There are two approaches to definition of sieved series accumulating to powers:  Positive then Negative

See also the method by which numbers accumulate.

Pascal's Triangle's row n is implicated for the nth power.  Deletions may be a rather concise definition of what Pascal's Triangle is about, in relation to power values.

Sieved accumulation to powers—definition by positive statements

Threads of series internal to each level of accumulation in the sieved accumulations to power values, are not "deleted" and accumulate.  Here is the fifth power as an example.  Go to a layout of the 5th power's accumulation for reference.  Any k in the below formulas will be from 1 to x in value—in order to fully describe the basis of x^5.

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FIRSTLY, RECURRENTLY SUM ONES,

THEN KEEP THE VALUES (in integer order) THAT SOLVE EQUATIONS A) through D)

A) 5k – 4, B) 5k – 3, C) 5k – 2, D) 5k – 1; DROP VALUES 5k – 5 [or 5(k - 1)]     

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RECURRENTLY SUM THE PREVIOUS VALUES,

THEN KEEP VALUES (in integer order) THAT SOLVE EQUATIONS A) through C)

A) 10k^2 – (5k – 1), B) 10k^2 – (5k – 1) – (5k – 2), C) 10k^2 – (5k – 1) – (5k – 2) – (5k – 3);

DROP VALUES 10k^2 – (5k – 1) – (5k – 2) – (5k – 3) – (5k – 4) [or 10(k^2) - 10(2k - 1)) which is equivalent to 10(k – 1)^2 since ((k^2) - (2k - 1)) defines a square value just previous to the kth.]

Or

A) 10k^2 – 5k + 1, B) 10k^2 – 10k + 3, C) 10k^2 – 15k + 6       

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RECURRENTLY SUM THE PREVIOUS,

THEN KEEP VALUES (in integer order) THAT SOLVE EQUATIONS A) through B)

A) 10k^3 – (10k^2 – 5k + 1), B) 10k^3 – (10k^2 – 5k + 1) – (10k^2 – 10k + 3);

DROP VALUES 10k^3 – (10k^2 – 5k + 1) – (10k^2 – 10k + 3) – (10k^2 – 15k + 6); [or 10(k^3 - (3k^2 - 3k + 1)) which is equivalent to 10(k – 1)^3 since k^3 - (3k^2 - 3k + 1) defines the cube value just previous to the kth.]

Or

A) 10k^3 – 10k^2 + 5k – 1, B) 10k^3 – 20k^2 + 15k – 4       

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RECURRENTLY SUM THE PREVIOUS,

THEN KEEP VALUES (in integer order) THAT SOLVE EQUATIONS A)

A) 5k^4 – (10k^3 – 10k^2 + 5k – 1);

DROP VALUES 5k^4 – (10k^3 – 10k^2 + 5k – 1) – (10k^3 – 20k^2 + 15k – 4) [or 5(k^4 - (4k^3 - 6k^2 + 4k - 1)) which is equivalent to 5(k – 1)^4]

Or

A) 5k^4 – 10k^3 + 10k^2 – 5k + 1       

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RECURRENTLY SUM THE ABOVE SHELL/NEXUS NUMBERS FOR VALUES OF THE POWER OF 5

x^5 values result

WHERE k^5 - (5k^4 – 10k^3 + 10k^2 – 5k + 1) = (k-1)^5 or the fifth power value just previous to k^5.  "Dropping" values from this sequence would leave 0's.

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See how shell/nexus number formulas are achieved logically and cumulatively following an initial decision to eliminate every 5th positive integer?  See how Pascal's Triangle is naturally (accumulatively) the template for generating series at a power's level?  A row of Pascal's Triangle (or binomial numbers) gives the variable term that is 1^n = 1 with the initial manifestation of coefficients for a power level.  The pattern that is the formula based on those coefficients carries forward across series as the variable x (no longer "1") is raised to a power according to the coefficient's implicated level of accumulation. 

Notice that descriptions of all of the "discarded" values ultimately match the shell/nexus numbers' descriptions of values—with the exception that the (x-1)th value rather than the xth value is given.

Sieved accumulation to powers—definition by negative statements

Here is the procedure of accumulation to powers by employing deletions within sequences of positive integer values.  The final "remove" leaves shell/difference series values which will, at last, accumulate to power series.

