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The questions asks:

Define a formula to yield the Kth N for which there exists an integer X less than or equal to N for which the sum of the integers from 0 to X (inclusive) is equal to the sum of the integers from X to N (inclusive). Through a bit of algebra, we can paraphrase the question as: Find the Kth N for which N(N+1)/2 equal to some perfect square.

I have defined a recursive formula for this:

f(1) = 1

f(2) = 8

f(3) = 49

f(k) = 4f(k-2)(f(k-2) + 1) for k > 3

So the numbers are: 1,8,49,288,9800...

My question is if it is possible to find an "implicit" function for yielding N.

J. W. Tanner
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BlueInfinite1729
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1 Answers1

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The answer to the title question is $f(k)=\dfrac{(3-2\sqrt2)^k+(3+2\sqrt2)^k-2}4$.

For further information, see this related question.

This sequence is in OEIS and includes $1681,$ which your formula doesn't have.

J. W. Tanner
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    Thank you for your response. I needed an implicit formula only. I understand there was some flaw in my reasoning which caused it to jump over numbers such as 1681. – BlueInfinite1729 Feb 11 '24 at 16:32