Repetend Rigidity at the Golden Scale

Divide 1 by 5 in base three.

1/5 => 0.|0121|   (base 3)

Base three uses the digits 0, 1, and 2. The long division works as usual. Multiply the remainder by the base, write down the quotient digit, and carry the new remainder into the next step. The calculations in this table are written in decimal.

Remainder	Multiply by 3 and divide by 5	Digit	New remainder
1	3 = 0 × 5 + 3	0	3
3	9 = 1 × 5 + 4	1	4
4	12 = 2 × 5 + 2	2	2
2	6 = 1 × 5 + 1	1	1

The remainder is back at 1. Everything from here repeats. It took four steps to return, so the shortest repeating block has four digits.

Change the base and the cycle can change. For division by five, only the base’s remainder after division by five matters. Here are all four possibilities for bases that are not multiples of five.

Base modulo 5	Remainders starting at 1, through the first return	Period
1	1 → 1	1
2	1 → 2 → 4 → 3 → 1	4
3	1 → 3 → 4 → 2 → 1	4
4	1 → 4 → 1	2

One, four, four, two. There is no three-digit cycle.

The golden-ratio calculation I am interested in produces three as a candidate period here. This small table gives us a way to test it.

Back to twelve

The question began with a shorter tail, in ordinary decimal.

1/12 => 0.08|333|

After 08, a single 3 repeats forever. I call an expansion with a nonrepeating prefix followed by a repeating tail a mixed cycle.

The three copies inside the bars come from my display convention. I repeat whole copies of the shortest block until their digit sum is divisible by one less than the base. In decimal that means nine. Here $3+3+3=9$, so I write |333|. The shortest repeating block is still one digit long. The prefix stays outside the sum.

Now write out all eleven fractions from $1/12$ through $11/12$. Advance each one past two decimal places. Their tails fall into three groups.

Numerators	Tail after two places	Number of rows
1, 4, 7, 10	repeating 3	4
2, 5, 8, 11	repeating 6	4
3, 6, 9	trailing zeros	3

Four rows match the reference tail. Giving the three terminating rows full credit as well makes the alignment score $7/11$. That credit is part of the definition of the score.

This is the count used in Three and the Golden Ratio. Replace twelve by $3m$, with $m$ built from factors of two and five, and the count becomes $m$ matching rows plus $m-1$ terminating rows, out of $3m-1$. The admissible scores lie on the curve

At $x=4$, this gives our $7/11$. We can also evaluate the curve between its admissible integer inputs. Write $\varphi=(1+\sqrt5)/2$ for the golden ratio. At its square, the curve takes the reciprocal value.

The crossing belongs to the curve. The eleven fractions themselves still have score $7/11$; there is no collection of rows at the irrational scale $\varphi^2$.

I wanted to test how much of that crossing depended on beginning with a one-digit tail. Keep $1/\varphi$ as the target and try the scales $\varphi,\varphi^2,\varphi^3,\ldots$. Allow other denominators, longer periods, and partial credit for terminating rows. These are chosen test points. The count does not require the golden ratio.

The same cycle, started elsewhere

Longer periods require a slightly different count. Look at two decimal sevenths.

1/7 => 0.|142857|
2/7 => 0.|285714|

Start the first block at its 2 and it becomes the second. The digits at corresponding positions differ, but the two rows follow the same cycle. I count both by accepting rotations of the reference tail.

That puts the period into the score. Each distinct rotation begins at a different remainder. If the period has $h$ digits, there are $h$ such starting points.

Write the denominator as $qm$. At twelve, $q=3$ and $m=4$. The factor $m$ must divide a power of the base, so advancing far enough clears it from the tail. The remaining denominator $q$ shares no factor with the base.

Among the $qm-1$ numerator rows, each of the $h$ accepted remainder classes occurs $m$ times. There are therefore $hm$ accepted rows, along with $m-1$ terminating rows. Give each terminating row a credit $\lambda$ between zero and one. The score is

Credit zero ignores terminating rows; credit one includes them fully. A rational credit can be implemented with ordinary copies of the rows. For half credit, for example, make two copies of every row and mark one copy of each terminating row, while marking both copies of every accepted rotating row.

We can extend each of these formulas to a curve just as we did at twelve. The question is whether any of them reach $1/\varphi$ at a positive integral power of $\varphi$, using a rational credit between zero and one. The length $h$ must be a period that long division can actually produce.

The missing three

The identity $\varphi^2=\varphi+1$ reduces every higher power to a whole number plus a whole-number multiple of $\varphi$. The first few are easy to work out.

The coefficients follow the Fibonacci sequence. This is classical arithmetic, described in Beard’s 1966 note.

Apply that reduction to the score equation. Because the credit is rational, the rational part and the coefficient of $\varphi$ must each vanish. They cannot cancel one another. The odd powers fail this test. At the fourth power, the two conditions reduce to

With zero credit for terminating rows, the calculation asks for denominator five and period three.

We have already listed every cycle modulo five. None has that period. The equations have a solution if $h$ is free to be any integer. Requiring it to be the length of an actual remainder cycle removes that solution.

The same obstruction extends to all the later even powers. Their Fibonacci coefficients require $q\geq5$ and put the period in one interval.

No remainder cycle starting at 1 can have a length in that interval. Its length must divide the number of invertible residues modulo $q$, which is less than $q$. A length at least $q/2$ would therefore have to use all of those residues.

For $q\geq5$, such a full cycle is possible only at an odd prime power or twice an odd prime power. In the first case, at least two thirds of the residues are invertible. In the second, fewer than half are. Neither fits the required interval. The paper gives the elementary proof.

Modulo five was the first instance of this gap. It is enough to rule out all the later even powers by the same argument.

Two cases at the square

Only the square remains. There the equations become

The denominator $q$ is an integer, and the credit lies between zero and one. That leaves two cases.

• At $q=2$, terminating rows receive no credit. The base must be odd.
• At $q=3$, terminating rows receive full credit. The base must leave remainder 1 on division by three, as bases 4, 7, and 10 do.

Both have a one-digit repeating tail. Across this family, allowing longer periods and intermediate rational credits adds no other solution to the golden-scale equation.

Decimal twelve belongs to the second case. We return to its single repeating 3, displayed as |333| under the cumulative-closure rule.

There is another way to see the three-digit closure. Write the numbers $1,11,111$. The first one divisible by three is $111$. In general, the closure length is the length of the shortest all-ones number, in the chosen base, divisible by the reduced tail denominator $q$. This connects the display rule to classical repunit arithmetic. Hasler gives the corresponding order formula in a 2025 SeqFan discussion; the paper proves the connection to whole-block digit sums.

At the two rational endpoints, the closure length is $q$. One case takes two copies of its repeating digit; the other takes three. These longer displayed groups leave the primitive period unchanged.

The line 0.08|333| already showed us the short tail. The new result is that, under the stated golden-scale condition with rational credits, allowing longer cycles always brings us back to a single repeating digit. The missing period modulo five is the first place we can see why.