|
| |
|
|
A359257
|
|
First differences of A002476.
|
|
1
|
|
|
|
6, 6, 12, 6, 6, 18, 6, 6, 6, 18, 6, 6, 18, 12, 12, 6, 6, 18, 12, 6, 12, 12, 6, 12, 30, 6, 6, 24, 6, 18, 6, 12, 18, 6, 6, 18, 12, 12, 12, 6, 18, 6, 24, 12, 24, 18, 6, 24, 6, 24, 6, 6, 6, 12, 12, 18, 12, 18, 18, 18, 6, 6, 12, 6, 12, 18, 24, 12, 6, 24, 6, 18, 6, 24, 12, 18, 30, 24, 6
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Each term of this sequence is a multiple of 6.
Since no term of A002476 ends in 5, the longest run of 6's in this sequence will have length 3 (e.g., 61, 67, 73, 79 in A002476), the longest run of 12's will have length 3 (e.g., 397, 409, 421, 433 in A002476), the longest run of 18's will have length 3 (e.g., 673, 691, 709, 727 in A002476), and the longest run of 24's will have length 3 (e.g., 14149, 14173, 14197, 14221 in A002476). This run limit of length 3 also extends to other multiples of 6 that are not divisible by 5.
For multiples of 6 that are divisible by 5, the length of the longest run does not appear to be bounded.
Of course there cannot be 6 or more consecutive 30s in this sequence because then one of the primes must be divisible by 7, but there could be up to 10 consecutive 210s. The first run of four 30s corresponds to the primes 320149, 320179, 320209, 320239, 320269 and the first run of five 30s corresponds to the primes 28204591, 28204621, 28204651, 28204681, 28204711, 28204741. - Charles R Greathouse IV, Jan 27 2023
|
|
|
LINKS
|
|
|
|
FORMULA
|
|
|
|
EXAMPLE
|
|
|
|
MATHEMATICA
|
|
|
|
PROG
|
(PARI) do(lim)=my(v=List(), p=7); forprimestep(q=13, lim\1, 6, listput(v, q-p); p=q); Vec(v) \\ Charles R Greathouse IV, Jan 27 2023
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
