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MathGroup Archive 2006

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Re: compound symmetrical primes

  • To: mathgroup at smc.vnet.net
  • Subject: [mg66495] Re: [mg66482] compound symmetrical primes
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Wed, 17 May 2006 03:29:26 -0400 (EDT)
  • References: <200605160349.XAA01092@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 16 May 2006, at 12:49, János wrote:

> Let's say that a prime is symmetrical if prime==FromDigits[Reverse
> [IntegerDigits[prime]]].
>
> I would say a prime is compound if two contiguous distinct subsets of
> its digits are summing up to the same number.  For example 211 is a
> compound prime because 2=1+1.  Similarly 15877 is a compound prime
> because 1+5+8=7+7.  May be there is a better definition in the prime
> literature.  The two distinct subset has to cover all the digits.
>
> Here is a little program that that looks for symmetrical compound
> primes up to an mx limit.
>
> In[14]:=
> lst = Timing[First[
>      Last[Reap[i = 1;
>         mx = 10^6; While[
>          i <= mx,
>          pr = Prime[i];
>           If[pr != FromDigits[
>            Reverse[
>            IntegerDigits[
>            pr]]], i++;
>            Continue[]; ];
>           prdig =
>            IntegerDigits[pr];
>           prlen = Length[
>            prdig]; j = 1;
>           While[j < prlen,
>            prLeft = Take[
>            prdig, {1, j}];
>            prRight = Take[
>            prdig, {j + 1,
>            prlen}];
>            If[Total[prLeft] !=
>            Total[prRight],
>            j++; Continue[],
>            Sow[pr]; Break[
>            ]]; ]; i++; ]; ]]]]
> Out[14]=
> {31.687534999999997*Second,
>    {11, 101, 16061, 31013,
>     35053, 38083, 73037,
>     74047, 91019, 94049,
>     1120211, 1150511, 1160611,
>     1180811, 1190911, 1250521,
>     1280821, 1300031, 1360631,
>     1390931, 1490941, 1520251,
>     1550551, 1580851, 1600061,
>     1630361, 1640461, 1660661,
>     1670761, 1730371, 1820281,
>     1880881, 1930391, 1970791,
>     3140413, 3160613, 3260623,
>     3310133, 3380833, 3400043,
>     3460643, 3470743, 3590953,
>     3670763, 3680863, 3970793,
>     7100017, 7190917, 7250527,
>     7300037, 7310137, 7540457,
>     7600067, 7630367, 7690967,
>     7750577, 7820287, 7850587,
>     7930397, 7960697, 9110119,
>     9200029, 9230329, 9280829,
>     9320239, 9400049, 9440449,
>     9470749, 9610169, 9620269,
>     9650569, 9670769, 9700079,
>     9770779, 9820289, 9980899}}
>
> Here are a few questions:
>
> Is there any compound symmetrical prime other than 11 whose length is
> even ?
>
> Is there any compound symmetrical prime where the middle digit is not
> zero ?
>
> Is there a much faster algorithm to find these numbers ?  /I am
> mostly procedural here :) because I could not find a functional check
> for compoundness. /
> I searched up to mx=10^8 and on my little iBook it took the whole  
> night.
>
> Let's say a number is periodical if it is a repetition of a subset of
> its digits .  For example 11 and 232323 are periodic numbers.
>
> Is 11 the only periodical symmetrical compound prime ?
>
> Thanks ahead,
>
> János
>


Consider a symmetrical number with an even number of digits. NTake  
the alternating sum of the digits e.g. if the number is abccba it  
will be a-b+c-c+b-a =0. In fact you always get zero. But an integer  
with this property must be divisible by 11, and hence unless it is 11  
itself it won't be prime. This answers your first question.

The second question: consider any symmetric and compound number with  
an odd number of digits. Let a will be its middle digit. The sums of  
the numbers to the left of a and to the right of will be equal, so  
unless a is 0, the number can't satisfy your definition of  
compoundedness (because if you add a non-zero number to one of two  
numbers that are equal, the resulting two numbers won't be equal any  
more. That is what non-zero means ;-)).

The question about primes that are periodic and symmetric is harder,  
but I think I can also show that the only one is 11. The proof again  
reduces to the same divisibility by 11 criterion as above. We use the  
fact that the middle number is 0. You write out the various ways in  
which a periodic number can be formed, and then use the fact that the  
first digit is the same as the last, the second the same as the last  
but one etc. Finally you take the alternating sum of digits and, I  
think, that will always come to 0.

Once one has answered your first two questions it is easy to  
construct a much faster code. Here is an example. One can make it  
much nicer, but I will leave that to others.


ls=Last[Last[n=
         1;Reap[While[n<
             10^4,If[PrimeQ[x=FromDigits[Join[IntegerDigits[n], {0},  
Reverse[
                      IntegerDigits[n]]]]],Sow[x]];n++]]]];//Timing


{0.650904 Second,Null}


Length[ls]


526


Take[ls,20]


{101,16061,31013,35053,38083,73037,74047,91019,94049,1120211,1150511,116 
0611,
1180811,1190911,1250521,1280821,1300031,1360631,1390931,1490941}

Note that the code does not construct the first number 11, which we  
know is rather "special".

One could also try to make use of the fact that there is no need to  
consider those  n whose first digit is 2 or 5, since that first digit  
will be the last digit of the symmetric number which we are  
constructing and numbers ending with 2 or 5 can't be prime. However,  
trying to make use of this in the code might make it more complex and  
may not result in any improved efficiency.

Andrzej Kozlowski
Tokyo, Japan


  


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