A003072 -id:A003072 - cp's OEIS frontend

A000578 The cubes: a(n) = n^3.

0, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507

Offset: 0

N. J. A. Sloane

a(n) is the sum of the next n odd numbers; i.e., group the odd numbers so that the n-th group contains n elements like this: (1), (3, 5), (7, 9, 11), (13, 15, 17, 19), (21, 23, 25, 27, 29), ...; then each group sum = n^3 = a(n). Also the median of each group = n^2 = mean. As the sum of first n odd numbers is n^2 this gives another proof of the fact that the n-th partial sum = (n(n + 1)/2)^2. - Amarnath Murthy, Sep 14 2002
Total number of triangles resulting from criss-crossing cevians within a triangle so that two of its sides are each n-partitioned. - Lekraj Beedassy, Jun 02 2004. See Propp and Propp-Gubin for a proof.
Also structured triakis tetrahedral numbers (vertex structure 7) (cf. A100175 = alternate vertex); structured tetragonal prism numbers (vertex structure 7) (cf. A100177 = structured prisms); structured hexagonal diamond numbers (vertex structure 7) (cf. A100178 = alternate vertex; A000447 = structured diamonds); and structured trigonal anti-diamond numbers (vertex structure 7) (cf. A100188 = structured anti-diamonds). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {4, 3}.
Least multiple of n such that every partial sum is a square. - Amarnath Murthy, Sep 09 2005
Draw a regular hexagon. Construct points on each side of the hexagon such that these points divide each side into equally sized segments (i.e., a midpoint on each side or two points on each side placed to divide each side into three equally sized segments or so on), do the same construction for every side of the hexagon so that each side is equally divided in the same way. Connect all such points to each other with lines that are parallel to at least one side of the polygon. The result is a triangular tiling of the hexagon and the creation of a number of smaller regular hexagons. The equation gives the total number of regular hexagons found where n = the number of points drawn + 1. For example, if 1 point is drawn on each side then n = 1 + 1 = 2 and a(n) = 2^3 = 8 so there are 8 regular hexagons in total. If 2 points are drawn on each side then n = 2 + 1 = 3 and a(n) = 3^3 = 27 so there are 27 regular hexagons in total. - Noah Priluck (npriluck(AT)gmail.com), May 02 2007
The solutions of the Diophantine equation: (X/Y)^2 - X*Y = 0 are of the form: (n^3, n) with n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^2k - XY = 0 are of the form: (m*n^(2k + 1), m*n^(2k - 1)) with m >= 1, k >= 1 and n >= 1. The solutions of the Diophantine equation: (m^2)*(X/Y)^(2k + 1) - XY = 0 are of the form: (m*n^(k + 1), m*n^k) with m >= 1, k >= 1 and n >= 1. - Mohamed Bouhamida, Oct 04 2007
Except for the first two terms, the sequence corresponds to the Wiener indices of C_{2n} i.e., the cycle on 2n vertices (n > 1). - K.V.Iyer, Mar 16 2009
Totally multiplicative sequence with a(p) = p^3 for prime p. - Jaroslav Krizek, Nov 01 2009
Sums of rows of the triangle in A176271, n > 0. - Reinhard Zumkeller, Apr 13 2010
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Numbers n for which order of torsion subgroup t of the elliptic curve y^2 = x^3 - n is t = 2. - Artur Jasinski, Jun 30 2010
The sequence with the lengths of the Pisano periods mod k is 1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, ... for k >= 1, apparently multiplicative and derived from A000027 by dividing every ninth term through 3. Cubic variant of A186646. - R. J. Mathar, Mar 10 2011
The number of atoms in a bcc (body-centered cubic) rhombic hexahedron with n atoms along one edge is n^3 (T. P. Martin, Shells of atoms, eq. (8)). - Brigitte Stepanov, Jul 02 2011
The inverse binomial transform yields the (finite) 0, 1, 6, 6 (third row in A019538 and A131689). - R. J. Mathar, Jan 16 2013
Twice the area of a triangle with vertices at (0, 0), (t(n - 1), t(n)), and (t(n), t(n - 1)), where t = A000217 are triangular numbers. - J. M. Bergot, Jun 25 2013
If n > 0 is not congruent to 5 (mod 6) then A010888(a(n)) divides a(n). - Ivan N. Ianakiev, Oct 16 2013
For n > 2, a(n) = twice the area of a triangle with vertices at points (binomial(n,3),binomial(n+2,3)), (binomial(n+1,3),binomial(n+1,3)), and (binomial(n+2,3),binomial(n,3)). - J. M. Bergot, Jun 14 2014
Determinants of the spiral knots S(4,k,(1,1,-1)). a(k) = det(S(4,k,(1,1,-1))). - Ryan Stees, Dec 14 2014
One of the oldest-known examples of this sequence is shown in the Senkereh tablet, BM 92698, which displays the first 32 terms in cuneiform. - Charles R Greathouse IV, Jan 21 2015
From Bui Quang Tuan, Mar 31 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ... 2*n-1 as follows. The first column contains all the integers 1, 2, 3, ... 2*n-1. Each succeeding column is the same as the previous column but without the first and last items. The last column contains only n. The sum of all the numbers in the triangle is n^3.
Here is the example for n = 4, where 1 + 2*2 + 3*3 + 4*4 + 3*5 + 2*6 + 7 = 64 = a(4):
  1
  2  2
  3  3  3
  4  4  4  4
  5  5  5
  6  6
  7
(End)
For n > 0, a(n) is the number of compositions of n+11 into n parts avoiding parts 2 and 3. - Milan Janjic, Jan 07 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
Number of inequivalent face colorings of the cube using at most n colors such that each color appears at least twice. - David Nacin, Feb 22 2017
Consider A = {a,b,c} a set with three distinct members. The number of subsets of A is 8, including {a,b,c} and the empty set. The number of subsets from each of those 8 subsets is 27. If the number of such iterations is n, then the total number of subsets is a(n-1). - Gregory L. Simay, Jul 27 2018
By Fermat's Last Theorem, these are the integers of the form x^k with the least possible value of k such that x^k = y^k + z^k never has a solution in positive integers x, y, z for that k. - Felix Fröhlich, Jul 27 2018

