A033430 -id:A033430 - cp's OEIS frontend

A050801 Numbers k such that k^2 is expressible as the sum of two positive cubes in at least one way.

3, 4, 24, 32, 81, 98, 108, 168, 192, 228, 256, 312, 375, 500, 525, 588, 648, 671, 784, 847, 864, 1014, 1029, 1183, 1225, 1261, 1323, 1344, 1372, 1536, 1824, 2048, 2187, 2496, 2646, 2888, 2916, 3000, 3993, 4000, 4200, 4225, 4536, 4563, 4644, 4704, 5184, 5324

Offset: 1

Patrick De Geest, Sep 15 1999

Analogous solutions exist for the sum of two identical cubes z^2 = 2*r^3 (e.g., 864^2 = 2*72^3). Values of 'z' are the terms in A033430, values of 'r' are the terms in A001105.
First term whose square can be expressed in two ways is 77976; 77976^2 = 228^3 + 1824^3 = 1026^3 + 1710^3. - Jud McCranie
First term whose square can be expressed in three ways is 3343221000; 3343221000^2 = 279300^3 + 2234400^3 = 790020^3 + 2202480^3 = 1256850^3 + 2094750^3.
First term whose square can be expressed in four ways <= 42794271007595289; 42794271007595289^2 = 14385864402^3 + 122279847417^3 = 55172161278^3 + 118485773289^3 = 64117642953^3 + 116169722214^3 = 96704977369^3 + 97504192058^3.
First term whose square can be expressed in five ways <= 47155572445935012696000; 47155572445935012696000^2 = 94405759361550^3 + 1305070263601650^3 = 374224408544280^3 + 1294899176535720^3 = 727959282778000^3 + 1224915311765600^3 = 857010857812200^3 + 1168192425418200^3 = 1009237516560000^3 + 1061381454915600^3.
After a(1) = 3 this is always composite, because factorization of the polynomial a^3 + b^3 into irreducible components over Z is a^3 + b^3 = (b+a)*(b^2 - ab + b^2). They may be semiprimes, as with 671 = 11 * 61, and 1261 = 13 * 97. The numbers can be powers in various ways, as with 32 = 2^5, 81 = 3^4, 256 = 2^8, 784 = 2^4 * 7^2 , 1225 = 5^2 * 7^2, and 2187 = 3^7. - Jonathan Vos Post, Feb 05 2011
If n is a term then n*b^3 is also a term for any b, e.g., 3 is a term hence 3*2^3 = 24, 3*3^3 = 81 and also 3*4^3 = 192 are terms. Sequence of primitive terms may be of interest. - Zak Seidov, Dec 11 2013
First noncubefree primitive term is 168 = 21*2^3 (21 is not a term of the sequence). - Zak Seidov, Dec 16 2013
From XU Pingya, Apr 10 2021: (Start)
Every triple (a, b, c) (with a^2 = b^3 + c^3) can produce a nontrivial parametric solution (x, y, z) of the Diophantine equation x^2 + y^3 + z^3 = d^4.
For example, to (1183, 65, 104), there is such a solution (d^2 - (26968032*d)*t^3 + 1183*8232^3*t^6, (376*d)*t - 65*8232^2*t^4, (92*d)*t - 104*8232^2*t^4).
To (77976, 228, 1824), there is (d^2 - (272916*d)*t^3 + 77976*57^3*t^6, (52*d)*t - 228*57^2*t^4, (74*d)*t - 1824*57^2*t^4).
Or to (77976, 1026, 1719), there is (d^2 - (25992*d)*t^3 + 77976*19^3*t^6, (37*d)*t - 1026*19^2*t^4, (11*d)*t - 1710*19^2*t^4). (End)

			1183^2 = 65^3 + 104^3.

Ian Stewart, "Game, Set and Math", Chapter 8, 'Close Encounters of the Fermat Kind', Penguin Books, Ed. 1991, pp. 107-124.

Chai Wah Wu, Table of n, a(n) for n = 1..10000 (terms 1..612 from T. D. Noe, terms 613..1000 from Harry J. Smith)

Cf. A050802, A000404, A033430, A001105, A038597, A050803, A106265, A217248.

