Robert Israel, Table of n, a(n) for n = 0..10000
Luciano Ancora, The 24-hedral Number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000447, A016755, A046142.

[(2*n+1)*(8*n^2+14*n+7): n in [0..40]]; // Bruno Berselli, Mar 27 2015

seq((2*n + 1)*(8*n^2 + 14*n + 7), n=0..100); # Robert Israel, Jan 11 2016

Table[(2 n + 1) (8 n^2 + 14 n + 7), {n, 0, 40}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {7, 87, 335, 847}, 40]

vector(40, n, n--; (2*n+1)*(8*n^2+14*n+7)) \\ Bruno Berselli, Mar 27 2015

[(2*n+1)*(8*n^2+14*n+7) for n in (0..40)] # Bruno Berselli, Mar 27 2015

G.f.: (7 + 59*x + 29*x^2 + x^3)/(1 - x)^4.
a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) - a(n-4).
a(n) = -A046142(-n) with A046142(0) = -7.
a(n) = 6*Sum_{k=0..n} (2*k+1)^2 + (2*n+1)^3. - Robert FERREOL, Nov 13 2023

A305290 Numbers k such that 4*k + 1 is a perfect cube, sorted by absolute values.

Original entry on oeis.org

0, -7, 31, -86, 182, -333, 549, -844, 1228, -1715, 2315, -3042, 3906, -4921, 6097, -7448, 8984, -10719, 12663, -14830, 17230, -19877, 22781, -25956, 29412, -33163, 37219, -41594, 46298, -51345, 56745, -62512, 68656, -75191, 82127, -89478, 97254, -105469, 114133, -123260, 132860

Offset: 1

Bruno Berselli, May 29 2018

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-3,-2,2,3,1).

Cf. A016755.
Cf. A000290: k such that 4*k is a square.
Cf. A002378: k such that 4*k+1 is a square.
Cf. A033431: k such that 4*k is a nonnegative cube.
Cf. A305291: k such that 4*k+3 is a cube.
Cf. A141046: k such that 4*k is a fourth power.
Cf. 4*A219086: k such that 4*k+1 is a fourth power.

seq(coeff(series(x^2*(-7+10*x-7*x^2)/((1-x)*(1+x)^4), x,50),x,n),n=1..45); # Muniru A Asiru, May 31 2018

LinearRecurrence[{-3, -2, 2, 3, 1}, {0, -7, 31, -86, 182}, 45] (* Jean-François Alcover, Jun 04 2018 *)

concat(0, Vec(-x^2*(7 - 10*x + 7*x^2) / ((1 - x)*(1 + x)^4) + O(x^40))) \\ Colin Barker, Jun 04 2018

G.f.: x^2*(-7 + 10*x - 7*x^2)/((1 - x)*(1 + x)^4).
a(n) = -3*a(n-1) - 2*a(n-2) + 2*a(n-3) + 3*a(n-4) + a(n-5).
a(n) = (-1 - A016755(n-1)*(-1)^n)/4.
a(n) + a(-n) = (-1)^n*2^((1-(-1)^n)/2).
(n - 2)*(4*n^2 - 16*n + 19)*a(n) + (12*n^2 - 36*n + 31)*a(n-1) - (n - 1)*(4*n^2 - 8*n + 7)*a(n-2) = 0.
From Colin Barker, May 30 2018: (Start)
a(n) = n*(4*n^2 + 6*n + 3)/2 for n even.
a(n) = -(n + 1)*(4*n^2 + 2*n + 1)/2 for n odd.
(End)

A305291 Numbers k such that 4*k + 3 is a perfect cube, sorted by absolute values.

Original entry on oeis.org

-1, 6, -32, 85, -183, 332, -550, 843, -1229, 1714, -2316, 3041, -3907, 4920, -6098, 7447, -8985, 10718, -12664, 14829, -17231, 19876, -22782, 25955, -29413, 33162, -37220, 41593, -46299, 51344, -56746, 62511, -68657, 75190, -82128, 89477, -97255, 105468, -114134, 123259, -132861

Offset: 1

Bruno Berselli, May 29 2018

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (-3,-2,2,3,1).

Cf. A016755, A305290.

seq(coeff(series(x*(-1+3*x-16*x^2+3*x^3-x^4)/((1-x)*(1+x)^4), x,50),x,n),n=1..45); # Muniru A Asiru, May 31 2018

LinearRecurrence[{-3, -2, 2, 3, 1}, {-1, 6, -32, 85, -183}, 45] (* Jean-François Alcover, Jun 04 2018 *)

Vec(-x*(1 - 3*x + 16*x^2 - 3*x^3 + x^4) / ((1 - x)*(1 + x)^4) + O(x^40)) \\ Colin Barker, Jun 04 2018

G.f.: x*(-1 + 3*x - 16*x^2 + 3*x^3 - x^4)/((1 - x)*(1 + x)^4).
a(n) = -3*a(n-1) - 2*a(n-2) + 2*a(n-3) + 3*a(n-4) + a(n-5).
a(n) = (-3 + A016755(n-1)*(-1)^n)/4.
a(n) = -A305290(n) - 1.
a(n) + a(-n) = 1 - 2^(1+(-1)^n).
(n - 2)*(4*n^2 - 16*n + 19)*a(n) + (12*n^2 - 36*n + 31)*a(n-1) - (n - 1)*(4*n^2 - 8*n + 7)*a(n-2) = 0.
From Colin Barker, May 30 2018: (Start)
a(n) = (4*n^3 - 6*n^2 + 3*n - 2)/2 for n even.
a(n) = -(4*n^3 - 6*n^2 + 3*n + 1)/2 for n odd.
(End)

A328408 G.f. A(x) satisfies: A(x) = A(x^2) + x * (1 + 4*x + x^2) / (1 - x)^4.