			For k=3, b(3) = 2 b(2) - b(1) = 4-1 = 3, so det(S(4,3,(1,1,-1))) = 3*3^2 = 27.
For n=3, a(3) = 3 + (3*0^2 + 3*0 + 3*1^2 + 3*1 + 3*2^2 + 3*2) = 27. - _Patrick J. McNab_, Mar 28 2016

Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See p. 191.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 43, 64, 81.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 292.
T. Aaron Gulliver, "Sequences from cubes of integers", International Mathematical Journal, 4 (2003), no. 5, 439 - 445. See http://www.m-hikari.com/z2003.html for information about this journal. [I expanded the reference to make this easier to find. - N. J. A. Sloane, Feb 18 2019]
J. Propp and A. Propp-Gubin, "Counting Triangles in Triangles", Pi Mu Epsilon Journal (to appear).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 6-7.
D. Wells, You Are A Mathematician, pp. 238-241, Penguin Books 1995.

N. J. A. Sloane, Table of n, a(n) for n = 0..10000
H. Bottomley, Illustration of initial terms
British National Museum, Tablet 92698
N. Brothers, S. Evans, L. Taalman, L. Van Wyk, D. Witczak, and C. Yarnall, Spiral knots, Missouri J. of Math. Sci., 22 (2010).
M. DeLong, M. Russell, and J. Schrock, Colorability and determinants of T(m,n,r,s) twisted torus knots for n equiv. +/-1(mod m), Involve, Vol. 8 (2015), No. 3, 361-384.
Ralph Greenberg, Math For Poets
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets [Cached version at the Wayback Machine]
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., 131 (2002), 65-75. - fixed by _Felix Fröhlich_, Jun 16 2014
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (8).
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003.
Ed Pegg, Jr., Sequence Pictures, Math Games column, Dec 08 2003 [Cached copy, with permission (pdf only)]
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
James Propp and Adam Propp-Gubin, Counting Triangles in Triangles, arXiv:2409.17117 [math.CO], 25 September 2024.
Kenneth A. Ross, First Digits of Squares and Cubes, Math. Mag. 85 (2012) 36-42. doi:10.4169/math.mag.85.1.36.
Eric Weisstein's World of Mathematics, Cubic Number, and Hex Pyramidal Number
Ronald Yannone, Hilbert Matrix Analyses
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for "core" sequences
Index entries for sequences related to Benford's law

