Problems & Puzzles: Puzzles

Puzzle 197. Always composite numbers?

Sequences of odd numbers that remain composite without any easily evident reason may be as attractive as primes.

Three well known examples are the sequences produced using the Sierpinski, the Riesel or the Brier numbers.

Less glamorous examples must be hidden here and there, for sure....

Very, recently while tackling the puzzle 195 (*), I faced one probably interesting example of that kind of sequences:

343
33
433
333
4333
3333
43333
...

• The members of the sequence can be obtained by 3*(10^(2*k+1)-1)/9+10^k, for k=1, 2, 3, ...
• The rate of growth (quantity of additional digits from one member of the sequence to the next one) is 2.
• According to  my old PRIMEFORM , these numbers remain composite up at least up to k=14000.

Questions:

1. Can you demonstrate that the members of that sequence are always composite, or can you find the first (probable) prime of them?

2. Can you report other interesting/beauty apparently always composite odd sequences ? (we prefer sequences with more number of members composites than the reported one above; this implicitly means a not very high rate of growth)

_____
* every member of this sequence is such that its square is a zero-free number

The Questions 1 & 2 of this puzzle was solved Ken Wilke, Jean Claude Rosa, Jim Howell, Jens Kruse Andersen and Faride FiroozBakht, basically the same way.

But only Patrick De Geest noticed that this question was posed and solved previously by Chris Caldwell and Harvey Dubner, in the Journal of Recreational Mathematics, Volume 28, Number 1, 1996-1997, pp. 1-9

Every term in our sequence is a composite of two factors according to the following pattern:

7 x 49 = 343
67 x 499 = 33433
667 x 4999 = 3334333
...
(6)
k7 x 4(9)k = (3)k4(3)k , for k=>1

Other similar sequences of odd composites similar to the ours, and studied by Caldwell & Dubner, are:

(1)k0(1)k, for k>1

(1)k2(1)k, for k>0

(3)k2(3)k, for k>0

In particular Ken Wilke wrote:

Here is an answer to Question 1 of Problem 197. All members of the sequence are composite. In the proof I use Tk to denote the nth term (to avoid using subscripts).

Proof: Let Tk = 3*(10^(2*k+1)+1)/9 + 10^k. then since Tk is never divisible by 3 we have

3*Tk = Ak*Bk where Ak = 2*10^k+1 and Bk = 5*10^k-1. Since Both Ak and BK are greater than 3 for k = 1,2,..., and Ak is clearly divisible by 3, we have

Tk = (Ak/3)*Bk yielding the desired factorization of Tk.

Perhaps an interesting follow-up question would be: For what values of k are both (Ak/3) and Bk prime?

Faride FiroozBakht, wrote:

The members of the sequence f(k)=3*(10^(2*k+1)-1)/9+10^k, for every natural number k is always composite numbers. That is because:

f(k)=(5*10^k-1)*(6*(10^k-1)/9+1)=10/3*10^(2k)+10^k-1/3

***

J. C. Rosa has gotten a better example than the one I posed in this puzzle: 7(1)k7, for k=>0, seems to be always composite and seems to be free of a factorizing pattern as before...

In order to save some time, I would like to anticipate that 7(1)k7 is composite in the following general cases:

a) k=even; in this case 11 is a divisor
b) (k-1)@6=0; in this case 3 is a divisor
b) (k+1)@6=0; in this case 13 is a divisor

In short the only "interesting" cases to search for possible primes are when k=6m+3, for m=>0.

Can you try again with this sequence detecting or a prime or the reason why this is a composite number for any k?

***

J. K. Andersen solved the follow-up posed by J.C. Rosa. Here is his email:

There were apparently no pattern covering all numbers so I started
Primeform/GW and it relatively quickly found the smallest prp for k=10905:
7*10^(10905+1)+(10^10905-1)/9*10+7 is 3-PRP!. It is also prp in many other bases. A Miller-Rabin test from the Miracl big integer library also finds it a probable prime. It is too big to be proved with current methods.

Primeform/GW found no other prp's for k<27000 but I guess there are infinitely many primes in the sequence. I have not computed when the first should have been expected statistically. With all the patterns in prime factors, I am not so surprised that the smallest prp was gigantic.

***
On May 18, 2022 Richard Chen wrote:

There are many examples of families that only contain composite numbers (only count the numbers > base).

