Is there prime number of the form$1101001000100001000001.....$after the trivial one $k(0)=11$?

Question

Let

• $k(0)=11$

• $k(1)=1101$

• $k(2)=1101001$

• $k(3)=11010010001$

• $k(4)=1101001000100001$

• And So on....

  I've checked it up to $k(120)$, and I did't find anymore prime of such form. Are there anymore prime numbers of that form ? (I just realized that only $k(6n+5)$ could be a prime (?))

Answer 1

This is not really an answer but might be of some help.

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According to the "Divisibility by $3$ Rule," if $n\equiv 1\pmod 3$ then $k(n)$ will not be prime as it will be divisible by $3$. And, it will be divisible by $11$ if $n$ is even.

That leaves only all the odd numbers for $n$ (since the congruence above is also even).

Answer 2

Here is a Mathematica search to answer your question:

b[1] = 1;
b[2] = 2;
b[n_] := b[n] = b[n - 1] + n - 1;
list[t_] := b /@ Range[t];
Reap@Do[a = ReplacePart[Array[0 &, b[t]], Transpose[{list[t]}] -> 1]; 
  c = FromDigits[a, 2]; If[PrimeQ[c], Sow@c], {t, 100}]

which yields

{Null, {{3, 13, 2713027506953773210808498184692097546276033420315106938029407997308\ 25845099036699701989532948734015220469369753358523432961}}}

Answer 3

The formation law is clearly

$$ n_k = 2^k n_{k-1}+1 $$

with $n_1=3$

n0 = 3; For[i = 2, i < 50, i++, n1 = 2^i n0 + 1; If[PrimeQ[n1], Print[n1, " ", IntegerString[n1, 2]]]; n0 = n1]

obtaining

n = 13 -- 1101

n = 271302750695377321080849818469209754627603342031510693802940799730825845099036699701989532948734015220469369753358523432961 -- 11010010001000010000010000001000000010000000010000000001000000000010000000000010000000000001000000000000010000000000000010000000000000001000000000000000010000000000000000010000000000000000001000000000000000000010000000000000000000010000000000000000000001000000000000000000000010000000000000000000000010000000000000000000000001000000000000000000000000010000000000000000000000000010000000000000000000000000001

If the number is considered in basis $10$ then the procedure is analogous. In this case we have $n_1 = 11$ and the recurrence equation is $n_k = 10^k n_{k-1}+1$ giving n = 1101001000100001000001000000100000001000000001000000000100000000001000000000001000000000000100000000000001000000000000001000000000000000100000000000000001000000000000000001000000000000000000100000000000000000001000000000000000000001000000000000000000000100000000000000000000001000000000000000000000001000000000000000000000000100000000000000000000000001000000000000000000000000001000000000000000000000000000100000000000000000000000000001000000000000000000000000000001000000000000000000000000000000100000000000000000000000000000001000000000000000000000000000000001000000000000000000000000000000000100000000000000000000000000000000001000000000000000000000000000000000001 -- 1101001000100001000001000000100000001000000001000000000100000000001000000000001000000000000100000000000001000000000000001000000000000000100000000000000001000000000000000001000000000000000000100000000000000000001000000000000000000001000000000000000000000100000000000000000000001000000000000000000000001000000000000000000000000100000000000000000000000001000000000000000000000000001000000000000000000000000000100000000000000000000000000001000000000000000000000000000001000000000000000000000000000000100000000000000000000000000000001000000000000000000000000000000001000000000000000000000000000000000100000000000000000000000000000000001000000000000000000000000000000000001