ExplorationsMoessner’s sieve turns the counting numbers into squares, cubes and higher powers

Strike Out, Add Up, and the Powers Appear

Exhaustively checkedsequencespowerspartial sumscurious

Write 1, 2, 3, 4, 5, 6, 7 and so on. Strike out every third number and add up what is left, keeping a running total: 1, 3, 7, 12, 19. Now strike out every second number of that row and add up again, which gives 1, 8, 27, 64. These are the cubes. Alfred Moessner noticed this in 1951 and Oskar Perron proved it the same year. Strike every second number once and you get the squares; strike every fourth, then third, then second and you get fourth powers. Change the order of the intervals and the pattern is gone.

Current view: Strike and sum

Follow each pass: which numbers are struck, and what the survivors add up to.

Moessner's sieve for power 2Intervals: 2

Start with the natural numbers

The natural numbers. Next pass: strike every second number.

Every row of the sieve, as text

Start: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
After pass 1 (strike every second number): 1, 4, 9, 16, 25, 36

a number: an outlined tile holding the number at that position of the row
a struck number: a tile with a vermilion line through it, left out of the sum
a sum: a tile in the lower row: the running total of the numbers that were not struck

Each tile is a number in a row. In a pass, the tiles with a vermilion line are struck and left out. The lower row is the running total of the tiles that remain: its first tile is the first survivor, its second tile is the first two survivors added, and so on. Tiles with a dashed outline in the last row are spare numbers beyond the terms being read.

The natural numbers. Next pass: strike every second number.

Passes done: 0 of 1
Numbers in this row: 11
Last number in this row: 11

How this page is laid out

Make a prediction

Change the assumption

What this assumes: Striking numbers out and adding up what is left always turns the counting numbers into powers.
What breaks it: The same intervals are used in the other order, every second number and then every third. The final row begins 1, 5, 21, 46 and is not a power sequence.

Where it holds

The intervals counting down from the power to 2, in that order: for example every third and then every second number gives the cubes.

The rule

Strike intervals

PowerStrike every nth number first, then every (n minus 1)th, down to every second.Terms to readHow many numbers of the final row the page reads off and checks.

Step by step

Run, Step, Reset and Show all passes are under the sieve. Speed sets how fast Run goes.

Speed

Cases worth seeing

Try this

Strike every second number and add up what is left, by hand, for the first five numbers of the result. What do the numbers you add have in common, and why does each total come out as a square?

1/3

What is going on

Each explanation begins with a worked example and follows the same operation through intuition, formal statements and proofs. Later sections distinguish what is established from questions that remain open, so you can follow the level of detail useful to you.

Worked by handThe cubes by hand. Start from 1, 2, 3, 4, 5, 6, 7, strike every third number and add up what is left, then strike every second number of the new row and add up again. Every number can be checked with a pencil.

1, 2, 3, 4, 5, 6, 7strike 3 and 6; add up 1, then 1 + 2, then 1 + 2 + 4, then 1 + 2 + 4 + 5, then 1 + 2 + 4 + 5 + 71, 3, 7, 12, 19The struck numbers are skipped, and each remaining number is added to the total so far.
1, 3, 7, 12, 19strike 3 and 12; add up 1, then 1 + 7, then 1 + 7 + 191, 8, 27The same two moves on the new row, now striking every 2nd number. The result is 1 × 1 × 1, 2 × 2 × 2 and 3 × 3 × 3.

The laboratory does exactly this for any power from 2 to 6, and shows the struck numbers with a line through them. Step to watch each pass, or show all the passes at once.

Nothing in either pass mentions cubes. Each pass only strikes numbers and adds, yet the last row is 1, 8, 27, which are 1, 2 and 3 multiplied by themselves three times.

What you are seeing

Two moves, repeated: strike, then add up

Begin with the counting numbers 1, 2, 3, 4, and so on. The first move is to strike out numbers at a fixed interval: every third number, say, which removes 3, 6, 9 and so on. The second move is to add up what is left, keeping a running total. The first total is the first survivor, the second is the first two survivors added together, and so on. The totals are a new row of numbers.

For Moessner’s theorem the two moves are repeated with a shorter interval each time. To get the cubes you strike every third number and add up, then strike every second number of the new row and add up again. Squares need one pass, striking every second number. Fourth powers need three: every fourth, then every third, then every second.

The tiles in the laboratory are the numbers of a row. Struck tiles carry a line through them and are left out of the sum. The lower row is the running total of the survivors. Tiles with a dashed outline in the final row are spare numbers beyond the terms being read; the page starts from just enough numbers that the terms you asked for can be worked out.

Why it starts to make sense

One pass is a sum of odd numbers

Take the simplest case. Strike every second number and the survivors are the odd numbers 1, 3, 5, 7, and so on. The running totals are 1, then 1 + 3 = 4, then 1 + 3 + 5 = 9, then 16. These are the squares, and the reason is a classic fact: the sum of the first n odd numbers is n squared.