A050801 := proc(n)
    option remember ;
    local a,x,y ;
    if n =1 then
        3
    else
        for a from procname(n-1)+1 do
            for x from 1 do
                if x^3 >= a^2 then
                    break ;
                end if;
                for y from 1 to x do
                    if x^3+y^3 = a^2 then
                        return a ;
                    end if;
                end do:
            end do:
        end do:
    end if;
end proc:
seq(A050801(n),n=1..20) ; # R. J. Mathar, Jan 22 2025

Select[Range[5350], Reduce[0 < x <= y && #^2 == x^3 + y^3, {x,y}, Integers] =!= False &] (* Jean-François Alcover, Mar 30 2011 *)
Sqrt[#]&/@Union[Select[Total/@(Tuples[Range[500],2]^3),IntegerQ[ Sqrt[ #]]&]] (* Harvey P. Dale, Mar 06 2012 *)
Select[Range@ 5400, Length@ DeleteCases[PowersRepresentations[#^2, 2, 3], w_ /; Times @@ w == 0] > 0 &] (* Michael De Vlieger, May 20 2017 *)

is(n)=my(N=n^2); for(k=sqrtnint(N\2,3),sqrtnint(N-1,3), if(ispower(N-k^3,3), return(n>1))); 0 \\ Charles R Greathouse IV, Dec 13 2013

a(n) = sqrt(A050802(n)). - Jonathan Sondow, Oct 28 2013

More terms from Michel ten Voorde and Jud McCranie

A184536 a(n) = floor(1/{(1+n^4)^(1/4)}), where {} = fractional part.

Original entry on oeis.org

5, 32, 108, 256, 500, 864, 1372, 2048, 2916, 4000, 5324, 6912, 8788, 10976, 13500, 16384, 19652, 23328, 27436, 32000, 37044, 42592, 48668, 55296, 62500, 70304, 78732, 87808, 97556, 108000, 119164, 131072, 143748, 157216, 171500, 186624, 202612, 219488, 237276, 256000, 275684, 296352, 318028, 340736, 364500, 389344, 415292, 442368, 470596, 500000, 530604, 562432, 595508, 629856, 665500, 702464, 740772, 780448, 821516, 864000, 907924, 953312, 1000188

Offset: 1

Clark Kimberling, Jan 16 2011

For n >= 1, the value of (n^4+1)^(1/4) is just slightly above n, so the fractional part is (n^4+1)^(1/4)-n. For n > 1, then, 4*n^3 < 1/((1+n^4)^(1/4)-n) < 4*n^3+1. [Proof that 4*n^3*((1+n^4)^(1/4)-n) < 1 follows easily by isolating the quartic root and raising to the 4th power; similarly, 1 < (4*n^3+1)*((1+n^4)^(1/4)-n) needs a sign estimation of an 8th-order polynomial.] In conclusion, a(n)=A033430(n) for n > 1. - Bruno Berselli, Jan 30 2011

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A184537. Essentially the same as A033430.

p[n_]:=FractionalPart[(n^4+1)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n],{n,1,80}]
LinearRecurrence[{4,-6,4,-1},{5,32,108,256,500},70] (* Harvey P. Dale, Dec 14 2023 *)

a(n) = floor[1/{(1+n^4)^(1/4)}], where {}=fractional part.

G.f.: x*(x^4-4*x^3+10*x^2+12*x+5)/(x-1)^4. - Colin Barker, Sep 21 2012

A244725 a(n) = 5*n^3.

Original entry on oeis.org

0, 5, 40, 135, 320, 625, 1080, 1715, 2560, 3645, 5000, 6655, 8640, 10985, 13720, 16875, 20480, 24565, 29160, 34295, 40000, 46305, 53240, 60835, 69120, 78125, 87880, 98415, 109760, 121945, 135000, 148955, 163840, 179685, 196520, 214375, 233280, 253265

Offset: 0

Vincenzo Librandi, Jul 05 2014

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. similar sequences of the type k*n^3: A000578 (k=1), A033431 (k=2), A117642 (k=3), A033430 (k=4), this sequence (k=5), A244726 (k=6), A244727 (k=7), A016743 (k=8), A244728 (k=9), A244729 (k=10), A016767 (k=27), A016803 (k=64), A016851 (k=125), A016911 (k=216), A016983 (k=343), A017067 (k=512), A017163 (k=729), A017271 (k=1000), A017391 (k=1331), A017523 (k=1728).

Magma
```
[5*n^3: n in [0..40]];
```

I:=[0,5,40,135]; [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..40]];

Table[5 n^3, {n, 0, 40}] (* or *) CoefficientList[Series[5 x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 40}], x]

a(n)=5*n^3 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: 5*x*(1 + 4*x + x^2)/(1 - x)^4.

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3.

