u/Few-Act-2519

10890 or 9999 Are Two Interesting Numbers

found an interesting number pattern and I'm curious whether it's already known.

Take any 4-digit number.

Reverse its digits.

Subtract the smaller number from the larger one.

If the subtraction result is still a 4-digit number, reverse that result.

Add the result to its reverse.

Examples:

1234 → 4321 − 1234 = 3087 → 3087 + 7803 = 10890

6283 → 6283 − 3826 = 2457 → 2457 + 7542 = 9999

From all the examples I've tried, whenever the subtraction gives a 4-digit result, the final answer is always either 10890 or 9999.

Has anyone seen this pattern before? Is there a mathematical proof for it, or can someone find a counterexample?

I'd love to hear your thoughts!

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Hey everyone,

I’ve been playing around with infinite series rearrangements and managed to set up an algebraic chain that links a grouped series of natural numbers to ζ(0) and ζ(−1) to output a finite constant (9).

I uploaded my handwritten work in the image. Formally, the algebra feels like it clicks together perfectly step-by-step, but when plugging in the standard values ζ(−1)=−1/12 and ζ(0)=−1/2, the equation collapses into a contradiction.

I suspect it has to do with index shifts (n=0 vs n=1) or the rules of splitting divergent series (∞−∞). I’d love to get your insights on exactly where the standard rules of arithmetic forbid this kind of manipulation! Thanks!

Hi everyone,

While exploring infinite nested radicals, I noticed an interesting pattern that seems to hold for several integers, but I’m not fully sure if it is already known or how to rigorously prove it.

The expression looks like this:

3 = √(1 + 2√(1 + 3√(1 + 4√(1 + ...))))

7 = √(1 + 6√(1 + 7√(1 + 8√(1 + ...))))

28 = √(1 + 27√(1 + 28√(1 + 29√(1 + ...))))

From these examples, I observed the general form:

k = √(1 + (k−1)√(1 + k√(1 + (k+1)√(1 + (k+2)√(...)))))

So the coefficients increase consecutively:

(k−1), k, (k+1), (k+2), ...

After testing several values, it consistently appears to evaluate to k.

I tried a simple recursive argument:

Let

R(n) = √(1 + n√(1 + (n+1)√(1 + (n+2)√(...))))

If we assume R(n+1) = n+2, then:

R(n) = √(1 + n(n+2))

= √(n² + 2n + 1)

= n + 1

So it seems to satisfy the recursion perfectly.

Is this identity already known?

How would one rigorously prove convergence of this infinite radical?

Does this hold for non-integers as well?

Would appreciate any references or insights