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Those integers $\binom{2n}n\ (n=0,1,2,\ldots)$ are called central binomial coefficients. By Stirling's formula, $$\binom{2n}n\sim \frac{4^n}{\sqrt{n\pi}}\ \ \ \ (n\to+\infty).$$ Of course, the central binomial coefficients are much sparser than the squares of integers.

Motivated by Lagrange's four-square theorem, here I ask the following question.

QUESTION: Is every integer $n>1$ the sum of two squares and two central binomial coefficients?

I conjecture that the answer is affirmative. I have verified this for $n$ up to $6\times10^9$; for example, $$2435=32^2+33^2+\binom{2\times4}4+\binom{2\times5}5.$$ For the number of ways to write a positive integer $n$ as $a^2+b^2+\binom{2c}c+\binom{2d}d$ with $a,b,c,d\in\mathbb N=\{0,1,2,\ldots\}$, $a\leqslant b$ and $c\leqslant d$, one may consult http://oeis.org/A303540.

A set $A\subseteq \mathbb N$ is called an additive basis of order two if $\{a+b:\ a,b\in A\}=\mathbb N$. My above conjecture essentially says that $$\left\{k^2+\binom{2m}m-1:\ k,m=0,1,2,\ldots\right\}$$ is an additive basis of order two.

I have some similar questions. I conjecture that each integer $n>1$ can be written as the sum of two squares and two Catalan numbers, and that $$\left\{a^2+b^2+\binom{2k+1}k+\binom{2m+1}m:\ a,b,k,m\in\mathbb N\right\}=\{2,3,\ldots\}.$$ See http://oeis.org/A303543 and http://oeis.org/A303639 for related data.

My above question looks quite challenging. Any helpful ideas?

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  • $\begingroup$ Although it appears $A303540(n) = 1$ only for $n=2$ and $2435$, $A303540(n)=2$ for $n = 3, 5, 6, 10, 14, 15, 18, 19, 50, 59, 64, 114, 159, 789, 1819, 12238, 41928$, and $99635$. Are those all? Moreover, in all these cases $c$ and $d$ are quite small, e.g. $99635 = 209^2 + 236^2 + {4 \choose 2} + {10 \choose 5} = 105^2 + 296^2 + {8 \choose 4} + {12 \choose 6}$. $\endgroup$ – Robert Israel May 30 '18 at 20:05
  • $\begingroup$ Another one with $A303540(n)=2$ is $510568 = 114^2 +514^2 + {18 \choose 9} + {20 \choose 10} = 244^2 + 516^2 + {6 \choose 3} + {20 \choose 10}$. $\endgroup$ – Robert Israel May 30 '18 at 22:16
  • $\begingroup$ This is virtually a copy of mathoverflow.net/questions/300832/… $\endgroup$ – Mark Sapir May 30 '18 at 23:13
  • $\begingroup$ @Mark Sapir Unlike my question 300832, the current one involves a sparse additive basis of order two. $\endgroup$ – Zhi-Wei Sun May 30 '18 at 23:52
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    $\begingroup$ @Zhi-WeiSun: Why do you think your conjecture can be proved (disproved) here? In that regard, the two conjectures are the same: you want to represent every natural number as a sum of two squares and two members of "exponentially sparse sets" and no evidence exists for the feasibility of any of the two questions. One can easily come up with 100 conjectures like that. $\endgroup$ – Mark Sapir May 31 '18 at 0:31

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