A027604 -id:A027604 - cp's OEIS frontend

A217843 Numbers which are the sum of one or more consecutive nonnegative cubes.

0, 1, 8, 9, 27, 35, 36, 64, 91, 99, 100, 125, 189, 216, 224, 225, 341, 343, 405, 432, 440, 441, 512, 559, 684, 729, 748, 775, 783, 784, 855, 1000, 1071, 1196, 1241, 1260, 1287, 1295, 1296, 1331, 1584, 1728, 1729, 1800, 1925, 1989, 2016, 2024, 2025, 2197

Offset: 1

T. D. Noe, Oct 23 2012

nonn

Contains A000578 (cubes), A005898 (two consecutive cubes), A027602 (three consecutive cubes), A027603 (four consecutive cubes) etc. - R. J. Mathar, Nov 04 2012
See A265845 for sums of consecutive positive cubes in more than one way. - Reinhard Zumkeller, Dec 17 2015
From Lamine Ngom, Apr 15 2021: (Start)
a(n) can always be expressed as the difference of the squares of two triangular numbers (A000217).
A168566 is the subsequence A000217(n)^2 - 1.
a(n) is also the product of two nonnegative integers whose sum and difference are both promic.
See example and formula sections for details. (End)

			From _Lamine Ngom_, Apr 15 2021: (Start)
Arrange the positive terms in a triangle as follows:
n\k |   1    2    3    4    5    6    7
----+-----------------------------------
  0 |   1;
  1 |   8,   9;
  2 |  27,  35,  36;
  3 |  64,  91,  99, 100;
  4 | 125, 189, 216, 224, 225;
  5 | 216, 341, 405, 432, 440, 441;
  6 | 343, 559, 684, 748, 775, 783, 784;
Column 1: cubes = A000217(n+1)^2 - A000217(n)^2.
The difference of the squares of two consecutive triangular numbers (A000217) is a cube (A000578).
Column 2: sums of 2 consecutive cubes (A027602).
Column 3: sums of 3 consecutive cubes (A027603).
etc.
Column k: sums of k consecutive cubes.
Row n: A000217(n)^2 - A000217(m)^2, m < n.
T(n,n) = A000217(n)^2 (main diagonal).
T(n,n-1) = A000217(n)^2 - 1 (A168566) (2nd diagonal).
Now rectangularize this triangle as follows:
n\k |   1    2     3     4    5     6   ...
----+--------------------------------------
  0 |   1,   9,   36,  100,  225,  441, ...
  1 |   8,  35,   99,  224,  440,  783, ...
  2 |  27,  91,  216,  432,  775, 1287, ...
  3 |  64, 189,  405,  748, 1260, 1989, ...
  4 | 125, 341,  684, 1196, 1925, 2925, ...
  5 | 216, 559, 1071, 1800, 2800, 4131, ...
  6 | 343, 855, 1584, 2584, 3915, 5643, ...
The general form of terms is:
T(n,k) = [n^4 + A016825(k)*n^3 + A003154(k)*n^2 + A300758(k)*n]/4, sum of n consecutive cubes after k^3.
This expression can be factorized into [n*(n + A005408(k))*(n*(n + A005408(k)) + 4*A000217(k))]/4.
For k = 1, the sequence provides all cubes: T(n,1) = A000578(k).
For k = 2, T(n,2) = A005898(k), centered cube numbers, sum of two consecutive cubes.
For k = 3, T(n,3) = A027602(k), sum of three consecutive cubes.
For k = 4, T(n,4) = A027603(k), sum of four consecutive cubes.
For k = 5, T(n,5) = A027604(k), sum of five consecutive cubes.
T(n,n) = A116149(n), sum of n consecutive cubes after n^3 (main diagonal).
For n = 0, we obtain the subsequence T(0,k) = A000217(n)^2, product of two numbers whose difference is 0*1 (promic) and sum is promic too.
For n = 1, we obtain the subsequence T(1,k) = A168566(x), product of two numbers whose difference is 1*2 (promic) and sum is promic too.
For n = 2, we obtain the subsequence T(2,k) = product of two numbers whose difference is 2*3 (promic) and sum is promic too.
etc.
For n = x, we obtain the subsequence formed by products of two numbers whose difference is the promic x*(x+1) and sum is promic too.
Consequently, if m is in the sequence, then m can be expressed as the product of two nonnegative integers whose sum and difference are both promic. (End)

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A034705, A217844-A217850, A062682, A131643, A240137.
Cf. A000578, A005898, A027602, A027603, A027604.
Cf. A265845 (subsequence).
Cf. A000217 (triangular numbers), A046092 (4*A000217).
Cf. A168566 (A000217^2 - 1).
Cf. A002378 (promics), A016825 (singly even numbers), A003154 (stars numbers).
Cf. A000330 (square pyramidal numbers), A300758 (12*A000330).
Cf. A005408 (odd numbers).