A036487 a(n) = floor((n^3)/2).

Original entry on oeis.org

0, 0, 4, 13, 32, 62, 108, 171, 256, 364, 500, 665, 864, 1098, 1372, 1687, 2048, 2456, 2916, 3429, 4000, 4630, 5324, 6083, 6912, 7812, 8788, 9841, 10976, 12194, 13500, 14895, 16384, 17968, 19652, 21437, 23328, 25326, 27436, 29659, 32000

Offset: 0

N. J. A. Sloane

Enrique Pérez Herrero, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-2,3,-1).

Cf. A033430 (bisection), A268201 (bisection).

Maple
```
[ seq(floor((n^3)/2), n=0..100) ];
```

A036487[n_]:=Floor[n^3/2]
Floor[Range[0,40]^3/2] (* or *) LinearRecurrence[{3,-2,-2,3,-1},{0,0,4,13,32},50] (* Harvey P. Dale, Jun 24 2018 *)

a(n)=n^3\2 \\ Charles R Greathouse IV, Jul 18 2014

[floor(n^3/2) for n in range(0,41)] # Zerinvary Lajos, Dec 02 2009

G.f. x^2*(4 + x + x^2)/((1 + x)*(1 - x)^4). - R. J. Mathar, Jan 29 2011
From Stefano Spezia, Sep 09 2022: (Start)
a(n) = ((-1)^n - 1 + 2*n^3)/4.
E.g.f.: (x*(1 + 3*x + x^2)*cosh(x) - (1 - x - 3*x^2 - x^3)*sinh(x))/2. (End)

Corrupted b-file corrected by Michael De Vlieger, Jul 18 2014

A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

Original entry on oeis.org

0, 1, 1, 8, 4, 8, 27, 15, 15, 27, 64, 40, 32, 40, 64, 125, 85, 65, 65, 85, 125, 216, 156, 120, 108, 120, 156, 216, 343, 259, 203, 175, 175, 203, 259, 343, 512, 400, 320, 272, 256, 272, 320, 400, 512, 729, 585, 477, 405, 369, 369, 405, 477, 585, 729

Offset: 0

Peter Luschny, Oct 26 2011

Read as an infinite symmetric square array, this is the table A(n,k)=(n+k)(n^2+k^2), cf. A321500 for the triangle with k <= n. - M. F. Hasler, Nov 22 2018

			[0]                   0
[1]                  1, 1
[2]                8, 4, 8
[3]             27, 15, 15, 27
[4]           64, 40, 32, 40, 64
[5]        125, 85, 65, 65, 85, 125
[6]   216, 156, 120, 108, 120, 156, 216
[7] 343, 259, 203, 175, 175, 203, 259, 343
From _M. F. Hasler_, Nov 22 2018: (Start)
Can also be seen as the square array A(n,k)=(n+k)*(n^2 + k^2) read by antidiagonals:
n | k: 0   1   2   3 ...
--+----------------------
0 |    0   1   8  27 ...
1 |    1   4  15  40 ...
2 |    8  15  32  65 ...
3 |   27  40  65 108 ...
...      ...     ...
(End)

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A057427, A003056, A073254, A198064, A198065.

[[2*k^2*n-2*k*n^2+n^3: k in [0..n]]: n in [0..12]]; // G. C. Greubel, Nov 23 2018

A198063 := (n,k) -> 2*k^2*n-2*k*n^2+n^3:

t[n_, k_] := 2 k^2*n - 2 k*n^2 + n^3; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 22 2018 *)

A198063(n,k)=2*k^2*n-2*k*n^2+n^3 \\ See also A321500. - M. F. Hasler, Nov 22 2018

[[ 2*k^2*n-2*k*n^2+n^3 for k in range(n+1)] for n in range(12)] # G. C. Greubel, Nov 23 2018

T(n,k) = 2*k^2*n - 2*k*n^2 + n^3.
T(n,0) = T(n,n) = n^m = n^3 = A000578(n).
T(2*n,n) = (m+1)n^m = 4*n^3 = A033430(n).
T(2*n+1,n+1) = (n+1)^(m+1) - n^(m+1) = (n+1)^4 - n^4 = A005917(n).
Sum{k=0..n} T(n,k) = (2*n^4 + 3*n^3 + n^2)/3 = A098077(n).
T(n+1,k+1)*C(n,k)^4/(k+1)^3 = A197653(n,k).

A332822 One part of a 3-way classification of the positive integers. Numbers n for which A048675(n) == 2 (mod 3).

