[seqfan] Re: Question about Three Integers Whose Sum is 2^N

[Arthur O'Dwyer]

Hi Andras,

I don't know that this phenomenon has a *name*, but it's pretty easy to see
why it happens.
Notice that you're removing digits from the left (high-order) end of the
bit-strings. So as long as you remove
exactly one 1-bit each time, the sum will decrease by exactly the
highest-remaining power of two each time;
and that means you'll stay at a power of two.
Here's a simpler set of bit-strings with the same property:
01001001001
00100100101
10010010010
The lowest-order bits (1,1,0) sum to 2. Then adding in the next-lower bit
increases the sum to 4; 8; 16; and so on.
Removing bits from the left-hand side reverses that process.
You can do the same thing in base-ten by picking columns that sum to 9 (ten
minus 1) instead of 1 (two minus 1):
51232701803
12530270183
36237028014
Removing or permuting *any* column out of these digit-strings (except for
the rightmost column, which sums to *base* instead of *base*-minus-1) will
preserve the property that all three digit-strings together sum to a power
of ten.

There's one more interesting case that works *only* in base-two: A single
1-bit in column k can be replaced with two 1-bits in column k-1, since 2^k
= 2^{k-1} + 2^{k-1}. So we can take our example above:
01001001001
00100100101
10010010010
and apply the trick in column 4:
01001001001
00100101101
10010001010
and again in column 3:
01001001101
00100100101
10010001110
We can do this arbitrarily many times to make our bit-strings look more
"interesting."
It reminds me of siteswaps in club-passing juggling: in four-count you can
"trade" an ordinary pass for a pass of twice the height one beat earlier
(an "early double"), or trade it for a pass of triple height two beats
earlier, etc.

–Arthur


On Sun, Feb 26, 2023 at 1:37 AM Andras via SeqFan <seqfan at list.seqfan.eu>
wrote:

>
> I have three ratioed integers whose sum is 524288:
>
> 111493, 374949, 37846
>
> If I delete the final digit from each binary string and add them up, I
> will always get a sum of exactly 2^N.
>
> Does anyone happen to know the exact name of this phenomenon ?
>
> 0011011001110000101 + 1011011100010100101 + 0001001001111010110 == 524288
> [2^19]
> 11011001110000101 + 11011100010100101 + 01001001111010110 == 262144 [2^18]
> 1011001110000101 + 1011100010100101 + 1001001111010110 == 131072 [2^17]
> 011001110000101 + 011100010100101 + 001001111010110 == 32768 [2^15]
> 1001110000101 + 1100010100101 + 1001111010110 == 16384 [2^14]
> 001110000101 + 100010100101 + 001111010110 == 4096 [2^12]
> 01110000101 + 00010100101 + 01111010110 == 2048 [2^11]
> 110000101 + 010100101 + 111010110  == 1024 [2^10]
> 10000101 + 10100101 + 11010110 == 512 [2^9]
> 0000101 + 0100101 + 1010110 == 128 [2^7]
> 000101 + 100101 + 010110 == 64 [2^6]
> 00101 + 00101 + 10110 == 32 [2^5]
> 0101 + 0101 + 0110 == 16 [2^4]
> 01 + 01 + 10 == 4 [2^2]
> 1 + 1 + 0 == 2 [2^1]
>