Republic of Mathematics blog

A curious number exploration

Posted by: Gary Ernest Davis on: May 6, 2011

Benjamin Vitale (@BenVitale) tweeted that he followed 203 people on Twitter and 203^2 + 203^0 + 203^3 = 8406637 is a prime number.

Alex Bogomolny (@CutTheKnotMath) observed that if n=abc (digits base 10) then abc^a+abc^b+abc^c has a factor of abc unless one of a, b, c =0.

A quick computational search (using Mathematica™) yields the following numbers n=d_k\ldots d_0 \leq 10000 (base 10 digits) for which n^{d_k}+\ldots +n^{d_0} is prime:

10, 20, 203, 230, 308, 309, 330, 350, 503, 603, 650, 960, 1068, 1110, 1206, 1350, 1404, 1480, 1730, 1802, 1860, 1910, 2032, 2038, 2044, 2054, 2250, 2320, 2502, 3044, 3082, 3402, 3970, 4032, 4046, 4072, 4120, 4340, 4450, 4540, 4650, 4908, 5204, 5310, 5402, 5530, 5770, 5820, 6038, 6076, 6120, 6206, 6490, 6520, 6830, 7110, 8074, 8106, 8120, 8380, 8450, 8610, 8690, 8902, 8960, 9064, 9330, 9402, 9408, 9950

How do these numbers increase?

Are they becoming relatively rarer?

Let’s denote by V(p) the number of n=d_k \ldots d_0 \textrm{ (base 10  digits) } \leq p such that n^{d_k}+\ldots + n^{d_0}  is prime.

How does the fraction \frac{V (p)}{p} vary with increasing p?

Here is a plot of that fraction up to p=200000:

It appears that \frac{V(p)}{p} \to 0 as p increases, although the decrease is highly exaggerated by the foreshortened plot.

Does \frac{V(p)}{p} \to 0 as p increases?

Postscript

Matt Henderson (@mattthen2) made the following useful observation:

Let Z(p) denote the number of n\leq p whose digits contain a 0 .

Then Z(10^n)< 10^n + 9/8 - \frac{9^{n+1}}{8} so \frac{Z(10^n)}{10^{nk}}\to 0\textrm{ as } n\to \infty for all k>1.

Because V(p)\leq Z(p) for all p we see that:

\frac{V(10^n)}{10^{nk}}\to 0\textrm{ as } n\to \infty for all k>1.

 

 

 

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