Original entry on oeis.org

3, 4, 7, 10, 13, 18, 19, 22, 24, 25, 29, 32, 34, 37, 42, 43, 45, 46, 53, 55, 56, 60, 61, 62, 71, 78, 79, 80, 81, 82, 85, 89, 94, 98, 99, 101, 104, 105, 107, 108, 113, 114, 115, 118, 121, 131, 132, 134, 139, 140, 144, 146, 150, 151, 152, 153, 155, 163, 166, 173, 174, 176, 181, 182, 187, 189, 192, 193, 194, 195, 199, 200, 204

Offset: 1

Antti Karttunen and Peter Munn, Feb 25 2020

nonn

The positive integers are partitioned between A332820, A332821 and this sequence.
For each prime p, the terms include exactly one of p and p^2. The primes alternate between this sequence and A332821. This sequence has the primes with even indexes, those in A031215.
The terms are the even numbers in A332820 halved. The terms are also the numbers m such that 5m is in A332820, and so on for alternate primes: 11, 17, 23 etc. Likewise, the terms are the numbers m such that 3m is in A332821, and so on for alternate primes: 7, 13, 19, 29 etc.
If we take each odd term of this sequence and replace each prime in its factorization by the next smaller prime, we get the same set of numbers as we get from halving the even terms of this sequence, and A332821 consists exactly of those numbers. The numbers that are one third of the terms that are multiples of 3 are in A332820, which consists exactly of those numbers. The numbers that are one fifth of the terms that are multiples of 5 constitute A332821, and for larger primes, an alternating pattern applies as described in the previous paragraph.
The product of any 2 terms of this sequence is in A332821, the product of any 3 terms is in A332820, and the product of a term of A332820 and a term of this sequence is in this sequence. So if a number k is present, k^2 is in A332821, k^3 is in A332820, and k^4 is in this sequence.
If k is an even number, exactly one of {k/2, k, 2k} is in the sequence (cf. A191257 / A067368 / A213258); and generally if k is a multiple of a prime p, exactly one of {k/p, k, k*p} is in the sequence.

Positions of terms valued -1 in A332823; equivalently, numbers in row 3k-1 of A277905 for some k >= 1.
Cf. A048675, A332820, A332821.
Comparable 2 or 3-way classifications: A000379/A000028, A001969/A000069, A003159/A036554, A005843/A005408, A028260/A026424, A191257/A067368/A213258, A325431/A325432, A329609/A329604/A332812.
Subsequences: intersection of A026478 and A066207, A031215 (prime terms), A033430\{0}, A117642\{0}, A169604, A244727\{0}, A244729\{0}, A338910 (semiprime terms).

Select[Range@ 204, Mod[Total@ #, 3] == 2 &@ Map[#[[-1]]*2^(PrimePi@ #[[1]] - 1) &, FactorInteger[#]] &] (* Michael De Vlieger, Mar 15 2020 *)

isA332822(n) =  { my(f = factor(n)); (2==((sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2)%3)); };

{a(n) : n >= 1} = {2 * A332821(k) : k >= 1} U {A003961(A332821(k)) : k >= 1}.

{a(n) : n >= 1} = {A332821(k)^2 : k >= 1} U {A331590(2, A332821(k)) : k >= 1}.

A103254 Positive integers x such that there exist positive integers y and z satisfying x^3 + y^3 = z^2.

Original entry on oeis.org

1, 2, 4, 7, 8, 9, 10, 11, 14, 16, 18, 21, 22, 23, 25, 26, 28, 32, 33, 34, 35, 36, 37, 38, 40, 44, 46, 49, 50, 56, 57, 63, 64, 65, 70, 72, 78, 81, 84, 86, 88, 90, 91, 92, 95, 98, 99, 100, 104, 105, 110, 112, 114, 121, 122, 126, 128, 129, 130, 132, 136, 140, 144, 148, 152, 154, 158, 160, 162, 169, 170, 175, 176, 177, 183, 184, 189, 190, 193, 196, 198, 200

