Questions:

a) Are there infinite of these numbers N?

b) Can you find other models and larger examples

10^n = 2^n*5^n works for all n.

If N is a solution without digits 2 or 5, and there are no 0's in the prime factorization of N, then N*10^n is also a solution. The first general example works as N.

There ought to be infinitely many prime numbers all of whose digits are 3 or 4. ... (because) The number of primes with n digits all of whom are 3 and 4 has the expectation 2^n/log(10^n)>1.99^n. Multiply any such number by 2

I'm sure everyone will head down the "pattern" route, however, the pattern route is a motorway to infinity(*) so I think I'll try to just find coincidences instead!

The largest powers of small primes: (ones <10^6 not included)

2^168 374144419156711147060143317175368453031918731001856
3^84 11972515182562019788602740026717047105681
7^39 909543680129861140820205019889143
13^6 4826809
37^14 9012061295995008299689
43^8 11688200277601
53^4 7890481
59^8 146830437604321
67^4 20151121
73^8 806460091894081
89^4 62742241
157^3 3869893
223^3 11089567
233^8 8686550888106661441
263^3 18191447
353^8 241100240228887100161
383^4 21517662721
487^3 115501303
577^3 192100033
587^3 202262003
739^4 298248146641
773^4 357040905841
823^3 557441767
839^4 495504774241
877^10 269150960295439095031490496649
887^4 619005459361
977^4 911125611841

and select others:

2213^6 117459590807554494409
7757^7 1689882622864926680809344293

Notice how many of those maxima are at powers which are multiples of 4 (168,84, a few 8s and many 4s). (Follow-up 2, PC) There _is_ a reason for this bias, I wonder if others would like to puzzle what it might be. ;-)

However, one can combine small primes that share at no extra cost to get a few more stabs at finding coincidences:

6^18 101559956668416
1454^4 4469486461456
1047^5 1258152857750007
9771^4 9114986990498481 (4 digits excluded)
10029^4 10116505576267281
10401^3 1125188511201 (4 digits excluded)
10599^4 12620006210117601
13119^4 29621219082801921
511^8 4649081944211090042881
5299^6 22139281232164381318201
8309^5 39604432524450440549
143^6 8550986578849
5359^5 4419964540656090799
369001^5 6841285512441040451611845001
18619^4 120178120515799921
4183^4 306162121305121
22513^4 256881888535258561
542609^4 86685802029100691588161

Not particularly impressive, but quite fun nonetheless. I particularly like the 10401^4 one, restricted to 6 digit, and 10401 being a palindrome.

I found several models, which one of them is the most interesting.

The most interesting model [almost pandigital when m>=n, CR]:

3(m).5.3(n)^2 = 1(m)2.4(m)8.4(n-m-1).2(m)0.8(n-1).9 ( for m < n )
3(m).5.3(n)^2 = 1(m).2.4(n).2(m-n).6.2(n-1).0.8(n-1).9 ( for m >= n )

Examples :

3(3).5.3(5) = 333533333 , 3(3).5.3(5)^2 = 111244484222088889
3(7).5.3(4) = 333333353333 , 3(7).5.3(4)^2 = 111111124444222622208889

Answer to part (a) :

Let f(m,n) = 3(m).5.3(n) ( m & n are non negative integers ), it seems that for each m there exist at least one n such that f(m,n) is prime, but so if this is true there exist infinite numbers N .

Other models :

3(n).77 ^2 = 1(n).40.2(n-2).4129
6(n).77 ^2 = 4(n).58.2(n-1).329
5.6(n).77 ^2 = 32.1(n-1).228.2(n-1).329
5.6(n).7 ^2 = 32.1(n).4.8(n).9
5.3(n).77 ^2 = 28.4(n-1).910.2(n-2).4129
6(n).73 ^2 = 4(n).52.8(n-1).929

N= f(7000,664)^2 is the largest N that I found. Since, N= 1(7000).2.4(664).2(7000-664).6.2(664-1).0.8(664-1).9 , (f(m,n)^2=1(m).2.4(n).2(m-n).6.2(n-1).0.8.(n-1).9 for m >= n)) N is 15330 digits long.

Another interesting model :

77.3(n).77 ^2 = 5980.4(n-3).51198.2(n-2).4129 (n > 2)

I also found four interesting [pandigital, CR] models with further property that the  union of the set of the digits in the decimal expansion of N and set of the digits present in all the prime factors of N is equal to {0,1,...,9}.

