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It was proven by Gauss that every integer is the sum of at most three triangle numbers, and by Lagrange that every integer is the sum of at most four squares.

The triangles are the set $\big\{1,3,6,10,\ldots\big\}$, and it's easy to see that the squares can be written as $\big\{1,1\!+\!3,3\!+\!6,6\!+\!10,\ldots\big\}$.
I wonder if it's generally the case that if a set $S$ (with $s_n$ the $n^{\mathrm{th}}$ element) is a $p$-additive base for the integers, then the set $S^*=\big\{s_1,s_1+s_2,s_2+s_3,\ldots\big\}$ is (at most?) $p+1$ additive.

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  • $\begingroup$ $\{0,1,3,5,7,9,\dots\}$ is a basis of order two, but $\{0,1,4,8,12,16,\dots\}$ is not a basis of order three. $\endgroup$ Commented Feb 26, 2023 at 0:04
  • $\begingroup$ Then probably it'll be a base of order at most 2p, but that's quite unimpressive of course. Thanks $\endgroup$ Commented Feb 26, 2023 at 1:04

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