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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
OFFSET
1,2
COMMENTS
From Bernard Schott, Feb 17 2021: (Start)
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(18) = 128, a(22) = 203433. [corrected by Jon E. Schoenfield, Nov 30 2023] (End)
From Jon E. Schoenfield, Dec 02 2023: (Start)
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)
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
Cf. A001694, A025487, A060355, A178810 (maximum size of such runs), A141621.
Sequence in context: A073888 A114642 A200980 * A099433 A373344 A051225
KEYWORD
more,nonn
AUTHOR
Will Nicholes, Jun 16 2010
EXTENSIONS
Minor edits by Ray Chandler, Jul 29 2010
a(6) corrected by Bobby Jacobs, Sep 25 2016
a(12) from Hugo van der Sanden, May 20 2019
a(13)-a(14) from Bernard Schott, Feb 16 2021
STATUS
approved