The four interesting models:

Let, s(m,n) be 77.3(m).6.3(n).77

s(m,m-2)^2 = 5980.4(m-2).908.4(m-5).511991.2(m-3).484.2(m-4).4129 (m > 4)
s(m , m)^2 = 5980.4(m-2).908.4(m-3).51288 .2(m-2).484.2(m-2).4129 (m >2)
s(m,m+1)^2 = 5980.4(m-2).908.4(m-2).52098 .2(m-2).484.2(m-1).4129 (m > 1)
s(m,m+3)^2 = 5980.4(m-2).908.4(m-1).541198.2(m-2).484.2(m+1).4129 (m > 1)

s(64,62)^2 , s(76,76)^2 , s(302,303)^2 & s(74,77)^2 are respectively the smallest N of the forms s(m,m-2)^2, s(m,m)^2, s(m,m+1)^2 & s(m,m+3)^2 .

Examples:

s(6,4)= 773333336333377 , s(6,4)^2 = 598044449084511991222484224129
s(4,4)= 7733336333377 , s(4,4)^2 = 59804490845128822484224129
s(5,6)= 7733333633333377 , s(5,6)^2 = 59804449084445209822248422224129
s(4,7)= 7733336333333377 , s(4,7)^2 = 59804490844454119822484222224129

(Follow up 3, FF) Can you find other such models?

In relation to Firoozbakht Conjecture in puzzle 244, I think its true but indecidable in this century. She says:"Let f(m,n)= 3(m).5.3(n), for each integer m there exists at least one n such that f(m,n) is a prime number."

3(m).5.3(n) means :

(10^m-1)x10^(n+1)/3 + 5x10^n +(10^n - 1)/3

I verified the conjecture for m =1 to 30, only with m=28 I could not find a solution. For m =1,2,3,..,30 the corresponding n were: 1,2,1,3,6,9,10,18,13,33,12,9,8,6,3,28,2,7,2,7,2,8,26,1,14,3,50,??,7,83

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On return Faride wrote:

The smallest solution For each m (m=1,2,...,100) are as follows:

1,2,1,3,6,9,10,18,13,33,12,9,2,5,3,28,2,7,2,8,26,1,14,3,13,3,50,
118,7,83,8,48,290,19,1,235,22,4,61,49,19,207,19,84,99,217,48,12,
14,183,1,63,61,17,28,738,6,60,18,48,1,3,14,19,9,29,2,9,8,9,36,24
,351,17,2174,54,70,11,15,10,72,1,62,123,166,135,333,145,16,13,21
,8,2,174,13,18,35,105,12,43.

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Luis Rodríguez wrote to Faride the following email and I got a copy of it with the following heading lines:

Dear Professor Farideh

Following your idea I found another sequence. In this case beginning with 3's and continuing with 1's. So : 31,331,3331,33331,333331,3333331,33333331,

3333333311111,33333333311111....etc

(Observe that for the first seven cases only one 1 was necessary)

The formula is: 10^n*(10^m - 1)/3 + (10^n - 1)/9

For the twenty first cases the necessary ones are:

1, 1, 1, 1, 1, 1, 1, 5, 5, 37, 8, 35, 7, 13, 8, 19, 1, 89, 34, 74

I suspect that this sequence, as yours, can continue indefinitely producing primes.

I replied with a single line "what are those heuristic reasons?". Rodríguez replied to me this:

En relación con las razones heurísticas de porqué la sucesión de Farideh puede producir primos indefinidamente, hasta ahora tengo las siguientes:

1.- El permitir que la búsqueda se prolongue indefinidamente. Por ejemplo para m = 56 se necesitó un n= 753 y no se sabe si llegará un momento cuando el algoritmo de primalidad usado no permita seguir adelante.

2.- Como la forma general de esos números es 3*(10^(m+n+1) - 1)/9 +

2*10^n = N = 3xA + B.   (A siendo un repunit)

   Resulta que como B es divisible por 2 y por 5 y A no lo es,  entonces N no es divisible ni por 2, ni por 3 , ni por 5.

 Por la ley de divisibilidad por 11 esos números no son divisibles por 11.

 De esta menear se aumenta mucho su posibilidad de ser primos.

 

 3.- Además, los números A por ser repunits, si m+n+1 es divisible por 6,  ellos son divisibles por 3,7,11,13,37  y  B no lo es, creciendo así  la probabilidad de  N de ser primo.

 Ahora bien, el que N tenga 6k cifras hace que cada vez 6k - 2 números N caigan dentro de esa categoría, pues el  5 puede recorrer los puestos del 2o  al penúltimo.

Ejemplo 353333, 335333,333533 y 333353 no pueden ser divisibles por los primos 2,3,5,7,11,13 y 37

 Pero en general los repunits son divisibles por muchos primos que no dividen a B, por ejemplo los repunits Rep(30k) tienen muchísimos  divisores. (Solo se conocen 5 repunits primos)

 Algunas de las leyes citadas  se pueden aplicar a la secuencia 31,331,3331,....,33333331,.3333333311111...  etc.

N = 3*10^n*(10^m - 1)/9 + (10^n - 1)/9

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