Jackpot numbers
Write
vertically a number (like 419) and try to win the jackpot (using the hereunder martingale):
4 – a – d –
g – j - m
1 – b - e – h – k - ...
9 – c – f –
i – l - ...
1 – b - e – h – k - ...
1) compute
the terms a, b, c, d, e… in the order shown above;
2) the sequence a, b, c, d, e… is always extended with the smallest prime not occurring before, such that…
3) … in each row, the first (yellow) digit of the next term is the rightmost digit of the preceding term;
4) a jackpot is won when all terms in a column have the same rightmost (blue) digit.
2) the sequence a, b, c, d, e… is always extended with the smallest prime not occurring before, such that…
3) … in each row, the first (yellow) digit of the next term is the rightmost digit of the preceding term;
4) a jackpot is won when all terms in a column have the same rightmost (blue) digit.
4 - 41 - 13 - 3 - 31 - 101
- 103 - 307 - 709
1 - 11 - 17 - 71 - 19 - 907 - 701 - 107 - 719
9 - 97 – 7 - 73 - 37
– 79 - 911 - 109
- 919 =
jackpot!
Indeed,
we have the same digit (9) ending all the terms of the last column.
And it takes 8 steps to win the jackpot with 419.
And it takes 8 steps to win the jackpot with 419.
We can compute the number of such winning steps for any integer k > 9 (and start accordingly a sequence S):
10 => -1 (no number that contains a digit 0 can win a jackpot; we assign -1 to 10 to mark this impossibility – see the two rows generated by 10 hereunder):
0 - ? (no prime starts with 0)
11 => 6
1 - 11 - 17 - 7 - 71 - 101 - 103
1 - 13 - 3 - 31 - 19 - 97 - 73
12 => 2
1 - 11 - 13
2 - 2 - 23
13 => 3
1 - 11 - 13 - 37
3 – 3 - 31 - 17
14 =>
1
1 - 11
4 - 41
15 => 2
1 - 11 - 13
5 - 5 - 53
16 => 1
1 - 11
6 - 61
17 => 10
1 – 11 – 13 – 3 - 31 – 19 – 97 - 701 – 101 – 103 - 307
7 – 7 - 71 – 17 – 73 – 37 – 79 – 907 – 709 – 911 – 107
18 => 2
1 - 11 - 13
8 - 83 - 3
19 => 6
1 - 11 – 13 – 3 - 31 – 19 - 907
9 - 97 – 7 - 71 – 17 – 73 – 37
20 => -1
21 => 2
2 – 2 - 23
1 – 11 - 13
...
100 to 110 => -1
111 => 5
1 - 11 – 19 – 97 – 73 - 37
1 - 13 – 3 - 31 – 101
– 107
1 - 17 – 7 - 71 – 103
- 307
112 =>
8
1 - 11 – 17 – 7 - 71 – 101 – 103 – 311 - 109
1 - 11
4 - 41
15 => 2
1 - 11 - 13
5 - 5 - 53
16 => 1
1 - 11
6 - 61
17 => 10
1 – 11 – 13 – 3 - 31 – 19 – 97 - 701 – 101 – 103 - 307
7 – 7 - 71 – 17 – 73 – 37 – 79 – 907 – 709 – 911 – 107
18 => 2
1 - 11 - 13
8 - 83 - 3
19 => 6
1 - 11 – 13 – 3 - 31 – 19 - 907
9 - 97 – 7 - 71 – 17 – 73 – 37
20 => -1
21 => 2
2 – 2 - 23
1 – 11 - 13
...
100 to 110 => -1
111 => 5
1 - 11 – 19 – 97 – 73 - 37
1 - 11 – 17 – 7 - 71 – 101 – 103 – 311 - 109
125
=> 2
1 - 11 - 13
2 - 2 - 23
5 - 5 - 53
...
Etc.
I guess
the sequence S of “winning steps” starts like that (thank you Carole!):
1 - 11 - 13
...
Etc.