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
For sums of cubes, cf. A000537 (partial sums), A003072, A003325, A024166, A024670, A101102 (fifth partial sums).
Cf. A001158 (inverse Möbius transform), A007412 (complement), A030078(n) (cubes of primes), A048766, A058645 (binomial transform), A065876, A101094, A101097.
Subsequence of A145784.
Cf. A260260 (comment). - Bruno Berselli, Jul 22 2015
Cf. A000292 (tetrahedral numbers), A005900 (octahedral numbers), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Cf. A098737 (main diagonal).

a000578 = (^ 3)
a000578_list = 0 : 1 : 8 : zipWith (+)
   (map (+ 6) a000578_list)
   (map (* 3) $ tail $ zipWith (-) (tail a000578_list) a000578_list)
-- Reinhard Zumkeller, Sep 05 2015, May 24 2012, Oct 22 2011

[ n^3 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014

I:=[0,1,8,27]; [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..45]]; // Vincenzo Librandi, Jul 05 2014

A000578 := n->n^3;
seq(A000578(n), n=0..50);
isA000578 := proc(r)
    local p;
    if r = 0 or r =1 then
        true;
    else
        for p in ifactors(r)[2] do
            if op(2, p) mod 3 <> 0 then
                return false;
            end if;
        end do:
        true ;
    end if;
end proc: # R. J. Mathar, Oct 08 2013

Table[n^3, {n, 0, 30}] (* Stefan Steinerberger, Apr 01 2006 *)
CoefficientList[Series[x (1 + 4 x + x^2)/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Jul 05 2014 *)
Accumulate[Table[3n^2+3n+1,{n,0,20}]] (* or *) LinearRecurrence[{4,-6,4,-1},{1,8,27,64},20](* Harvey P. Dale, Aug 18 2018 *)

A000578(n):=n^3$
makelist(A000578(n),n,0,30); /* Martin Ettl, Nov 03 2012 */

A000578(n)=n^3 \\ M. F. Hasler, Apr 12 2008

is(n)=ispower(n,3) \\ Charles R Greathouse IV, Feb 20 2012

A000578_list, m = [], [6, -6, 1, 0]
for _ in range(10**2):
    A000578_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

(define (A000578 n) (* n n n)) ;; Antti Karttunen, Oct 06 2017

a(n) = Sum_{i=0..n-1} A003215(i).
Multiplicative with a(p^e) = p^(3e). - David W. Wilson, Aug 01 2001
G.f.: x*(1+4*x+x^2)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
Dirichlet generating function: zeta(s-3). - Franklin T. Adams-Watters, Sep 11 2005, Amarnath Murthy, Sep 09 2005
E.g.f.: (1+3*x+x^2)*x*exp(x). - Franklin T. Adams-Watters, Sep 11 2005 - Amarnath Murthy, Sep 09 2005
a(n) = Sum_{i=1..n} (Sum_{j=i..n+i-1} A002024(j,i)). - Reinhard Zumkeller, Jun 24 2007
a(n) = lcm(n, (n - 1)^2) - (n - 1)^2. E.g.: lcm(1, (1 - 1)^2) - (1 - 1)^2 = 0, lcm(2, (2 - 1)^2) - (2 - 1)^2 = 1, lcm(3, (3 - 1)^2) - (3 - 1)^2 = 8, ... - Mats Granvik, Sep 24 2007
Starting (1, 8, 27, 64, 125, ...), = binomial transform of [1, 7, 12, 6, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
a(n) = A007531(n) + A000567(n). - Reinhard Zumkeller, Sep 18 2009
a(n) = binomial(n+2,3) + 4*binomial(n+1,3) + binomial(n,3). [Worpitzky's identity for cubes. See. e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n + 6*binomial(n+1,3) = binomial(n,1)+6*binomial(n+1,3). - Ron Knott, Jun 10 2019
A010057(a(n)) = 1. - Reinhard Zumkeller, Oct 22 2011
a(n) = A000537(n) - A000537(n-1), difference between 2 squares of consecutive triangular numbers. - Pierre CAMI, Feb 20 2012
a(n) = A048395(n) - 2*A006002(n). - J. M. Bergot, Nov 25 2012
a(n) = 1 + 7*(n-1) + 6*(n-1)*(n-2) + (n-1)*(n-2)*(n-3). - Antonio Alberto Olivares, Apr 03 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6. - Ant King Apr 29 2013
a(n) = A000330(n) + Sum_{i=1..n-1} A014105(i), n >= 1. - Ivan N. Ianakiev, Sep 20 2013
a(k) = det(S(4,k,(1,1,-1))) = k*b(k)^2, where b(1)=1, b(2)=2, b(k) = 2*b(k-1) - b(k-2) = b(2)*b(k-1) - b(k-2). - Ryan Stees, Dec 14 2014
For n >= 1, a(n) = A152618(n-1) + A033996(n-1). - Bui Quang Tuan, Apr 01 2015
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Jon Tavasanis, Feb 21 2016
a(n) = n + Sum_{j=0..n-1} Sum_{k=1..2} binomial(3,k)*j^(3-k). - Patrick J. McNab, Mar 28 2016
a(n) = A000292(n-1) * 6 + n. - Zhandos Mambetaliyev, Nov 24 2016
a(n) = n*binomial(n+1, 2) + 2*binomial(n+1, 3) + binomial(n,3). - Tony Foster III, Nov 14 2017
From Amiram Eldar, Jul 02 2020: (Start)
Sum_{n>=1} 1/a(n) = zeta(3) (A002117).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*zeta(3)/4 (A197070). (End)
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)/Pi.
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)/(3*Pi). (End)
a(n) = Sum_{d|n} sigma_3(d)*mu(n/d) = Sum_{d|n} A001158(d)*A008683(n/d). Moebius transform of sigma_3(n). - Ridouane Oudra, Apr 15 2021

