A027603 -id:A027603 - 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

A265845 Numbers that are sums of consecutive (positive) cubes in more than one way.

Original entry on oeis.org

216, 8000, 33075, 64000, 89559, 105525, 164800, 188784, 189189, 216000, 343000, 353241, 443456, 608391, 1271600, 2370816, 3132116, 3132675, 3184236, 5821200, 5832000, 9018000, 9769375, 11437525, 20793591, 22153600, 24359616, 28685440, 35937000, 47651373

Offset: 1

Reinhard Zumkeller, Dec 16 2015

nonn

A131643 (cubes that are also sums of three or more consecutive positive cubes) is a sparse subsequence: only 17 of its terms appear in the first 1000 terms of A265845. - Jonathan Sondow, Jan 10 2016

			a(1) = 216 = 6^3 = 3^3 + 4^3 + 5^3;
a(2) = 8000 = 20^3 = 11^3 + 12^3 + 13^3 + 14^3;
a(3) = 33075 = 11^3 + 12^3 + 13^3 + 14^3 + 15^3 + 16^3 + 17^3 + 18^3 + 19^3 = 15^3 + 16^3 + 17^3 + 18^3 + 19^3 + 20^3.

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Subsequence of A217843; subsequences: A000578, A005898, A027602, A027603, A062682.

Supersequence of A131643.

import Data.Set (singleton, deleteFindMin, insert, Set)
a265845 n = a265845_list !! (n-1)
a265845_list = f (singleton (1, (1, 1))) 0 0 where
   f s z z' = if y == z && z' /= z then y : f s'' y z else f s'' y z
              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

A027604 a(n) = n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3 + (n+4)^3.

Original entry on oeis.org

100, 225, 440, 775, 1260, 1925, 2800, 3915, 5300, 6985, 9000, 11375, 14140, 17325, 20960, 25075, 29700, 34865, 40600, 46935, 53900, 61525, 69840, 78875, 88660, 99225, 110600, 122815, 135900, 149885, 164800, 180675, 197540

Offset: 0

Patrick De Geest

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, A027603.

Cf. A008587, A114949.

[5*n^3+30*n^2+90*n+100: n in [0..40]]; // Vincenzo Librandi, Jun 04 2011

Table[n^3 +(n+1)^3 +(n+2)^3 +(n+3)^3 +(n+4)^3, {n, 0, 60}] (* Vladimir Joseph Stephan Orlovsky, Jun 27 2011 *)
Table[100+90n+30n^2+5n^3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{100,225,440,775},40] (* Harvey P. Dale, Dec 19 2022 *)
Total/@Partition[Range[0,40]^3,5,1] (* Harvey P. Dale, Mar 28 2025 *)

a(n)=5*((n+6)*n+18)*n+100 \\ Charles R Greathouse IV, Jun 28 2011

[n^3+(n+1)^3+(n+2)^3+(n+3)^3+(n+4)^3 for n in range(0,35)] # 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.
From Bruno Berselli, Jan 24 2011: (Start)
G.f.: 5*(20 - 35*x + 28*x^2 - 7*x^3)/(1-x)^4.
a(n) = 5*n^3 + 30*n^2 + 90*n + 100 = A008587(n+2)*A114949(n+2). (End)
E.g.f.: 5*(4+x)*(5+5*x+x^2)*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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A265845 Numbers that are sums of consecutive (positive) cubes in more than one way.

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A027604 a(n) = n^3 + (n+1)^3 + (n+2)^3 + (n+3)^3 + (n+4)^3.

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