cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 41 results. Next

A141806 Terms of A024670 that are not in A141805.

Original entry on oeis.org

730, 737, 756, 793, 854, 945, 1072, 1241, 2060, 2457, 2926, 3473, 4825, 5642, 6561, 7588, 8729, 9990, 11377, 12896, 14553, 16354, 18305, 20412, 21953, 21960, 21979, 22016, 22077, 22168, 22295, 22464, 22681, 22952, 23283, 23680, 24149, 24696
Offset: 1

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Author

Klaus Brockhaus, Jul 16 2008

Keywords

Comments

A024670 gives the sums of cubes of two distinct positive integers. A141805, the complement of A031980, is a subsequence of A024670.

Examples

			1072 is the sum of two distinct nonzero cubes in exactly one way: 1072 = 7^3 + 9^3. 9 is not in A031980, so 1072 is not the sum of cubes of two distinct earlier terms of A031980 and hence 1072 is in A031980. Therefore 1072 is in not in A141805 and so a term of this sequence.
1729 is the sum of two distinct nonzero cubes in exactly two ways: 1729 = 9^3 + 10^3 = 1^3 + 12^3. 1 and 12 are in A031980, so 1729 is the sum of cubes of two distinct earlier terms of A031980 and hence 1729 is in not A031980. Therefore 1729 is in A141805 and so not a term of this sequence.
		

Crossrefs

Cf. A024670, A141805, A031980 (smallest number not occurring earlier and not the sum of cubes of two distinct earlier terms).

A024672 Positions of odd numbers in A024670 (distinct sums of cubes of distinct positive integers).

Original entry on oeis.org

1, 3, 4, 6, 8, 10, 11, 13, 15, 17, 19, 21, 23, 24, 26, 29, 31, 33, 34, 35, 37, 40, 42, 44, 45, 47, 50, 52, 54, 56, 57, 60, 62, 65, 67, 68, 69, 71, 73, 75, 77, 80, 81, 82, 85, 87, 88, 91, 94, 95, 96, 97, 99, 101, 103, 104, 108, 110, 112, 113, 116, 117, 119, 120, 124, 126, 128, 129, 131, 132
Offset: 1

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Keywords

A024673 Position of 1 + n^3 in A024670 (distinct sums of cubes of distinct positive integers).

Original entry on oeis.org

1, 2, 4, 7, 11, 16, 21, 28, 35, 43, 52, 61, 73, 84, 97, 109, 124, 140, 155, 173, 190, 210, 228, 250, 271, 292, 316, 338, 365, 391, 417, 444, 472, 501, 531, 560, 593, 626, 659, 694, 728, 765, 802, 842, 880, 919, 960, 1001, 1045, 1086, 1131, 1176, 1222, 1268, 1313
Offset: 2

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Author

Keywords

Programs

  • Maple
    with(priqueue):
    initialize(PQ):
    insert([-9,1,2],PQ): v:= 0:
    R:= NULL: count:= 0: i:= 0:
    while count < 100 do
      t:= extract(PQ);
      w:= -t[1];
      if floor(w) > v then i:= i+1; v:= floor(w)  fi;
      if t[2] = 1 then R:= R, i; count:= count+1 fi;
      wp:= w + (t[3]+1)^3 - t[3]^3;
      insert([-wp,t[2],t[3]+1], PQ);
      if t[3] = t[2]+1 then
         wp:= t[3]^3 + (t[3]+1)^3 + 1/10;
         insert([-wp, t[3], t[3]+1],PQ);
      fi
    od:
    R; # Robert Israel, Jan 21 2024

Extensions

Offset corrected by Sean A. Irvine, Jul 19 2019

A024671 Positions of even numbers in A024670 (distinct sums of cubes of distinct positive integers).

Original entry on oeis.org

2, 5, 7, 9, 12, 14, 16, 18, 20, 22, 25, 27, 28, 30, 32, 36, 38, 39, 41, 43, 46, 48, 49, 51, 53, 55, 58, 59, 61, 63, 64, 66, 70, 72, 74, 76, 78, 79, 83, 84, 86, 89, 90, 92, 93, 98, 100, 102, 105, 106, 107, 109, 111, 114, 115, 118, 121, 122, 123, 125, 127, 130, 133, 135, 138, 139, 140
Offset: 1

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Keywords

A024674 a(n) = position of n^3 + (n+1)^3 in A024670 (distinct sums of cubes of distinct positive integers).

Original entry on oeis.org

1, 3, 6, 10, 15, 24, 33, 42, 52, 67, 80, 95, 112, 131, 150, 170, 195, 220, 243, 268, 300, 328, 357, 388, 425, 459, 492, 528, 571, 612, 652, 692, 743, 788, 833, 879, 937, 985, 1035, 1092, 1150, 1206, 1262, 1324, 1388, 1448, 1507, 1579, 1647, 1712, 1777, 1857
Offset: 1

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Author

Keywords

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

			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
		

References

  • 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.

Crossrefs

(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).

Programs

  • Haskell
    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
    
  • Magma
    [ n^3 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 14 2014
    
  • Magma
    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
    
  • Maple
    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
  • Mathematica
    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 *)
  • Maxima
    A000578(n):=n^3$
    makelist(A000578(n),n,0,30); /* Martin Ettl, Nov 03 2012 */
    
  • PARI
    A000578(n)=n^3 \\ M. F. Hasler, Apr 12 2008
    
  • PARI
    is(n)=ispower(n,3) \\ Charles R Greathouse IV, Feb 20 2012
    
  • Python
    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
    
  • Scheme
    (define (A000578 n) (* n n n)) ;; Antti Karttunen, Oct 06 2017

Formula

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

Views

Author

Keywords

Comments

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

References

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

Crossrefs

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).

