# Positive integers written as $\frac{a(a+1)}2+\frac{b(b+1)}2+4^c5^d$

Let $$\mathbb N=\{0,1,2,\ldots\}$$. Those $$T_n:=n(n+1)/2$$ with $$n\in\mathbb N$$ are called triangular numbers. It is well known that $$\{T_a+T_b+T_c:\ a,b,c\in\mathbb N\}=\mathbb N\tag{1}$$ which was conjectured by Fermat and proved by Gauss. Euler observed that \begin{aligned}&\{(2x)^2+(2y+1)^2=4(x^2+2T_y)+1:\ x,y\in\mathbb N\} \\=&\{(a-b)^2+(a+b+1)^2=4(T_a+T_b)+1:\ a,b\in\mathbb N\}.\end{aligned}\tag{2}

Motivated by $$(1)$$ and $$(2)$$, I formulated the following conjecture on June 7, 2019.

Conjecture. Let $$n$$ be any positive integer. Then we can write $$n$$ as $$T_a+T_b+4^c5^d$$ with $$a,b,c,d\in\mathbb N$$; equivalently, $$n$$ can be written as $$u^2+2T_v+4^x5^y$$ with $$u,v,x,y\in\mathbb N$$ (or $$4n+1=a^2+b^2+4^c5^d$$ for some $$a,b,c,d\in\mathbb N$$ with $$c>0$$).

I have verified this for all $$n=1,\ldots,5\times10^8$$. For example, $$585$$ has a unique required representation: $$585 = T_{10}+T_{20}+4^35^1=5^2 +2T_{15} + 4^35^1.$$

QUESTION. Can one find a counterexample to the conjecture?

• For a positive integer $n$, let $a(n)$ be the number of ways to write $n$ as $w^2+x(x+1)+4^y5^z$ with $w,x,y,z\in\mathbb N$. The sequence $a(1),a(2),a(3),\ldots$ is available from oeis.org/A308566. – Zhi-Wei Sun Jun 8 at 7:00
• Giovanni Resta just reported on OEIS that he had verified my above conjecture for $n$ up to $10^{10}$. – Zhi-Wei Sun Jun 8 at 12:44
• I also conjecture that any positive integer can be written as $T_a+T_b+5^c8^d$ with $a,b,c,d\in\mathbb N$. See oeis.org/A308584. – Zhi-Wei Sun Jun 9 at 4:16

The following is a heuristics showing that the conjecture is likely to be true. For a given $$n$$ there are around $$\log^2{n}$$ positive numbers of the form $$n-4^c5^d$$. Each number has "probability" around $$1/\sqrt{\log{n}}$$ of being of the form $$a^2+b^2$$. If we further regard these events as independent, the probability of $$4n+1$$ being not of the form $$a^2+b^2+4^c5^d$$ is $$\approx (1-1/\sqrt{\log{n}})^{\log^2{n}}\approx e^{-\log^{3/2}{n}}$$. Since sum of these probabilities over all $$n>10^{10}$$ is tiny, we should expect all numbers larger than $$10^{10}$$ to be of this form.
Clearly, this argument doesn't take into account possible congruence constraints but it seems like there is none given that all numbers up to $$10^{10}$$ are of this form.