# $4494410$ 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?

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I typed 3883544142410 into Google, and the only hit was this question. –  Gerry Myerson Jan 30 '13 at 23:54
The inequality $16^{m−1}>10^{n−1}$ may be wrong, e.g., when the first hex digit is F. –  Egor Skriptunoff Jan 31 '13 at 8:06
@EgorSkriptunoff: I think of the two decimal digits coming from one hex letter as one, with the place value of the lower, so in this case $n$ would be one less than the number of decimal digits. The idea is that the leading two digits have to almost cancel so the lower digits can handle the difference. In that case the hex leading with a letter just changes the number of letters we expect, raising it by 1. –  Ross Millikan Jan 31 '13 at 16:13
@GerryMyerson: I tried 4494410 (I would bet you did, too) and found phone numbers and part numbers. –  Ross Millikan Feb 2 '13 at 5:21
... and then there are near-misses such as $3AF6A8FE786169D1B0224A5ECDEEE9A8BD991E86C949757BE711AA1F4E0EC94145577316156DBF5‌​DF3C3781C72E5D1498B6E0D206F65F5B3932F242C1C5440D79EC908CD9EE8C6CA2C8300D6DC881D9A‌​F0C8B0AE66946DE80_{16}=3101561081514786169131110224105141213141414910811139911486‌​129497571114711101011541401412941455773161561311155131531237811272145131498116140‌​13206156515511393215242121125440137914129081213914148126121021283001360000_{10}$, just a few digits short –  Hagen von Eitzen Aug 5 '13 at 5:12

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)

So,

$y = 6.25*x$ ....(3)

Let's assume,

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

or,

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

So,

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

Also from (3),

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

So,

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

Using up to 6 digits for x,

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

or,

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

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,

10-15
4494410-4494415
5660810-5660815
6784010-6784015
7950410-7950415

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$\LaTeX\text{ify}$ please. ${}{}$ –  user93957 Nov 14 '13 at 12:58