import Data.Set (singleton, deleteFindMin, insert, Set)
a217843 n = a217843_list !! (n-1)
a217843_list = f (singleton (0, (0,0))) (-1) where
   f s z = if y /= z then y : f s'' y else f s'' y
              where s'' = (insert (y', (i, j')) $
                           insert (y' - i ^ 3 , (i + 1, j')) s')
                    y' = y + j' ^ 3; j' = j + 1
                    ((y, (i, j)), s') = deleteFindMin s
-- Reinhard Zumkeller, Dec 17 2015, May 12 2015

nMax = 3000; t = {0}; Do[k = n; s = 0; While[s = s + k^3; s <= nMax, AppendTo[t, s]; k++], {n, nMax^(1/3)}]; t = Union[t]

lista(nn) = {my(list = List([0])); for (i=1, nn, my(s = 0); forstep(j=i, 1, -1, s += j^3; if (s > nn^3, break); listput(list, s););); Set(list);} \\ Michel Marcus, Nov 13 2020

a(n) >> n^2. Probably a(n) ~ kn^2 for some k but I cannot prove this. - Charles R Greathouse IV, Aug 07 2013

a(n) is of the form [x*(x+2*k+1)*(x*(x+2*k+1)+2*k*(k+1))]/4, sum of n consecutive cubes starting from (k+1)^3. - Lamine Ngom, Apr 15 2021

Name edited by N. J. A. Sloane, May 24 2021

A027602 a(n) = n^3 + (n+1)^3 + (n+2)^3.

Original entry on oeis.org

9, 36, 99, 216, 405, 684, 1071, 1584, 2241, 3060, 4059, 5256, 6669, 8316, 10215, 12384, 14841, 17604, 20691, 24120, 27909, 32076, 36639, 41616, 47025, 52884, 59211, 66024, 73341, 81180, 89559, 98496, 108009, 118116, 128835, 140184

Offset: 0

Patrick De Geest

a(3) = 216 = 6^3 (a cube). - Howard Berman (howard_berman(AT)hotmail.com), Nov 07 2008
Pairs [n,a(n)] for n<=10^7 such that a(n) is a perfect power are [0, 9], [1, 36], [3, 216], [23, 41616]. - Joerg Arndt, Jan 25 2011
Sums of three consecutive cubes. - Al Hakanson (hawkuu(AT)gmail.com), May 20 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..750
Patrick De Geest, Palindromic Sums of Cubes of Consecutive Integers
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000537, A003215, A006527, A008585, A059100.

Cf. A000578, A005898, A027603, A027604.

[3*n^3 + 9*n^2 + 15*n + 9: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

f[n_]:=n^3; Table[f[n]+f[n+1]+f[n+2], {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Jan 03 2009 *)
Table[3n^3+9n^2+15n+9,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{9,36,99,216},40] (* Harvey P. Dale, Nov 27 2024 *)

a(n)=3*(n^3 + 3*n^2 + 5*n + 3) \\ Charles R Greathouse IV, Jun 11 2015

A027602_list, m = [], [18, 0, 9, 9]
for _ in range(10**2):
    A027602_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

[i^3+(i+1)^3+(i+2)^3 for i in range(0,48)] # Zerinvary Lajos, Jul 03 2008

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - 1*a(n-4) for n>=4.
a(n) = 9*A006527(n+1). - Lekraj Beedassy, Feb 01 2007
a(n) = 3*n^3 + 9*n^2 + 15*n + 9.
G.f.: 9*(1+x^2)/(1-x)^4. - Bruno Berselli, Jan 21 2011
a(n) = A008585(n+1)*A059100(n+1). - Bruno Berselli, Jan 21 2011
E.g.f.: 3*(3 + 9*x + 6*x^2 + x^3)*exp(x). - G. C. Greubel, Aug 24 2022
Sum_{n>=0} 1/a(n) = (2*gamma + polygamma(0, 1-i*sqrt(2)) + polygamma(0, 1+i*sqrt(2)))/12 = 0.161383557127191633050394086192620963436504... where i denotes the imaginary unit. - Stefano Spezia, Aug 31 2023

A126203 Middle number of a set of 5 consecutive integers whose sum of cubes is a square.