A003325 Numbers that are the sum of 2 positive cubes.

Original entry on oeis.org

2, 9, 16, 28, 35, 54, 65, 72, 91, 126, 128, 133, 152, 189, 217, 224, 243, 250, 280, 341, 344, 351, 370, 407, 432, 468, 513, 520, 539, 559, 576, 637, 686, 728, 730, 737, 756, 793, 854, 855, 945, 1001, 1008, 1024, 1027, 1064, 1072, 1125, 1216, 1241, 1332, 1339, 1343

Offset: 1

N. J. A. Sloane

It is conjectured that this sequence and A052276 have infinitely many numbers in common, although only one example (128) is known. [Any further examples are greater than 5 million. - Charles R Greathouse IV, Apr 12 2020] [Any further example is greater than 10^12. - M. F. Hasler, Jan 10 2021]
A113958 is a subsequence; if m is a term then m+k^3 is a term of A003072 for all k > 0. - Reinhard Zumkeller, Jun 03 2006
From James R. Buddenhagen, Oct 16 2008: (Start)
(i) N and N+1 are both the sum of two positive cubes if N=2*(2*n^2 + 4*n + 1)*(4*n^4 + 16*n^3 + 23*n^2 + 14*n + 4), n=1,2,....
(ii) For n >= 2, let N = 16*n^6 - 12*n^4 + 6*n^2 - 2, so N+1 = 16*n^6 - 12*n^4 + 6*n^2 - 1.
Then the identities 16*n^6 - 12*n^4 + 6*n^2 - 2 = (2*n^2 - n - 1)^3 + (2*n^2 + n - 1)^3 16*n^6 - 12*n^4 + 6*n^2 - 1 = (2*n^2)^3 + (2*n^2 - 1)^3 show that N, N+1 are in the sequence. (End)
If n is a term then n*m^3 (m >= 2) is also a term, e.g., 2m^3, 9m^3, 28m^3, and 35m^3 are all terms of the sequence. "Primitive" terms (not of the form n*m^3 with n = some previous term of the sequence and m >= 2) are 2, 9, 28, 35, 65, 91, 126, etc. - Zak Seidov, Oct 12 2011
This is an infinite sequence in which the first term is prime but thereafter all terms are composite. - Ant King, May 09 2013
By Fermat's Last Theorem (the special case for exponent 3, proved by Euler, is sufficient), this sequence contains no cubes. - Charles R Greathouse IV, Apr 03 2021

C. G. J. Jacobi, Gesammelte Werke, vol. 6, 1969, Chelsea, NY, p. 354.

N. J. A. Sloane, Table of n, a(n) for n = 1..20000 (first 1000 terms from T. D. Noe)
F. Beukers, The Diophantine equation Ax^p+By^q=Cz^r, Duke Math. J. 91 (1998), 61-88.
Kevin A. Broughan, Characterizing the sum of two cubes, J. Integer Seqs., Vol. 6, 2003.
Nils Bruin, On powers as sums of two cubes, in Algorithmic number theory (Leiden, 2000), 169-184, Lecture Notes in Comput. Sci., 1838, Springer, Berlin, 2000.
C. G. J. Jacobi, Gesammelte Werke.
Michael Penn, 1674 is not a perfect cube, 2020 video
N. J. A. Sloane, Table of n, a(n) for n = 1..59562
D. Tournes, A Glance on Indian Mathematician Srinivasa Ramanujan(1887-1920). [Text in French]
Eric Weisstein's World of Mathematics, Cubic Number
Index entries for sequences related to sums of cubes

Subsequence of A004999 and hence of A045980; supersequence of A202679.