Beginning with the positive integers, from 1 to x:

For the 2nd power:  Remove every 2nd integer (that is, of 2k value, k=1 to x) add what is not removed cumulatively.

For the 3rd power:  Remove every 3rd value (that is, of 3k value, k=1 to x) add untouched integers cumulatively; next, remove every 2nd value (incidentally that is, of 3k^2 value, k=1 to x) add remaining values cumulatively.

For the 4th power:  Remove every 4th integer (that is, of 4k value, k=1 to x) add remaining integers cumulatively; next, remove every 3rd value (incidentally that is, of 6k^2 value, k=1 to x) add remaining values cumulatively; next, remove every 2nd value (incidentally that is, of 4k^3 value, k=1 to x) add remaining values cumulatively.

For the 5th power:  Remove every 5th integer (that is, of 5k value, k=1 to x) add remaining integers cumulatively; next, remove every 4th value (incidentally that is, of 10k^2 value, k=1 to x) add remaining values cumulatively; next, remove every 3rd value (incidentally that is, of 10k^3 value, k=1 to x) add remaining values cumulatively; next, remove every 2nd value (incidentally that is, of 5k^4 value, k=1 to x) add remaining values cumulatively.

For the 6th power:  Remove every 6th integer (that is of 6k value, k=1 to x) add remaining integers cumulatively; next, remove every 5th value (incidentally that is, of 15k^2 value, k=1 to x) add remaining values cumulatively; next, remove every 4th value (incidentally that is, of 20k^3 value, k=1 to x) add remaining values cumulatively; next, remove every 3rd value (incidentally that is, of 15k^4 value, k=1 to x) add remaining values cumulatively; next, remove every 2nd value (incidentally that is, of 6k^5 value, k=1 to x) add remaining values cumulatively.

Et cetera.

Here are some formulaic conclusions from the layouts of powers accumulating:

1      = (k+1)^0

1k + 1       = (k+1)^1

1k^2 + 2k + 1       = (k+1)^2

1k^3 + 3k^2 + 3k + 1       = (k+1)^3

1k^4 + 4k^3 + 6k^2 + 4k + 1       = (k+1)^4

1k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1       = (k+1)^5

1k^6 + 6k^5 + 15k^4 + 20k^3 + 15k^2 + 6k + 1       = (k+1)^6

etc.

            for all x,      1     = x^0

for k = 1 to x,        1      = x^1

for k = 1 to x,        2k – 1       = x^2

for k = 1 to x,        3k^2 – 3k + 1       = x^3

for k = 1 to x,        4k^3 – 6k^2 + 4k – 1       = x^4

for k = 1 to x,         5k^4 – 10k^3 + 10k^2 – 5k + 1    = x^5

for k = 1 to x,        6k^5 – 15k^4 + 20k^3 – 15k^2 + 6k – 1     = x^6

for k = 1 to x,         7k^6 – 21k^5 + 35k^4 – 35k^3 + 21k^2 – 7k + 1    = x^7

etc.

1       = (k-1)^0

 1k - (1)       = (k-1)^1

1k^2 – (2k – 1)       = (k-1)^2

1k^3 – (3k^2 – 3k + 1)       = (k-1)^3

1k^4 – (4k^3 – 6k^2 + 4k – 1)       = (k-1)^4

1k^5 – (5k^4 – 10k^3 + 10k^2 – 5k + 1)       = (k-1)^5

1k^6 – (6k^5 – 15k^4 + 20k^3 – 15k^2 + 6k – 1)       = (k-1)^6

etc.

0       = "Shell" difference formula S₀ for x^0

1       = "Shell" difference formula S₁ for x^1

2k – 1       = "Shell" difference formula S₂ for x^2

3k^2 – 3k + 1       = "Shell" difference formula S₃ for x^3

4k^3 – 6k^2 + 4k – 1       = "Shell" difference formula S₄ for x^4

5k^4 – 10k^3 + 10k^2 – 5k + 1       = "Shell" difference formula S₅ for x^5

6k^5 – 15k^4 + 20k^3 – 15k^2 + 6k – 1       = "Shell" difference formula S₆ for x^6

etc.

See the Euler/Pascal Cube for more about power series and the Pascal's Triangle about numbers that recur in these formulas.

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Last modified: 12/16/05