For the cubes the same thing happens in two stages. The first pass gives 1, 3, 7, 12, 19, 27. The second pass strikes the numbers in the even positions (3, 12, 27) and adds up what remains: 1, then 1 + 7 = 8, then 1 + 7 + 19 = 27. The totals have jumped to the cubes, but only because the first pass left exactly the right numbers behind.

The key idea

The whole list of intervals does the work

The counting numbers are the first powers: n to the power one. The rows between the first pass and the last are not powers at all. For the cubes the first pass gives 1, 3, 7, 12, 19, 27, which no one would call a pattern, and only the final pass turns the row into 1, 8, 27. That is why a table of the rows looks arbitrary until the end. The page offers a text version of every row so the whole sequence can be read in order.

What matters is the list of intervals as a whole, not any single pass. Counting down from the power to 2 is what Moessner’s theorem names. The same numbers in a different order, or the same interval repeated, give sequences that are not powers; the boundary card on this page shows two of them.

Formal statement

A theorem about a list of intervals

Fix a whole number $k \geq 1$ . Start from the sequence $1, 2, 3, \dots$ . For $j = 1, \dots, k - 1$ , strike every $(k + 1 - j)$ th term of the current sequence and replace what remains by its sequence of partial sums. Moessner’s theorem says the final sequence is

1^{k}, 2^{k}, 3^{k}, \dots

The first pass strikes every $k$ th term and the last strikes every second term. For $k = 1$ there are no passes and the counting numbers are the first powers.

The laboratory runs the passes in exact whole-number arithmetic with no rounding. It reads off a stated number of terms and compares them with $n^{k}$ computed by multiplication.

What is proved, and where

The squares here; the general theorem by citation

The case of squares is proved on this page. The survivors of one pass are the odd numbers, and the sum of the first n odd numbers is $\sum_{m = 1}^{n} (2 m - 1) = n (n + 1) - n = n^{2}$ .

For general $k$ the page relies on the published theorem. Moessner stated the construction in 1951 and Perron proved it that year. Kozen and Silva (2013) prove a more general theorem that contains it. The citations are recorded here as unresolved, which means their bibliographic details could not be confirmed when the page was written. The page also checks the statement in a finite range: every power from 1 to 8 and each of the first 100 terms, with a second implementation of the sieve alongside the first. That check is evidence for the cited proof being applied correctly, not a replacement for it.

Limits and further questions

Other intervals, and other sequences

Changing the intervals changes the answer. Striking every second number and then every third gives 1, 5, 21, 46, while the other order gives the cubes. The page has checked that none of its four non-Moessner interval lists gives a power sequence in its first twelve terms. That is a statement about those four lists, not about every other list of intervals.

Generalisations by Paasche and by Long, treated algebraically by Kozen and Silva, start from sequences other than the counting numbers or use other deletion patterns. The sources also report a related procedure that removes the triangular numbers in the first pass and leads to the factorials. This page does not implement that variant and does not claim it: its exact rules were not confirmed here, so it is a pointer to the literature rather than a result of the laboratory.

What is actually established

Every statement on this page, with its status, its exact scope, and the date that status was last checked.

Proved

Striking every second number of 1, 2, 3, ... and adding up what remains gives the squares 1, 4, 9, 16, ... .

Scope: Every term. This is the case of a single pass and is the only case the page proves itself.
Why: The survivors are the odd numbers 1, 3, 5, ... and the nth running total is the sum of the first n odd numbers. Each odd number is 2m − 1 for some m, so the sum of the first n is n(n + 1) − n = n². The kernel also checks this exactly for the first hundred terms.
Status checked: 17 October 2026

Proved

For every power k, striking every kth number of 1, 2, 3, ... and adding up, then every (k − 1)th number of the result and adding up, and so on down to every second number, gives the numbers 1, 2ᵏ, 3ᵏ, 4ᵏ, ... .

Scope: Every whole number k from 1 up and every term of the final row, as a published theorem. The page does not reproduce the proof; the computation below checks the statement in a finite range and is not a proof of it.
Why: Moessner stated the construction in 1951 and Perron proved it that year. Kozen and Silva give a short algebraic proof of a more general theorem that contains it. The citations are recorded as unresolved: the records could not be retrieved when this page was written.
Status checked: 17 October 2026
Sources: Eine Bemerkung über die Potenzen der natürlichen Zahlen citation unresolved; Beweis des Moessnerschen Satzes citation unresolved; On Moessner’s Theorem citation unresolved

Exhaustively checked

The laboratory’s sieve gives exactly n to the power k for every power k from 1 to 8 and every one of the first 100 terms.

Scope: Powers 1 to 8 and positions n from 1 to 100 only. It says nothing about larger powers or later terms, which the cited theorem covers.
Why: The unit tests run every power from 1 to 8 over all 100 terms. The last term of the eighth power is 10¹⁶, which is beyond the range where ordinary floating-point numbers are exact. A second implementation that never builds the list of survivors agrees with the first on every interval list tried.
Computation (exact): Exact BigInt arithmetic throughout, run with the kernel’s intervals k, k − 1, ..., 2 on the counting numbers, and compared term by term with n multiplied by itself k times, and with an independent second implementation of the sieve. Algorithm moessner-sieve-1.
Status checked: 17 October 2026