Original entry on oeis.org

0, 2, 3, 27, 98, 120

Offset: 1

Nick Hobson, Mar 10 2007

That is, numbers n >= 0 such that 5n(n^2+6) is a square.
According to Dickson, Lucas stated that the only terms are 2, 3, 98 and 120 (missing 27).
The Mordell reference shows that there are only finitely many solutions. - Allan Wilks, Mar 10 2007
From Max Alekseyev, Mar 10 2007: (Start)
It can be shown that all such numbers n can be obtained from elements that are perfect squares in the following 3 recurrent sequence:
1) x(0)=0, x(1)=4, x(k+1) = 98*x(k) - x(k-1). If x(k) is a square then n = 30*x(k). In particular: for k=0, we have n=30*x(0)=0, for k=1, we have n=30*x(1)=120.
2) x(0)=1, x(1)=49, x(k+1) = 38*x(k) - x(k-1). If x(k) is a square then n = 2*x(k). In particular: for k=1, we have n=2*x(1)=2, for k=2, we have n=2*x(1)=98.
3) x(0)=1, x(1)=9, x(k+1) = 8*x(k) - x(k-1). If x(k) is a square then n = 3*x(k). In particular: for k=1, we have n=3*x(1)=3, for k=2, we have n=3*x(1)=27.
It also follows that for any such n one of n/2, n/3, or n/30 is a perfect square. I have tested 10^5 terms of each of the recurrent sequences above and found no new perfect squares. (End)
From Warut Roonguthai, Apr 28 2007: (Start)
The sequence is finite because the number of integral points on an elliptic curve is finite; in this case the curve is m^2 = 5n^3 + 30n. Multiplying the equation by 25 and letting y = 5m and x = 5n, we have y^2 = x^3 + 150x. According to Magma, the integral points on this curve are (x, y) = (0, 0), (10, 50), (15, 75), (24, 132), (135, 1575), (490, 10850), (600, 14700). So the list is complete. (End)
This was also confirmed (using Sage) by Jaap Spies, May 27 2007
About the second comment: Lucas, however, did actually include 27 in his original note, which can be seen at the link cited. The mistake appears to have originated with Dickson. - Matt Westwood, Mar 05 2022

L. E. Dickson, History of the Theory of Numbers, Volume 2, Chapter 21, page 587.
L. J. Mordell, Diophantine Equations, Ac. Press; see Th. 1, Chap. 27, p. 255.

L. E. Dickson, History of the Theory of Numbers, Volume 2, Chapter 21, page 587.
Edouard Lucas, Recherches sur l'analyse indéterminée, Bull. Soc. d'Emulation du Département de l'Allier, 12, 1873, 532.

Cf. A000290, A027604.

q:= n-> issqr((n^2+6)*5*n):
select(q, [$0..150])[];  # Alois P. Heinz, Mar 08 2022

Select[Partition[Range[-5,130],5,1],IntegerQ[Sqrt[Total[#^3]]]&][[All,3]] (* Harvey P. Dale, Jan 31 2017 *)

for(n=1, 10^8, if(issquare(5*n*(n*n+6)), print(n)))

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

Zak Seidov, Apr 14 2007

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

Cf. A024670, A027602, A027604, A116108.

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 *)

A027603 a(n) = n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3.

Original entry on oeis.org

36, 100, 224, 432, 748, 1196, 1800, 2584, 3572, 4788, 6256, 8000, 10044, 12412, 15128, 18216, 21700, 25604, 29952, 34768, 40076, 45900, 52264, 59192, 66708, 74836, 83600, 93024, 103132, 113948, 125496, 137800, 150884, 164772

Offset: 0

Patrick De Geest

Sums of four consecutive cubes. - Wesley Ivan Hurt, Dec 16 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Patrick De Geest, Palindromic Sums of Cubes of Consecutive Integers
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000578, A005898, A027602, A027604.

Cf. A016825, A027689, A173965.

[4*n^3 + 18*n^2 + 42*n + 36: n in [0..40]]; // Vincenzo Librandi, Jun 04 2011

A027603:=n->n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3: seq(A027603(n), n=0..50); # Wesley Ivan Hurt, Dec 16 2015

Table[n^3 +(n+1)^3 +(n+2)^3 +(n+3)^3, {n, 0, 33}] (* or *)
Table[Plus@@(Range[n, n + 3]^3), {n, 0, 33}] (* Alonso del Arte, Jan 24 2011 *)

Vec(-4*(-9+11*x-10*x^2+2*x^3)/(1-x)^4 + O(x^100)) \\ Altug Alkan, Dec 16 2015

A027603_list, m = [], [24, 12, 28, 36]
for _ in range(10**2):
    A027603_list.append(m[-1])
    for i in range(3):
        m[i+1] += m[i] # Chai Wah Wu, Dec 15 2015

[n^3+(n+1)^3+(n+2)^3+(n+3)^3 for n in range(0,40)] # Zerinvary Lajos, Jul 03 2008

a(n) = 4*a(n-1) -6*a(n-2) +4*a(n-3) -a(n-4) for n>=4.
a(n) = 4*n^3 + 18*n^2 + 42*n + 36 = 4*A173965(n+2).
From Bruno Berselli, Jan 24 2011: (Start)
G.f.: 4*(9 - 11*x + 10*x^2 - 2*x^3)/(1-x)^4.
a(n) = A027689(n+1) * A016825(n+1). (End)
E.g.f.: 2*(18 + 32*x + 15*x^2 + 2*x^3)*exp(x). - G. C. Greubel, Aug 24 2022

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A217843 Numbers which are the sum of one or more consecutive nonnegative cubes.

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A027602 a(n) = n^3 + (n+1)^3 + (n+2)^3.

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A126203 Middle number of a set of 5 consecutive integers whose sum of cubes is a square.

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

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A027603 a(n) = n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3.

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