Offset: 1

Cino Hilliard, Mar 20 2005

nonn

A001105 is a subset (excluding 0), since (x, y, z) = (A001105(k), A001105(k), A033430(k)) satisfies x^3 + y^3 = z^2. - R. J. Mathar, Sep 11 2006
A parametric solution: {x,y,z} = {g*(4*e + g)*(4*e^2 + 8*e*g + g^2), 2*g*(4*e + g)*(-2*e^2 +2*e*g + g^2), 3*g^2*(4*e + g)^2*(4*e^2 + 2*e*g + g^2)}, provided (-2*e^2 +2*e*g + g^2) > 0. - James Mc Laughlin, Jan 27 2007
Allowing y = 0 would give the same sequence, since x^3 = z^2 implies x is a square, and all squares are terms since (t^2)^3 + (2*t^2)^3 = (3*t^3)^2. On the other hand, allowing y to be negative would introduce new terms: 71, 74, and 155 would be terms since 71^3 + (-23)^3 = 588^2, 74^3 + (-47)^3 = 549^2, and 155^3 + (-31)^3 = 1922^2. See A356720. - Jianing Song, Aug 24 2022

			x=7, y=21, 7^3 + 21^3 = 98^2. 7 is the 4th term in the list.
Other solutions are (x, y, z)=(1, 2, 3), (4, 8, 24), (7, 21, 98), (9, 18, 81), (10, 65, 525), (11, 37, 228), (14, 70, 588), (16, 32, 192), (21, 7, 98), (22, 26, 168), (23, 1177, 40380), ...

Fritz Beukers, The Diophantine equation Ax^p+By^q=Cz^r, Duke Math. J. 91 (1998), 61-88.

See A103255 for another version.

[ k : k in [1..200] | exists{P : P in IntegralPoints(EllipticCurve([0,k^3])) | P[1] gt 0 and P[2] ne 0 } ]; // Geoff Bailey, Jan 28 2007

Recomputed and extended to 48 terms by Geoff Bailey (geoff(AT)maths.usyd.edu.au) using Magma, Jan 28 2007

Terms 104..200 added by Joerg Arndt, Sep 29 2012

A317297 a(n) = (n - 1)(4n^2 - 8*n + 5).

Original entry on oeis.org

0, 5, 34, 111, 260, 505, 870, 1379, 2056, 2925, 4010, 5335, 6924, 8801, 10990, 13515, 16400, 19669, 23346, 27455, 32020, 37065, 42614, 48691, 55320, 62525, 70330, 78759, 87836, 97585, 108030, 119195, 131104, 143781, 157250, 171535, 186660, 202649, 219526, 237315, 256040, 275725, 296394, 318071

Offset: 1

Omar E. Pol, Sep 01 2018

Conjecture: For n > 1, a(n) is the maximum eigenvalue of a 2*(n-1) X 2*(n-1) square matrix M defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even (see A317614). - Stefano Spezia, Dec 27 2018

Connections can be made to A022144 and A010014. Namely, a formula for A022144 is (2*n+1)^2 - (2*n-1)^2. A formula for A010014 is (2*n+1)^3 - (2*n-1)^3. The general form can be represented by (2*n+1)^d - (2*n-1)^d, where d designates the number of dimensions. When d is 4, a(n) = ((2*(n-1)+1)^4 - (2*(n-1)-1)^4)/16, namely the general form shifted by 1 and divided by 16 is a(n). - Yigit Oktar, Aug 16 2024

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

First bisection of A006003.
Nonzero terms give the row sums of A007607.
Conjecture: 0 together with a bisection of A246697.
Cf. A219086 (partial sums).
Cf. A008595, A033430, A135453, A317614.
Cf. A010014, A022144 (see comments)

Table[(n - 1) (4 n^2 - 8 n + 5), {n, 1, 50}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 5, 34, 111}, 50] (* or *) CoefficientList[Series[x (5 + 14 x + 5 x^2)/(1 - x)^4, {x, 0, 50}], x] (* Stefano Spezia, Sep 01 2018 *)

PARI
```
a(n) = (n - 1)*(4*n^2 - 8*n + 5)
```

concat(0, Vec(x^2*(5 + 14*x + 5*x^2)/(1 - x)^4 + O(x^50))) \\ Colin Barker, Sep 01 2018

a(n) = 4*n^3 - 12*n^2 + 13*n - 5 = A033430(n) - A135453(n) + A008595(n) - 5.
G.f.: x^2*(5 + 14*x + 5*x^2)/(1 - x)^4. - Colin Barker, Sep 01 2018
a(n) = 4*a(n - 1) - 6*a(n - 2) + 4*a(n - 3) - a(n - 4) for n > 4. - Stefano Spezia, Sep 01 2018
E.g.f.: exp(x)*(5*x + 12*x^2 + 4*x^3). - Stefano Spezia, Jan 15 2019
a(n) = ((2*(n-1)+1)^4 - (2*(n-1)-1)^4)/16. - Yigit Oktar, Aug 16 2024

A321491 Numbers of the form (x+y)(x^2+y^2), with integers x > y > 0.