Cf. A024670 (2 distinct cubes), A003072, A001235, A011541, A003826, A010057, A000578, A027750, A010052, A085323 (n such that a(n+1)=a(n)+1).

a003325 n = a003325_list !! (n-1)
a003325_list = filter c2 [1..] where
   c2 x = any (== 1) $ map (a010057 . fromInteger) $
                       takeWhile (> 0) $ map (x -) $ tail a000578_list
-- Reinhard Zumkeller, Mar 24 2012

nn = 2*20^3; Union[Flatten[Table[x^3 + y^3, {x, nn^(1/3)}, {y, x, (nn - x^3)^(1/3)}]]] (* T. D. Noe, Oct 12 2011 *)
With[{upto=2000},Select[Total/@Tuples[Range[Ceiling[Surd[upto,3]]]^3,2],#<=upto&]]//Union (* Harvey P. Dale, Jun 11 2016 *)

cubes=sum(n=1, 11, x^(n^3), O(x^1400)); v = select(x->x, Vec(cubes^2), 1); vector(#v, k, v[k]+1) \\ edited by Michel Marcus, May 08 2017

isA003325(n) = for(k=1,sqrtnint(n\2,3), ispower(n-k^3,3) && return(1)) \\ M. F. Hasler, Oct 17 2008, improved upon suggestion of Altug Alkan and Michel Marcus, Feb 16 2016

T=thueinit('z^3+1); is(n)=#select(v->min(v[1],v[2])>0, thue(T,n))>0 \\ Charles R Greathouse IV, Nov 29 2014

list(lim)=my(v=List()); lim\=1; for(x=1,sqrtnint(lim-1,3), my(x3=x^3); for(y=1,min(sqrtnint(lim-x3,3),x), listput(v, x3+y^3))); Set(v) \\ Charles R Greathouse IV, Jan 11 2022

from sympy import integer_nthroot
def aupto(lim):
  cubes = [i*i*i for i in range(1, integer_nthroot(lim-1, 3)[0] + 1)]
  sum_cubes = sorted([a+b for i, a in enumerate(cubes) for b in cubes[i:]])
  return [s for s in sum_cubes if s <= lim]
print(aupto(1343)) # Michael S. Branicky, Feb 09 2021

Error in formula line corrected by Zak Seidov, Jul 23 2009

A000408 Numbers that are the sum of three nonzero squares.

Original entry on oeis.org

3, 6, 9, 11, 12, 14, 17, 18, 19, 21, 22, 24, 26, 27, 29, 30, 33, 34, 35, 36, 38, 41, 42, 43, 44, 45, 46, 48, 49, 50, 51, 53, 54, 56, 57, 59, 61, 62, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 77, 78, 81, 82, 83, 84, 86, 88, 89, 90, 91, 93, 94, 96, 97, 98, 99, 101, 102, 104

Offset: 1

N. J. A. Sloane and J. H. Conway

nonn

a(n) !== 7 (mod 8). - Boris Putievskiy, May 05 2013
A025427(a(n)) > 0. - Reinhard Zumkeller, Feb 26 2015
According to Halter-Koch (below), a number n is a sum of 3 squares, but not a sum of 3 nonzero squares (i.e., is in A000378 but not A000408), if and only if it is of the form 4^j*s, where j >= 0 and s in {1,2,5,10,13,25,37,58,85,130,?}, where ? denotes at most one unknown number that, if it exists, is > 5*10^10. - Jeffrey Shallit, Jan 15 2017

L. E. Dickson, History of the Theory of Numbers, vol. II: Diophantine Analysis, Dover, 2005, p. 267.
Savin Réalis, Answer to question 25 ("Toute puissance entière de 3 est une somme de trois carrés premiers avec 3"), Mathesis 1 (1881), pp. 87-88. (See also p. 73 where the question is posed.)

