[ October 23, 2016 ] [ Last update January 16, 2023 ]
Search for the smallest numbers that when appended by the same
amount of digits 1, 3, 7 and 9 generates primes in all four cases
Abhiram R. Devesh (email)
Please find below the list of primes generated by
appending n_1s, n_3s, n_7s and n_9s to one number k.
Appending 1 time [1,3,7,9] to k=1 generates the primes [11, 13, 17, 19]
Appending 2 times [1,3,7,9] to k=23 generates the primes [2311, 2333, 2377, 2399]
so on and so forth
Can you find more cases whereby the number k
is smaller or equal than the appended amount of digits d
(in the table below highlighted in light blue) ?
( '^^' is symbol for concatenation )
n  [k][list of primes] 
1  [1][d^^n] 
2  [23][d^^n] 
3  [2][d^^n] 
4  [325][d^^n] 
5  [2084][d^^n] 
6  [685][d^^n] 
7  [18166][d^^n] 
8  [2855][d^^n] 
9  [28253][d^^n] 
10  [45964][d^^n] 
11  [90452][d^^n] 
12  [6239][d^^n] 
13  [116863][d^^n] 
14  [13103][d^^n] 
15  [9116][d^^n] 
16  [1798][d^^n] 
17  [91643][d^^n] 
18  [32519][d^^n] 
19  [72313][d^^n] 
20  [35900][d^^n] 
21  [31552][d^^n] 
22  [276187][d^^n] 
23  [736202][d^^n] 
24  [30889][d^^n] 
25  [32947][d^^n] 
26  [505229][d^^n] 
27  [26888][d^^n] 
28  [139621][d^^n] 
29  [2426324][d^^n] 
30  [381580][d^^n] 
[ October 24, 2022 ]
Xinyao Chen (email)
extended the listing up to n=100 without finding a k ⩽ n.
Following his table he explains why this is very unlikely to happen.
[1999756]][d^^31] 
[68744][d^^32] 
[30256][d^^33] 
[15067][d^^34] 
[2660075][d^^35] 
[213308][d^^36] 
[2469904][d^^37] 
[5738156][d^^38] 
[2571430][d^^39] 
[2850424][d^^40] 
[2608031][d^^41] 
[67645][d^^42] 
[6483622][d^^43] 
[2847830][d^^44] 
[605489][d^^45] 
[3514927][d^^46] 
[391529][d^^47] 
[144311][d^^48] 
[3613165][d^^49] 
[5651627][d^^50] 
[1768552][d^^51] 
[1096552][d^^52] 
[3533255][d^^53] 
[317594][d^^54] 
[2955328][d^^55] 
[6591881][d^^56] 
[9760955][d^^57] 
[3026032][d^^58] 
[7864037][d^^59] 
[323464][d^^60] 
[5343502][d^^61] 
[7584278][d^^62] 
[5324609][d^^63] 
[2424883][d^^64] 
[5000798][d^^65] 
[1679383][d^^66] 
[1569067][d^^67] 
[54363140][d^^68] 
[5672354][d^^69] 
[2865547][d^^70] 
[29634728][d^^71] 
[5281154][d^^72] 
[29132692][d^^73] 
[10738940][d^^74] 
[9063986][d^^75] 
[6959539][d^^76] 
[113159675][d^^77] 
[36346][d^^78] 
[3424051][d^^79] 
[33153794][d^^80] 
[47493881][d^^81] 
[34472227][d^^82] 
[111577325][d^^83] 
[1911443][d^^84] 
[24559321][d^^85] 
[38051264][d^^86] 
[29569556][d^^87] 
[6448315][d^^88] 
[55506611][d^^89] 
[4825087][d^^90] 
[25787977][d^^91] 
[151551920][d^^92] 
[7632794][d^^93] 
[60076228][d^^94] 
[31619447][d^^95] 
[44733034][d^^96] 
[22352758][d^^97] 
[17516795][d^^98] 
[52268882][d^^99] 
[15687688][d^^100] 
Of course, such k cannot be divisible by either 3 or 7
(or all numbers [k][3^^n] or [k][7^^n] will be divisible by 3
or 7, respectively), i.e. such k must be coprime to 21.