Original entry on oeis.org

15, 40, 65, 85, 120, 156, 175, 203, 259, 272, 320, 369, 400, 405, 477, 520, 580, 585, 671, 680, 715, 803, 820, 888, 935, 960, 1080, 1105, 1111, 1157, 1248, 1261, 1400, 1417, 1464, 1484, 1624, 1625, 1695, 1755, 1820, 1875, 1885, 2055, 2072, 2080, 2176, 2295, 2336, 2380, 2465

Offset: 1

Geoffrey B. Campbell and M. F. Hasler, Nov 22 2018

nonn

If y = 0 is allowed, this adds the cubes A000578; if x = y is allowed, this adds A033430 = numbers of the form 4*x^3. None of these variants is in the OEIS yet.

			Let f(x,y) = (x+y)(x^2+y^2) = A321490(x,y), then:
a(1) = f(2,1) = 3*5 = 15,a(2) = f(3,1) = 4*10 = 40, a(3) = f(3,2) = 5*13 = 65,a(4) = f(4,1) = 5*17 = 85,a(5) = f(4,2) = 6*20 = 120, etc.

Cf. A321490, A321492.

list_A321491(L=1e4,S=[])={for(m=2, sqrtnint(L, 3), for(n=1, m-1, if(L

A321500 Triangular table T(n,k) = (n+k)*(n^2+k^2), n >= k >= 0; read by rows n = 0, 1, 2, ...

Original entry on oeis.org

0, 1, 4, 8, 15, 32, 27, 40, 65, 108, 64, 85, 120, 175, 256, 125, 156, 203, 272, 369, 500, 216, 259, 320, 405, 520, 671, 864, 343, 400, 477, 580, 715, 888, 1105, 1372, 512, 585, 680, 803, 960, 1157, 1400, 1695, 2048

Offset: 0

M. F. Hasler, Nov 22 2018

			The table starts:
n | T(n,k), k = 0..n:
0 |   0;
1 |   1,   4;
2 |   8,  15,  32;
3 |  27,  40,  65, 108;
4 |  64,  85, 120, 175, 256;
5 | 125, 156, 203, 272, 369,  500;
6 | 216, 259, 320, 405, 520,  671,  864;
7 | 343, 400, 477, 580, 715,  888, 1105, 1372;
8 | 512, 585, 680, 803, 960, 1157, 1400, 1695, 2048;
etc.

M. F. Hasler, Rows n = 0..141 of triangle, flattened

Cf. A000578 (column 0: the cubes), A033430 (diagonal: 4*n^3), A053698 (column 1).

Cf. A198063 (read as A(n,k)=(n+k)*(n^2+k^2)).

[[(n+k)*(n^2+k^2): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Nov 23 2018

t[n_, k_] := (n + k) (n^2 + k^2); Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, Nov 22 2018 *)

A321500(n, k)=(n+k)*(n^2+k^2)
A321500_row(n)=vector(n+1, k, (n+k--)*(n^2+k^2))
A321500_list(N=11)=concat(apply(A321500_row, [0..N]))

[[(n+k)*(n^2+k^2) for k in range(n+1)] for n in range(12)] # G. C. Greubel, Nov 23 2018

Sum_{k=0..n} T(n,k) = 5*n^2*(n+1)*(5*n+1)/12 = 5*A117066(n). - G. C. Greubel, Nov 23 2018

cp's OEIS Frontend

A050801 Numbers k such that k^2 is expressible as the sum of two positive cubes in at least one way.

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A184536 a(n) = floor(1/{(1+n^4)^(1/4)}), where {} = fractional part.

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A244725 a(n) = 5*n^3.

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A036487 a(n) = floor((n^3)/2).

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A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)*C(i,j)*n^j*k^(m-j).

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A332822 One part of a 3-way classification of the positive integers. Numbers n for which A048675(n) == 2 (mod 3).

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A103254 Positive integers x such that there exist positive integers y and z satisfying x^3 + y^3 = z^2.

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A198063 Triangle read by rows (n >= 0, 0 <= k <= n, m = 3); T(n,k) = Sum{j=0..m} Sum{i=0..m} (-1)^(j+i)C(i,j)n^j*k^(m-j).

A317297 a(n) = (n - 1)(4n^2 - 8*n + 5).