Ray Chandler, Table of n, a(n) for n = 1..10000
Alexander Berkovich and Will Jagy, On representation of an integer as the sum of three squares and the ternary quadratic forms with the discriminants p^2, 16p^2, arXiv:1101.2951 [math.NT], 2011.
B. Farhi, On the Representation of the Natural Numbers as the Sum of Three Terms of the Sequence floor(n^2/a), J. Int. Seq. 16 (2013) #13.6.4.
Franz Halter-Koch, Darstellung natürlicher Zahlen als Summe von Quadraten, Acta Arithmetica 42 (1982), pp. 11-20.
S. Mezroui, A. Azizi, and M'hammed Ziane, On a Conjecture of Farhi , J. Int. Seq. 17 (2014) #14.1.8.
Index entries for sequences related to sums of squares

Cf. A000378, A000404, A000414, A003072, A004214, A024795, A024796, A025321, A025427.

a000408 n = a000408_list !! (n-1)
a000408_list = filter ((> 0) . a025427) [1..]
-- Reinhard Zumkeller, Feb 26 2015

N:= 1000: # to get all terms <= N
S:= series((JacobiTheta3(0,q)-1)^3,q,1001):
select(t -> coeff(S,q,t)>0, [$1..N]); # Robert Israel, Jan 14 2016

f[n_] := Flatten[Position[Take[Rest[CoefficientList[Sum[x^(i^2), {i, n}]^3, x]], n^2], ?Positive]];f[11] (* _Ray Chandler, Dec 06 2006 *)
pr[n_] := Select[ PowersRepresentations[n, 3, 2], FreeQ[#, 0] &]; Select[ Range[104], pr[#] != {} &] (* Jean-François Alcover, Apr 04 2013 *)
max = 1000; s = (EllipticTheta[3, 0, q] - 1)^3 + O[q]^(max+1); Select[ Range[max], SeriesCoefficient[s, {q, 0, #}] > 0 &] (* Jean-François Alcover, Feb 01 2016, after Robert Israel *)

is(n)=for(x=sqrtint((n-1)\3)+1,sqrtint(n-2), for(y=1,sqrtint(n-x^2-1), if(issquare(n-x^2-y^2), return(1)))); 0 \\ Charles R Greathouse IV, Apr 04 2013

is(n)= my(a, b) ; a=1 ; while(a^2+1Altug Alkan, Jan 18 2016

def aupto(lim):
  squares = [k*k for k in range(1, int(lim**.5)+2) if k*k <= lim]
  sum2sqs = set(a+b for i, a in enumerate(squares) for b in squares[i:])
  sum3sqs = set(a+b for a in sum2sqs for b in squares)
  return sorted(set(range(lim+1)) & sum3sqs)
print(aupto(104)) # Michael S. Branicky, Mar 06 2021

a(n) = 6n/5 + O(log n). - Charles R Greathouse IV, Mar 14 2014; error term improved Jul 05 2024

A003336 Numbers that are the sum of 2 positive 4th powers.

Original entry on oeis.org

2, 17, 32, 82, 97, 162, 257, 272, 337, 512, 626, 641, 706, 881, 1250, 1297, 1312, 1377, 1552, 1921, 2402, 2417, 2482, 2592, 2657, 3026, 3697, 4097, 4112, 4177, 4352, 4721, 4802, 5392, 6497, 6562, 6577, 6642, 6817, 7186, 7857, 8192, 8962, 10001, 10016, 10081, 10256, 10625

Offset: 1

N. J. A. Sloane

Numbers k such that k = x^4 + y^4 has a solution in positive integers x, y.
There are no squares in this sequence. - Altug Alkan, Apr 08 2016
As the order of addition doesn't matter we can assume terms are in nondecreasing order. - David A. Corneth, Aug 01 2020

			From _David A. Corneth_, Aug 01 2020: (Start)
16378801 is in the sequence as 16378801 = 43^4 + 60^4.
39126977 is in the sequence as 39126977 = 49^4 + 76^4.
71769617 is in the sequence as 71769617 = 19^4 + 92^4. (End)

Sean A. Irvine, Table of n, a(n) for n = 1..20000 (terms 1..1000 from T. D. Noe, terms 1001..10000 from David A. Corneth)
A. Bremner and P. Morton, A new characterization of the integer 5906, Manuscripta Math. 44 (1983) 187-229; Math. Rev. 84i:10016.
S. R. Finch, On a generalized Fermat-Wiles equation [broken link]
Steven R. Finch, On Generalized Fermat-Wiles Equation [From the Wayback Machine]
Samuel S. Wagstaff, Jr., Equal Sums of Two Distinct Like Powers, J. Int. Seq., Vol. 25 (2022), Article 22.3.1.
Eric Weisstein's World of Mathematics, Biquadratic Number.