Such k always exist if Dickson's conjecture is true. In fact,
there are infinitely many such k since [k][1^^n], [k][3^^n], [k][7^^n], [k][9^^n]
for fixed n and variable k equals to these polynomials of k: k*10^n+(10^n1)/9,
k*10^n+3*(10^n1)/9, k*10^n+7*(10^n1)/9, k*10^n+9*(10^n1)/9, all of them
have degree 1, thus they are covered by Dickson's conjecture. In fact, this
proof works in all bases b, and furthermore, even we require [k][d^^r] is
composite for all 1 ⩽ r < n, such k still always exist and there are still
infinitely many such k, since we can change the polynomials to make [k][d^^r]
divisible by small primes for all 1 ⩽ r < n, by Chinese remainder theorem.
Unlike WONplate 197 and WONplate 219, there are many k such that the prime
does not exist, however, there may be no more cases (besides the cases
[1][d^^1] and [2][d^^3]) whereby the number k is smaller or equal than
the appended amount of digits d, since it is unlikely that the corresponding
numbers can be primes simultaneously for large n (heuristically, the number of
such n is finite, like the cases in the 'Semiprimes' section in WONplate 219), e.g.
the numbers n such that [1][1^^n] is prime: A0040231
the numbers n such that [1][3^^n] is prime: A056698
the numbers n such that [1][7^^n] is prime: A089147
the numbers n such that [1][9^^n] is prime: A002957
and the intersection for these four sequences is very likely to be {1},
in fact, the intersection for any two of these four sequences is likely to be finite.
the numbers n such that [2][1^^n] is prime: A056700
the numbers n such that [2][3^^n] is prime: A056701
the numbers n such that [2][7^^n] is prime: A056702
the numbers n such that [2][9^^n] is prime: A056703
and the intersection for these four sequences is very likely to be {0, 3},
in fact, the intersection for any two of these four sequences is likely to be finite.
the numbers n such that [4][1^^n] is prime: A056706
the numbers n such that [4][3^^n] is prime: A056707
the numbers n such that [4][7^^n] is prime: A056708
the numbers n such that [4][9^^n] is prime: A056712
and the intersection for these four sequences is very likely to be empty,
in fact, the intersection for any two of these four sequences is likely to be finite.
the numbers n such that [5][1^^n] is prime: A056713
the numbers n such that [5][3^^n] is prime: A056714
the numbers n such that [5][7^^n] is prime: A056715
the numbers n such that [5][9^^n] is prime: A056716
and the intersection for these four sequences is very likely to be {0},
in fact, the intersection for any two of these four sequences is likely to be finite.
the numbers n such that [8][1^^n] is prime: A056722
the numbers n such that [8][3^^n] is prime: A056723
the numbers n such that [8][7^^n] is prime: A056724
the numbers n such that [8][9^^n] is prime: A056725
and the intersection for these four sequences is very likely to be empty,
in fact, the intersection for any two of these four sequences is likely to be finite.
[ November 21, 2022 ]
Xinyao Chen proved that n = 1 is the only n such that [k][d^^n] are primes
for k = 1 and all of d = {1,3,7,9}, and there is no n such that [k][d^^n] are primes
for k = 8 and all of d = {1,3,7,9}, the case for k = 2, 4, 5 are still open questions.
The case k = 1
n is odd: [k][1^^n] is divisible by 11
(the only possible prime case is the prime 11 itself, i.e. n = 1)
n is even: [k][3^^n] cannot be prime, since the formula of [1][3^^n] is
(4*10^n1)/3, if n = 2*m is even, then (4*10^n1)/3 can be factored as
(2*10^m1) * (2*10^m+1) / 3, and since if n ⩾ 2, then m ⩾ 1,
2*10^m1 ⩾ 2*10^11 = 19 > 3, 2*10^m+1 ⩾ 2*10^1+1 = 21 > 3,
this factorization is nontrivial, thus no number of the form [1][3^^n] is prime if n is even
(note that we can exclude n = 0, since for n = 0 the number [1][3^^n] is 1, and 1 is not prime).
