|
| |
|
|
A178811
|
|
The smallest integer that begins the longest run of consecutive integers with the prime signature of A025487(n).
|
|
1
|
|
|
|
1, 2, 4, 33, 8, 10093613546512321, 16, 28375, 1309, 32, 36, 7939375, 932537185321, 64
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
The corresponding lengths of these longest runs of consecutive integers are in A178810.
If a(n) = 2^k for some k <> 1, then a(n) = A025487(n) and A178810(n) = 1; for k = 1, a(2) = A025487(2) = A178810(2) = 2 because there exists a run of two consecutive primes (2,3).
a(16) = 3302209375 = 5^5 * 1056707. (3302209376 = 2^5 * 103194043, 3302209377 = 3^5 * 13589339.) No run of four consecutive integers of the form p^5 * q with p,q distinct primes can exist: one of the two even numbers would be 32*q and the other would be 2*p^5, and they would differ by 2, yielding either (1) 2*p^5 + 2 = 32*q -> p^5 + 1 = 16*q -> (p^4 - p^3 + p^2 - p + 1)*(p+1) = 16*q, so p+1 = 16, but then p = 15 would not be a prime, or (2) 2*p^5 - 2 = 32*q -> p^5 - 1 = 16*q -> (p^4 + p^3 + p^2 + p + 1)*(p-1) = 16*q, so p-1 = 16, so p = 17, but then the odd number between 2*p^5 and 32*q would be 2*17^5 - 1 = 2839713 = 3 * 37 * 25583 (which would not have the required prime signature).
a(17) <= 921198089181020748838245 (which starts a run of seven consecutive integers of the form p^3*q*r; no run of eight or more can exist, as any set of eight consecutive integers includes an odd multiple of 4).
a(n) = A025487(n) if A025487(n) is a proper power (i.e., a number of the form b^e where b,e > 1). (Thus a(3) = 4, a(5) = 8, a(7) = 16, a(10) = 32, a(11) = 36, a(14) = 64, a(18) = 128, a(19) = 144, a(23) = 216, a(25) = 256, a(32) = 512, a(33) = 576, a(38) = 900, a(40) = 1024, a(44) = 1296, a(48) = 1728, a(51) = 2048, a(53) = 2304.)
Conjecture: a(n) = A025487(n) if A025487(n) is a powerful number (A001694); i.e., if A025487(n) is a powerful number, then there exists no run of two or more consecutive integers with the same prime signature as that of A025487(n). (E.g., if this conjecture holds, a(15) = 72 (cf. A367781), a(26) = 288, a(30) = 432, a(37) = 864, a(42) = 1152, a(49) = 1800.) (End)
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
For n = 3, A025487(3) = 4, corresponding to a prime signature of {2}. Since the maximum number of consecutive integers with that prime signature is 1, a(3) is 4, the smallest integer that starts a "run" of 1.
A025487(4) = 6 whose prime signature is {1,1}; a(4) = 33 because 33 is the smallest integer where starts a run of A178810(4) = 3 consecutive integers with prime signature {1,1}: (33=3*11, 34=2*17, 35=5*7). - Bernard Schott, Feb 16 2021
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
more,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