Example 1: base 10, family 4{6}9 (formula: (14*10^(n+1)+7)/3)

49 = 7 * 7
469 = 7 * 67
4669 = 7 * 667
46669 = 7 * 6667
466669 = 7 * 66667
4666669 = 7 * 666667

Example 2: base 10, family 28{0}7 (formula: 28*10^(n+1)+7)

287 = 7 * 41
2807 = 7 * 401
28007 = 7 * 4001
280007 = 7 * 40001
2800007 = 7 * 400001
28000007 = 7 * 4000001

Example 3: base 9, family {1} (formula: (9^n-1)/8)

11 = 2 * 5
111 = 7 * 14
1111 = 22 * 45
11111 = 67 * 144
111111 = 222 * 445
1111111 = 667 * 1444
11111111 = 2222 * 4445
111111111 = 6667 * 14444
1111111111 = 22222 * 44445
11111111111 = 66667 * 144444
111111111111 = 222222 * 444445
1111111111111 = 666667 * 1444444

Example 4: base 9, family 3{8} (formula: 4*9^n-1)

38 = 5 * 7
388 = 18 * 21
3888 = 58 * 61
38888 = 188 * 201
388888 = 588 * 601
3888888 = 1888 * 2001
38888888 = 5888 * 6001
388888888 = 18888 * 20001
3888888888 = 58888 * 60001
38888888888 = 188888 * 200001
388888888888 = 588888 * 600001
3888888888888 = 1888888 * 2000001

Example 5: base 8, family 1{0}1 (formula: 8^(n+1)+1)

11 = 3 * 3
101 = 5 * 15
1001 = 11 * 71
10001 = 21 * 361
100001 = 41 * 1741
1000001 = 101 * 7701
10000001 = 201 * 37601
100000001 = 401 * 177401
1000000001 = 1001 * 777001
10000000001 = 2001 * 3776001
100000000001 = 4001 * 17774001
1000000000001 = 10001 * 77770001

Example 6: base 9, family {8}5 (formula: 9^(n+1)-4)

85 = 7 * 12
885 = 87 * 102
8885 = 887 * 1002
88885 = 8887 * 10002
888885 = 88887 * 100002
8888885 = 888887 * 1000002
88888885 = 8888887 * 10000002
888888885 = 88888887 * 100000002
8888888885 = 888888887 * 1000000002
88888888885 = 8888888887 * 10000000002
888888888885 = 88888888887 * 100000000002
8888888888885 = 888888888887 * 1000000000002

Example 7: base 11, family 2{5} (formula: (5*11^n-1)/2)

25 = 3 * 9
255 = 2 * 128
2555 = 3 * 919
25555 = 2 * 12828
255555 = 3 * 91919
2555555 = 2 * 1282828
25555555 = 3 * 9191919
255555555 = 2 * 128282828
2555555555 = 3 * 919191919
25555555555 = 2 * 12828282828
255555555555 = 3 * 91919191919
2555555555555 = 2 * 1282828282828

Example 8: base 12, family {B}9B (formula: 12^(n+2)-25)

9B = 7 * 15
B9B = 11 * AB
BB9B = B7 * 105
BBB9B = 11 * B0AB
BBBB9B = BB7 * 1005
BBBBB9B = 11 * B0B0AB
BBBBBB9B = BBB7 * 10005
BBBBBBB9B = 11 * B0B0B0AB
BBBBBBBB9B = BBBB7 * 100005
BBBBBBBBB9B = 11 * B0B0B0B0AB
BBBBBBBBBB9B = BBBBB7 * 1000005
BBBBBBBBBBB9B = 11 * B0B0B0B0B0AB

Example 9: base 14, family B{0}1 (formula: 11*14^(n+1)+1)

B1 = 5 * 23
B01 = 3 * 395
B001 = 5 * 22B3
B0001 = 3 * 39495
B00001 = 5 * 22B2B3
B000001 = 3 * 3949495
B0000001 = 5 * 22B2B2B3
B00000001 = 3 * 394949495
B000000001 = 5 * 22B2B2B2B3
B0000000001 = 3 * 39494949495
B00000000001 = 5 * 22B2B2B2B2B3
B000000000001 = 3 * 3949494949495

Example 10: base 13, family 3{0}95 (formula: 3*13^(n+2)+122)