The case k = 8
n == 1 mod 3: [k][1^^n] and [k][7^^n] are divisible by 3
n == 2 mod 6: [k][3^^n] is divisible by 7
n == 5 mod 6: [k][1^^n] is divisible by 7
n == 3 mod 6: [k][3^^n] is divisible by 13
n == 0 mod 2: [k][3^^n] and [k][9^^n] cannot be primes, since the formula of [8][3^^n]
is (25*10^n1)/3, if n = 2*m is even, then (25*10^n1)/3 can be factored as
(5*10^m1) * (5*10^m+1) / 3, and since if n ⩾ 0,
then m ⩾ 0, 5*10^m1 ⩾ 5*10^01 = 4 > 3, 5*10^m+1 ⩾ 5*10^0+1 = 6 > 3,
this factorization is nontrivial, thus no number of the form [8][3^^n] is prime
if n is even, and since the formula of [8][9^^n] is 9*10^n1, if n = 2*m
is even, then 9*10^n1 can be factored as (3*10^m1) * (3*10^m+1), and since if n ⩾ 0,
then m ⩾ 0, 3*10^m1 ⩾ 3*10^01 = 2 > 1, 3*10^m+1 ⩾ 3*10^0+1 = 4 > 1, this
factorization is nontrivial, thus no number of the form [8][9^^n] is prime if n is even.
[ December 28, 2022 ]
Xinyao Chen adds
The cases for 2, 4, 5 are unlikely to have covering sets, since:
for the k = 2 case,
none of the four numbers [k][d^^93] has a prime factor < 20747,
none of the four numbers [k][d^^195] has a prime factor < 12377,
none of the four numbers [k][d^^216] has a prime factor < 15263,
none of the four numbers [k][d^^273] has a prime factor < 8281454161,
none of the four numbers [k][d^^666] has a prime factor < 5683,
none of the four numbers [k][d^^990] has a prime factor < 183119,
besides, the formula of [2][1^^n], [2][3^^n], [2][7^^n], [2][9^^n] are
(19*10^n1)/9, (7*10^n1)/3, (25*10^n7)/9, 3*10^n1, respectively,
and none of them has algebraic factorization since there is no n such
that 19*10^n, 7*10^n, or 3*10^n is perfect power (after all, 19*10^n
is divisible by 19 but not 19^2, 7*10^n is divisible by 7 but not 7^2,
and 3*10^n is divisible by 3 but not 3^2), and for the case of
(25*10^n7)/9, although 25 is square, 7 is not perfect power
(however, to avoid be divisible by 3, 7, 13, n must be == 0 mod 3,
and n must not == 6 mod 18 to avoid be divisible by 19).

for the k = 4 case,
none of the four numbers [k][d^^192] has a prime factor < 4451,
none of the four numbers [k][d^^348] has a prime factor < 6079,
none of the four numbers [k][d^^360] has a prime factor < 4999,
none of the four numbers [k][d^^924] has a prime factor < 2351,
none of the four numbers [k][d^^960] has a prime factor < 46439,
none of the four numbers [k][d^^1368] has a prime factor < 145487,
none of the four numbers [k][d^^1500] has a prime factor < 1443647,
besides, the formula of [4][1^^n], [4][3^^n], [4][7^^n], [4][9^^n] are
(37*10^n1)/9, (13*10^n1)/3, (43*10^n7)/9, 5*10^n1, respectively,
and none of them has algebraic factorization since there is no n such
that 37*10^n, 13*10^n, 43*10^n, or 5*10^n is perfect power (after all,
37*10^n is divisible by 37 but not 37^2, 13*10^n is divisible by 13 but
not 13^2, and 43*10^n is divisible by 43 but not 43^2), and for the case
of 5*10^n1, 5*10^n = 2^n*5^(n+1), and n and n+1 must be coprime,
thus 5*10^n cannot be perfect power (however, to avoid be
divisible by 3 or 7, n must be == 0 mod 6).

for the k = 5 case,
note that the formula of [5][3^^n] is (16*10^n1)/3, thus n == 2*m
has algebraic factorization (factored to (4*10^m1) * (4*10^m+1) / 3),
thus we only need to consider odd n, and
none of the four numbers [k][d^^533] has a prime factor < 853,
none of the four numbers [k][d^^1133] has a prime factor < 1609,
none of the four numbers [k][d^^1373] has a prime factor < 1955887,
none of the four numbers [k][d^^1733] has a prime factor < 101159,
besides, the formula of [5][1^^n], [5][7^^n], [5][9^^n] are
(46*10^n1)/9, (52*10^n7)/9, 6*10^n1, respectively,
and none of them has algebraic factorization since there is no n such
that 46*10^n, 52*10^n, or 6*10^n is perfect power (after all, 46*10^n
is divisible by 23 but not 23^2, 52*10^n is divisible by 13 but not 13^2,
and 6*10^n is divisible by 3 but not 3^2) (however, to avoid be
divisible by 3 or 7 as well as the algebraic factorization of [5][3^^n],
n must be == 5 mod 6).