5906 is the first term in A060387 but not in this sequence. Cf. A020897.
Cf. A088687 (2 distinct 4th powers).
A###### (x, y): Numbers that are the form of x nonzero y-th powers.
Cf. A000404 (2, 2), A000408 (3, 2), A000414 (4, 2), A003072 (3, 3), A003325 (3, 2), A003327 (4, 3), A003328 (5, 3), A003329 (6, 3), A003330 (7, 3), A003331 (8, 3), A003332 (9, 3), A003333 (10, 3), A003334 (11, 3), A003335 (12, 3), A003336 (2, 4), A003337 (3, 4), A003338 (4, 4), A003339 (5, 4), A003340 (6, 4), A003341 (7, 4), A003342 (8, 4), A003343 (9, 4), A003344 (10, 4), A003345 (11, 4), A003346 (12, 4), A003347 (2, 5), A003348 (3, 5), A003349 (4, 5), A003350 (5, 5), A003351 (6, 5), A003352 (7, 5), A003353 (8, 5), A003354 (9, 5), A003355 (10, 5), A003356 (11, 5), A003357 (12, 5), A003358 (2, 6), A003359 (3, 6), A003360 (4, 6), A003361 (5, 6), A003362 (6, 6), A003363 (7, 6), A003364 (8, 6), A003365 (9, 6), A003366 (10, 6), A003367 (11, 6), A003368 (12, 6), A003369 (2, 7), A003370 (3, 7), A003371 (4, 7), A003372 (5, 7), A003373 (6, 7), A003374 (7, 7), A003375 (8, 7), A003376 (9, 7), A003377 (10, 7), A003378 (11, 7), A003379 (12, 7), A003380 (2, 8), A003381 (3, 8), A003382 (4, 8), A003383 (5, 8), A003384 (6, 8), A003385 (7, 8), A003387 (9, 8), A003388 (10, 8), A003389 (11, 8), A003390 (12, 8), A003391 (2, 9), A003392 (3, 9), A003393 (4, 9), A003394 (5, 9), A003395 (6, 9), A003396 (7, 9), A003397 (8, 9), A003398 (9, 9), A003399 (10, 9), A004800 (11, 9), A004801 (12, 9), A004802 (2, 10), A004803 (3, 10), A004804 (4, 10), A004805 (5, 10), A004806 (6, 10), A004807 (7, 10), A004808 (8, 10), A004809 (9, 10), A004810 (10, 10), A004811 (11, 10), A004812 (12, 10), A004813 (2, 11), A004814 (3, 11), A004815 (4, 11), A004816 (5, 11), A004817 (6, 11), A004818 (7, 11), A004819 (8, 11), A004820 (9, 11), A004821 (10, 11), A004822 (11, 11), A004823 (12, 11), A047700 (5, 2).
Cf. A000583 (4th powers).

nn=12; Select[Union[Plus@@@(Tuples[Range[nn],{2}]^4)], # <= nn^4&] (* Harvey P. Dale, Dec 29 2010 *)
Select[Range@ 11000, Length[PowersRepresentations[#, 2, 4] /. {0, } -> Nothing] > 0 &] (* _Michael De Vlieger, Apr 08 2016 *)

list(lim)=my(v=List()); for(x=1, sqrtnint(lim\=1,4), for(y=1, min(sqrtnint(lim-x^4,4), x), listput(v, x^4+y^4))); Set(v) \\ Charles R Greathouse IV, Apr 24 2012; updated July 13 2024

T=thueinit('x^4+1,1);
is(n)=#thue(T,n)>0 && !issquare(n) \\ Charles R Greathouse IV, Feb 26 2017

def aupto(lim):
  p1 = set(i**4 for i in range(1, int(lim**.25)+2) if i**4 <= lim)
  p2 = set(a+b for a in p1 for b in p1 if a+b <= lim)
  return sorted(p2)
print(aupto(10625)) # Michael S. Branicky, Mar 18 2021

{i: A216284(i) > 0}. - R. J. Mathar, Jun 04 2021

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