Original entry on oeis.org

1, 9, 27, 73, 125, 243, 343, 585, 729, 1125, 1331, 1971, 2197, 3087, 3375, 4681, 4913, 6561, 6859, 9125, 9261, 11979, 12167, 15795, 15625, 19773, 19683, 25039, 24389, 30375, 29791, 37449, 35937, 44217, 42875, 53217, 50653, 61731, 59319, 73125, 68921, 83349, 79507, 97163, 91125

Offset: 1

Ilya Gutkovskiy, Oct 14 2019

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000578, A001511, A016755, A023872, A059376, A129527, A209229, A328407.

[n eq 1 select 1 else IsOdd(n) select n^3 else Self(n div 2)+n^3 :n in [1..45]]; // Marius A. Burtea, Oct 15 2019

nmax = 45; CoefficientList[Series[Sum[x^(2^k) (1 + 4 x^(2^k) + x^(2^(k + 1)))/(1 - x^(2^k))^4, {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] // Rest
a[n_] := If[EvenQ[n], a[n/2] + n^3, n^3]; Table[a[n], {n, 1, 45}]
Table[DivisorSum[n, Boole[IntegerQ[Log[2, n/#]]] #^3 &], {n, 1, 45}]
f[p_, e_] :=p^(3*e); f[2, e_] := (8^(e+1)-1)/7; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Oct 23 2023 *)

G.f.: Sum_{k>=0} x^(2^k) * (1 + 4*x^(2^k) + x^(2^(k+1))) / (1 - x^(2^k))^4.
G.f.: (1/7) * Sum_{k>=1} J_3(2*k) * x^k / (1 - x^k), where J_3() is the Jordan function (A059376).
Dirichlet g.f.: zeta(s-3) / (1 - 2^(-s)).
a(2*n) = a(n) + 8*n^3, a(2*n+1) = (2*n + 1)^3.
a(n) = Sum_{d|n} A209229(n/d) * d^3.
Product_{n>=1} (1 + x^n)^a(n) = g.f. for A023872.
Sum_{k=1..n} a(k) ~ 4*n^4/15. - Vaclav Kotesovec, Oct 15 2019
Multiplicative with a(2^e) = (8^(e+1)-1)/7, and a(p^e) = p^(3*e) for an odd prime p. - Amiram Eldar, Oct 23 2023

A371532 Centered cuboctahedral numbers: the number of integer triples (x,y,z) such that max(|x|,|y|,|z|) <= n and |x|+|y|+|z| <= 2n.

Original entry on oeis.org

1, 19, 93, 263, 569, 1051, 1749, 2703, 3953, 5539, 7501, 9879, 12713, 16043, 19909, 24351, 29409, 35123, 41533, 48679, 56601, 65339, 74933, 85423, 96849, 109251, 122669, 137143, 152713, 169419, 187301, 206399, 226753, 248403, 271389, 295751, 321529, 348763

Offset: 0

Peter Kagey, Mar 26 2024

			The a(1) = 19 lattice points are all permutations of the points (0,0,0), (0,0,1), and (0,1,1), where any number of the coordinates can also be made negative (e.g., (1,-1,0)).

Paolo Xausa, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Cuboctahedron.
Eric Weisstein's World of Mathematics, Figurate Number.
Wikipedia, Ehrhart polynomial.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A016755, A130809, A371515.

Array[(20*#^3 + 24*#^2 + 10*# + 3)/3 &, 50, 0] (* or *)
LinearRecurrence[{4, -6, 4, -1}, {1, 19, 93, 263}, 50] (* Paolo Xausa, Apr 02 2024 *)

def A371532(n): return n*(n*(5*n+6<<2)+10)//3+1 # Chai Wah Wu, Apr 02 2024

a(n) = (20*n^3 + 24*n^2 + 10*n + 3)/3.
a(n) =  A016755(n) - A130809(n-2).
G.f.: (x^3 + 23*x^2 + 15*x + 1)/(x-1)^4. - Paolo Xausa, Apr 02 2024
From Elmo R. Oliveira, Aug 22 2025: (Start)
E.g.f.: exp(x)*(3 + 54*x + 84*x^2 + 20*x^3)/3.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)

cp's OEIS Frontend

A057781 a(n) = n^4+4 = (n^2-2*n+2)*(n^2+2*n+2) = ((n-1)^2+1)*((n+1)^2+1).

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A287091 Expansion of Product_{k>=1} 1/(1 - x^((2*k-1)^3)).

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A290276 Numbers that are the sum of distinct odd positive cubes.

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A309335 a(n) = n^3 if n odd, 7*n^3/8 if n even.

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A383368 Number of intercalates in pine Latin squares of order 2n.

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A254473 24-hedral numbers: a(n) = (2*n + 1)*(8*n^2 + 14*n + 7).

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A305290 Numbers k such that 4*k + 1 is a perfect cube, sorted by absolute values.

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A305291 Numbers k such that 4*k + 3 is a perfect cube, sorted by absolute values.

A057781 a(n) = n^4+4 = (n^2-2n+2)(n^2+2n+2) = ((n-1)^2+1)((n+1)^2+1).

A254473 24-hedral numbers: a(n) = (2n + 1)(8n^2 + 14n + 7).