$4494410$ and friends

The number $4494410$ has the property that when converted to base $16$ it is $44944A_{16}$, then if the $A$ is expanded to $10$ in the string we get back the original number.

$3883544142410_{10}=3883544E24A_{16}$ is another.

These numbers are in OEIS A187829. They come in blocks of $6$ or $10$, depending on whether the one’s digit in hex is $A-F$ or $0-9$.

I suspect the list is complete but have not proven it. The largest is $806123145829415507126939101294137128298625241370656314360169_{10}=\\806C3E58294F507C6939AC94D7C829862524D706563E360169_{16}$

If the number has $m$ hex digits and $n$ base $10$ digits, we must have $16^{m-1} \gt 10^{n-1}$ and $16^{m-2} \lt 10^{n-2}$ which leads the hunt to $m=6,n=7;\ m=11,n=13;\ m=16, n=19;\ m=50,n=60$ guided by the convergents of $\frac {\log 16}{\log 10}$.

We can view finding these numbers as finding solutions to the subset-sum problem, where each hex digit contributes the difference between its value in base $16$ and base $10$ (depending on how many base $16$ digits to the right are $A-F$ and counting the two base $10$ digits coming from one hex digit together). The sum then has to be zero.

My search program ran reasonably quickly even for the next convergent, $m=535, n=644$ and didn’t find any. I believe they just have too many ways to fail as the number gets long.

Can we prove that there are no more, or at least that there are no more with very high probability, in the sense of “proofs” of Goldbach that if the primes are “random” the chance of any large even number having no solution is very low?

Solutions Collecting From Web of "$4494410$ and friends"

I can prove a very narrow form.

Let’s consider numbers of the form:

$(D_{n-1}D_{n-2}…..D_{2}D_{1}D_0)_{10} = (D_{n-1}D_{n-2}…..D_2[A..F])_{16}$.

Here 2 least significant digits in decimal representation change
to [A..F]. For these numbers, conditions that need to satisfy are,

$100*x + 10 = n$ ….(1)

$16*y + 10 = n$ ….(2)


$y = 6.25*x$ ….(3)

Let’s assume,

$x_{10}$ is of form $….N_{k-1}N_{k-2}…..N_{2}N_{1}N_{0}$


$x_{10} = ….+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ….+ (N_{2}10^2) + (N_{1}10^1) + N_{0}$


$y_{10} = ….+ (N_{k-1}16^{k-1}) + (N_{k-2}16^{k-2}) + ….+ (N_{2}16^2) + (N_{1}16^1) + N_{0}$

Also from (3),

$y_{10} = 6.25(….+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ….+ (N_{2}10^2) + (N_{1}10^1) + N_{0})$


$(….+ (N_{k-1}16^{k-1}) + (N_{k-2}16^{k-2}) + ….+ (N_{2}16^2)
+ (N_{1}16) + N_{0}) =
6.25(….+ (N_{k-1}10^{k-1}) + (N_{k-2}10^{k-2}) + ….+ (N_{2}10^2)
+ (N_{1}10^1) + N_{0})$

Using up to 6 digits for x,

$(1048576N_5 + 65536N_4 + 4096N_3 + 256N_2 + 16N_1 + N_0) =
(625000N_5 + 62500N_4 + 6250N_3 + 625N_2 + 62.5N_1 + 6.25N_0)$


$42357600N_5 + 303600N_4 – 215400N_3 – 36900N_2 – 4650N_1 – 525N_0 = 0$

Value in Hex falls behind till N3 because initial modulus was 100 for decimal
and 16 for hex. But at and after N4, hex value overtakes decimal forever.

Only solution for N5 (and above) between 0 and 9 is 0. Also, there are no solutions possible less
than 5 digits for x in (1).

So essentially numbers of these forms are only 7-digits or 2 digits,
And only possible solutions are,