Programs

  • Haskell
    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
    
  • Mathematica
    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 *)
  • PARI
    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
    
  • PARI
    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
    
  • PARI
    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
    
  • PARI
    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
    
  • Python
    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

Extensions

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

A120398 Sums of two distinct prime cubes.

Original entry on oeis.org

35, 133, 152, 351, 370, 468, 1339, 1358, 1456, 1674, 2205, 2224, 2322, 2540, 3528, 4921, 4940, 5038, 5256, 6244, 6867, 6886, 6984, 7110, 7202, 8190, 9056, 11772, 12175, 12194, 12292, 12510, 13498, 14364, 17080, 19026, 24397, 24416, 24514
Offset: 1

Views

Author

Tanya Khovanova, Jul 24 2007

Keywords

Comments

If an element of this sequence is odd, it must be of the form a(n)=8+p^3, else it is a(n)=p^3+q^3 with two primes p>q>2. - M. F. Hasler, Apr 13 2008

Examples

			2^3+3^3=35=a(1), 2^3+5^3=133=a(2), 3^3+5^3=152=a(3), 2^3+7^3=351=a(4).
		

Crossrefs

Subsequence of A024670.

Programs

  • Mathematica
    Select[Sort[ Flatten[Table[Prime[n]^3 + Prime[k]^3, {n, 15}, {k, n - 1}]]], # <= Prime[15^3] &]
  • PARI
    isA030078(n)=n==round(sqrtn(n,3))^3 && isprime(round(sqrtn(n,3)))  \\ M. F. Hasler, Apr 13 2008
    
  • PARI
    isA120398(n)={ n%2 & return(isA030078(n-8)); n<35 & return; forprime( p=ceil( sqrtn( n\2+1,3)),sqrtn(n-26.5,3), isA030078(n-p^3) & return(1))} \\ M. F. Hasler, Apr 13 2008
    
  • PARI
    for( n=1,10^6, isA120398(n) & print1(n",")) \\ - M. F. Hasler, Apr 13 2008
    
  • PARI
    list(lim)=my(v=List()); lim\=1; forprime(q=3,sqrtnint(lim-8,3), my(q3=q^3); forprime(p=2,min(sqrtnint(lim-q3,3),q-1), listput(v,p^3+q3))); Set(v) \\ Charles R Greathouse IV, Mar 31 2022

Formula

A120398 = (A030078 + A030078) - 2*A030078 = 8+(A030078\{8}) U { A030078(m)+A030078(n) ; 1M. F. Hasler, Apr 13 2008

A116145 Squares that are equal to the sum of 5 consecutive cubes.

Original entry on oeis.org

0, 100, 225, 99225, 4708900, 8643600
Offset: 1

Views

Author

Zak Seidov, Apr 14 2007

Keywords

Comments

m^3+(m+1)^3+(m+2)^3+(m+3)^3+(m+4)^3=5*(2+m)*(10+4*m+m^2). Corresponding values of m are -2,0,1,25,96,118.
Using the theory of elliptic curves one can show that there are no other terms. - Jaap Spies, May 28 2007

Crossrefs

Programs

  • Mathematica
    Select[Table[n^3 +(n+1)^3 +(n+2)^3 +(n+3)^3 +(n+4)^3, {n, -2, 1000}] ,IntegerQ[Sqrt[#]]&] (* James C. McMahon, Aug 18 2024 *)

A155468 Numbers that are sums of 8th powers of 2 distinct positive integers.

Original entry on oeis.org

257, 6562, 6817, 65537, 65792, 72097, 390626, 390881, 397186, 456161, 1679617, 1679872, 1686177, 1745152, 2070241, 5764802, 5765057, 5771362, 5830337, 6155426, 7444417, 16777217, 16777472, 16783777, 16842752, 17167841, 18456832, 22542017, 43046722, 43046977, 43053282
Offset: 1

Views

Author

Keywords

Examples

			1^8 + 2^8 = 257, 1^8 + 3^8 = 6562, 2^8 + 3^8 = 6817, ...
		

Crossrefs

Cf. A003380, A088719 (distinct 7th), A088677 (distinct 6th), A088703, A088687, A024670 (distinct 3rd), A004431 (distinct 2nd).

Programs

  • Mathematica
    lst={};e=8;Do[Do[x=a^e;Do[y=b^e;If[x+y==n,Print[n,",",Date[]];AppendTo[lst,n]],{b,Floor[(n-x)^(1/e)],a+1,-1}],{a,Floor[n^(1/e)],1,-1}],{n,4*8!}];lst
  • PARI
    list(lim)=my(v=List(),t); lim\=1; for(m=2,sqrtnint(lim-1,8), t=m^8; for(n=1,min(sqrtnint(lim-t,8),m-1), listput(v,t+n^8))); Set(v) \\ Charles R Greathouse IV, Nov 05 2017

Extensions

8 more terms. - R. J. Mathar, Sep 07 2017
More terms from Chai Wah Wu, Nov 05 2017
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