Fibonacci parity pattern: why every third term is even

The Fibonacci parity pattern can be read without expanding a very large term. For the sequence defined here, the first two terms are 1 and 1, and every following term is the sum of the previous two. Its odd-even labels repeat odd, odd, even. The useful question is why that pattern continues, rather than merely whether the first few examples display it.

Keep the definition and position convention together. Call the first term F1 and the second F2, so F1 = 1 and F2 = 1. For each positive whole-number position n, define F(n + 2) = F(n) + F(n + 1). The conclusion below concerns those starting values and that indexing. No term is a product measurement, supplied worksheet entry or prediction about a real event.

Write a short beginning with positions attached

The first twelve values are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89 and 144. Their positions run from 1 through 12 in the same order. For example, the value at position 6 is 8; the value at position 8 is 21. Keeping position and value separate prevents a later divisibility test from being applied to the wrong number.

Adding the first two values gives the third, 2. Then 1 + 2 gives 3, and 2 + 3 gives 5. Continue with the two most recent values each time. The next sum after 55 and 89 is 144. This calculation establishes the displayed entries. It does not, by itself, establish a statement about every later position.

Label those twelve terms odd, odd, even, odd, odd, even, odd, odd, even, odd, odd, even. The even values occur at positions 3, 6, 9 and 12. That suggests every third term is even. We still need a reason covering positions beyond the written list, including ones whose full integer values we have not calculated.

Retain the information addition needs

An even integer has the form 2a for an integer a. An odd integer has the form 2a + 1. Adding two even integers gives 2(a + b), which is even. Adding an even and an odd integer gives 2(a + b) + 1, which is odd. Adding two odd integers gives 2(a + b + 1), which is even.

These calculations justify the three kinds of parity sum used here: even plus even is even, unlike parities sum to odd, and odd plus odd is even. The actual sizes of a and b do not enter those conclusions. We may retain the parity labels when the question asks only whether a recurrence term is odd or even.

One current label is not enough to determine the next label. The first term is odd and is followed by another odd term. The second term is also odd but is followed by an even term. A record saying only current term odd would omit the distinction needed for the next step. Keep two consecutive labels in their stated order.

Define the ordered pair state

At position n, let the state record the parities of F(n) and F(n + 1). Its first entry describes the term at the named position. Its second describes the following term. To advance one position, retain the old second entry as the new first, then use the sum of the old pair for the new second.

Begin at position 1. The pair of values is 1 and 1, so the state is odd–odd. Advance to the values 1 and 2 at position 2: odd–even. At position 3, the values 2 and 3 give even–odd. At position 4, the values 3 and 5 give odd–odd again.

The pair order matters. Odd–even and even–odd each contain one odd and one even entry, but they describe different positions. In the former, the current term is odd; in the latter, it is even. Treating the pair as an unordered collection would discard exactly the information needed to answer the current-term question.

Derive each transition rather than guessing its arrow

From odd–odd, the old second entry is odd and the sum is even. The next state is therefore odd–even. From odd–even, the old second entry is even and the sum is odd, giving even–odd. From even–odd, the old second entry is odd and the sum is odd, returning to odd–odd.

Thus the three reached states form the directed cycle odd–odd to odd–even to even–odd to odd–odd. Each arrow follows the same recurrence and the same parity-sum calculations. Once the first state returns, its next state is again determined. There is no new choice to make at the next repetition.

For completeness, the fourth possible pair is even–even. Its old second entry and its sum are both even, so it advances to even–even. That state exists in the transition rule but is not reached from our starting odd–odd pair. Listing all four possibilities prevents the three-state result from being misreported as a rule for every possible starting pair.

Explain why the cycle settles every later term

The state at position 1 is odd–odd. The three verified transitions take it through odd–even and even–odd and back to odd–odd at position 4. Applying the identical rule for three more steps repeats the same three states. Repeating that reasoning establishes the cycle for every following block, rather than relying on how many terms happen to fit on the page.

In each block of three positions, the first entries are odd, odd and even. The first even entry is at position 3, and the cycle puts the next at 6, then 9, and so on. Therefore F(n) is even exactly when n is divisible by 3. When n leaves remainder 1 or 2 on division by 3, the term is odd.

The word exactly includes both directions. A position divisible by 3 has the even–odd state and hence an even current term. A position not divisible by 3 has one of the two states whose first entry is odd. We have not merely shown that some every-third positions work while leaving other even positions unexplained.

Use large positions without replacing them by values

For position 2026, write 2026 = 3 × 675 + 1. It belongs to the first position of a three-position block, so F2026 is odd. We have not calculated its full value. The remainder selects a parity position; it is not itself the value of the term.

For position 2028, write 2028 = 3 × 676. The zero remainder selects the third position in our one-through-three cycle, whose current term is even. Thus F2028 is even. It would be a mistake to read zero remainder as a claim that the term equals zero.

Check the convention against a small case. Position 6 has zero remainder on division by 3, and its value is 8, which is even. The value 8 is not divisible by 3. The theorem tests the position to determine the value's parity; it does not claim that every even term must itself be a multiple of 3.

Keep starting values and labels with the result

Change the starting values to 2 and 2 while retaining the same sum rule. The first six values become 2, 2, 4, 6, 10 and 16. Their parity pair begins even–even and remains there. Every term is even. The recurrence wording alone therefore does not justify the every-third-only conclusion; the original starting values matter.

Likewise, relabelling the same original list 1, 1, 2, 3 and onward with first index 0 changes the index names attached to its entries. Under that relabelling, its even entries have indices 2, 5, 8 and onward. Keep this separate from any other definition that supplies a different term at index 0. A position statement must retain the convention used to prove it.

Keep nearby cycle questions separate

The repeated rule application guide compares first and second outputs of complete unary tables, asking whether a result stays fixed or a start returns. This recurrence instead derives an ordered two-term parity state and locates even terms. Neither of those table-wide two-application properties is the question answered here.

The last digit power-sum guide retains decimal ones digits under multiplication and combines selected cycle entries. Our sum recurrence retains two parity labels under addition. Its three-state result is not a four-position power cycle, and it supplies no final decimal digit beyond the odd-even information requested.

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The Unbranded Clear clipboard offer, SKU 17055816187 describes a fifteen-board pack with low-profile clips and assorted transparent colours. Its sole gallery shows coloured boards arranged in three rows of five. Keep the Clear name beside that wording and appearance; no colourless-only variant, independently tested transparency, delivered count or particular paper-fit result is established here.

Keep a possible purchase outside the recurrence

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