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
Keywords
Examples
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)
Links
- T. D. Noe, Table of n, a(n) for n = 1..1000
Programs
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Haskell
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 -
Mathematica
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] -
PARI
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
Formula
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
Extensions
Name edited by N. J. A. Sloane, May 24 2021
Comments