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EXAMPLE
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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)
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PROG
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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
(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
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