395 = 14 * 2B
3095 = 7 * 58A
30095 = 5 * 7A71
300095 = 7 * 5758A
3000095 = 14 * 23A92B
30000095 = 7 * 575758A
300000095 = 5 * 7A527A71
3000000095 = 7 * 57575758A
30000000095 = 14 * 23A923A92B
300000000095 = 7 * 5757575758A
3000000000095 = 5 * 7A527A527A71
30000000000095 = 7 * 575757575758A

Example 11: base 16, family {4}D (formula: (4*16^(n+1)+131)/15)

4D = 7 * B
44D = 3 * 16F
444D = D * 541
4444D = 7 * 9C0B
44444D = 3 * 16C16F
444444D = D * 540541
4444444D = 7 * 9C09C0B
44444444D = 3 * 16C16C16F
444444444D = D * 540540541
4444444444D = 7 * 9C09C09C0B
44444444444D = 3 * 16C16C16C16F
444444444444D = D * 540540540541

Example 12: base 16, family {C}D (formula: (4*16^(n+1)+1)/5)

CD = 5 * 29
CCD = 71 * 1D
CCCD = 1E1 * 6D
CCCCD = 18D * 841
CCCCCD = 64D * 2081
CCCCCCD = 7F01 * 19CD
CCCCCCCD = 1FE01 * 66CD
CCCCCCCCD = 198CD * 80401
CCCCCCCCCD = 664CD * 200801
CCCCCCCCCCD = 7FF001 * 199CCD
CCCCCCCCCCCD = 1FFE001 * 666CCD
CCCCCCCCCCCCD = 1998CCD * 8004001

Example 13: base 17, family 1{9} (formula: (25*17^n-9)/16)

19 = 2 * D
199 = B * 27
1999 = 2 * D4D
19999 = AB * 287
199999 = 2 * D4D4D
1999999 = AAB * 2887
19999999 = 2 * D4D4D4D
199999999 = AAAB * 28887
1999999999 = 2 * D4D4D4D4D
19999999999 = AAAAB * 288887
199999999999 = 2 * D4D4D4D4D4D
1999999999999 = AAAAAB * 2888887

Example 14: base 19, family 1{6} (formula: (4*19^n-1)/3)

16 = 5 * 5
166 = D * 1I
1666 = 5 * 515
16666 = CD * 1II
166666 = 5 * 51515
1666666 = CCD * 1III
16666666 = 5 * 5151515
166666666 = CCCD * 1IIII
1666666666 = 5 * 515151515
16666666666 = CCCCD * 1IIIII
166666666666 = 5 * 51515151515
1666666666666 = CCCCCD * 1IIIIII

Example 15: base 25, family 2{1} (formula: (49*25^n-1)/24)

21 = 3 * H
211 = 14 * 1J
2111 = 2N * HC
21111 = 144 * 1IJ
211111 = 2MN * HCC
2111111 = 1444 * 1IIJ
21111111 = 2MMN * HCCC
211111111 = 14444 * 1IIIJ
2111111111 = 2MMMN * HCCCC
21111111111 = 144444 * 1IIIIJ
211111111111 = 2MMMMN * HCCCCC
2111111111111 = 1444444 * 1IIIIIJ

Example 16: base 36, family O{Z} (formula: 25*36^n-1)

OZ = T * V
OZZ = 4Z * 51
OZZZ = TZ * U1
OZZZZ = 4ZZ * 501
OZZZZZ = TZZ * U01
OZZZZZZ = 4ZZZ * 5001
OZZZZZZZ = TZZZ * U001
OZZZZZZZZ = 4ZZZZ * 50001
OZZZZZZZZZ = TZZZZ * U0001
OZZZZZZZZZZ = 4ZZZZZ * 500001
OZZZZZZZZZZZ = TZZZZZ * U00001
OZZZZZZZZZZZZ = 4ZZZZZZ * 5000001

However, there are also many families, which have no known prime (or PRP) > base, neither can be proven that they only contain composite numbers (only count the numbers > base), e.g.

base 11, family 5{7}
base 13, family 9{5}
base 13, family A{3}A
base 16, family {3}AF
base 16, family {4}DD
base 17, family 1{7}
base 17, family F1{9}
base 18, family C{0}C5

You can try to find a prime (or PRP) > base in these families.

***

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