S = -1, 6, 2, 3, 1, 2, 1, 10, 2, 6, -1, 2, 6, 17, 3, 2, 3, 5, 2, 4, -1, 3, 17, 16, 2, 17, 2, 2, 1, 3, -1, 1, 3, 2, 4, 3, 1, 2, 4, 8, -1, 2, 2, 17, 3, 6, 3, 5, 2, 4, -1, 1, 3, 2, 1, 3, 11, 2, 4, 8, -1, 10, 5, 2, 2, 5, 2, 3, 3, 1, -1, 2, 2, 1, 4, 2, 4, 3, 4, 3, -1, 6, 4, 3, 8, 4, 8, 1, 3, 1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 5, 8, 19, 34, 8, 34, 6, 11, 8, -1, 8, 37, 3, 20, 2, 20, 50, 2, 9, -1, 19, 3, 26, 11, 3, 11, 28, 3, 36, -1, 34, 20, 11, 8, 20, 1, 73, 8, 8, -1, 8, 2, 3, 20, 37, 20, 50, 2, 9, -1, 34, 20, 11, 1, 20, 19, 73, 8, 8, -1, 6, 50, 28, 73, 50, 73, 17, 5, 123, -1, 11, 2, 3, 8, 2, 8, 5, 9, 55, -1, 8, 9, 36, 8, 9, 8, 123, 55, 12, ...
Another sequence (T) would be "Smallest integers needing n steps to win a jackpot"; I guess T would start like this:
T = 14, 12, 13, 29, 27, 11, ...
n = 1 2 3 4 5 6 ...
(no n = 7 yet in the first 190 terms of S)
P.-S.
Carole and a friend of mine say that S could start with nine 1s, provided you accept the result 1 for 0 < k < 10. We would then have for S':
S' = 1, 1, 1, 1, 1, 1, 1, 1, 1, -1, 6, 2, 3, 1, 2, 1, 10, 2, 6, -1, 2, 6, 17, 3, 2, 3, 5, 2, 4, -1, 3, 17, 16, 2, 17, 2, 2, 1, 3, -1, 1, 3, 2, 4, 3, 1, 2, 4, 8, -1, 2, 2, 17, 3, 6, 3, 5, 2, 4, -1, 1, 3, 2, 1, 3, 11, 2, 4, 8, -1, 10, 5, 2, 2, 5, 2, 3, 3, 1, -1, 2, 2, 1, 4, 2, 4, 3, 4, 3, -1, 6, 4, 3, 8, 4, 8, 1, 3, 1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 5, 8, 19, 34, 8, 34, 6, 11, 8, -1, 8, 37, 3, 20, 2, 20, 50, 2, 9, -1, 19, 3, 26, 11, 3, 11, 28, 3, 36, -1, 34, 20, 11, 8, 20, 1, 73, 8, 8, -1, 8, 2, 3, 20, 37, 20, 50, 2, 9, -1, 34, 20, 11, 1, 20, 19, 73, 8, 8, -1, 6, 50, 28, 73, 50, 73, 17, 5, 123, -1, 11, 2, 3, 8, 2, 8, 5, 9, 55, -1, 8, 9, 36, 8, 9, 8, 123, 55, 12, ...
... and for T':
T' = 1, 12, 13, 29, 27, 11, ...
n = 1 2 3 4 5 6 ...
What do you think? S or S'? T or T'?
What would T'(7) be?
Ce commentaire a été supprimé par l'auteur.
RépondreSupprimeryes, I think one should let S(n) = 1 when n < 10 is not a prime (for the one iteration to reach the jackpot : all terms (here just 1) are primes ending in the same digit).
Supprimerbut maybe S(n) = 0 when n < 10 is a prime, because then we have this jackpot already at the beginning. For a(k) = smallest n which has S(n) = k, I get your values T'(1..6) and
a(7) = 235, (which you didn't list above)
a(8) = 49,
a(9) = 129,
a(10) = 17,
a(11) = 66,
a(12) = 199,
a(13) = 239,
a(14) = 225,
a(15) = 227,
a(16) = 33,
a(17) = 23,
a(18) = ??? (must be very large, > 10^6 I think !),
a(19) = 113,
a(20) = 124, ...