OFFSET
1,1
COMMENTS
a(17) > 10^7.
a(18) = 127890361; a(n) > 4*10^8 for n=17 and for n >= 19. - Jon E. Schoenfield, Jul 16 2017
a(n) > 6*10^12 for n >= 25. - Giovanni Resta, Jul 16 2019
REFERENCES
M. F. Hasler, Posting to Sequence Fans Mailing List, May 06 2017
EXAMPLE
We have omega(10) = 2, omega(11) = 1, omega(12) = 2, omega(13) = 1. Therefore 11 starts a run of exactly 2 consecutive integers (11, 12) which have nondecreasing (here: strictly increasing) values of omega.
The 6 numbers from 1 through 6 yield values (0, 1, ..., 1, 2) for omega, therefore a(6) = 1. The 4 numbers from 7 through 10 yield values (1, 1, 1, 2) for omega, therefore a(4) = 7.
A run of length 1 is a single number n such that omega(n-1) > omega(n) > omega(n+1). (If we had "<=" in one of the cases, it would be part of a run of at least 2 numbers with nondecreasing omega.) This first happens for a(1) = 106.
MATHEMATICA
Prepend[#, Module[{k = 2}, While[Sign@ Differences@ PrimeNu[k + {-1, 0, 1}] != {-1, -1}, k++]; k]] &@ Function[s, Function[r, If[Length@ # > 0, #[[1, 1]], -1] &@ Select[s, Length@ # == r &]] /@ Range@ Max@ Map[Length, s]]@ DeleteCases[SplitBy[MapIndexed[Function[k, (2 Boole[#1 <= #2] - 1) k & @@ #1]@ First@ #2 &, Partition[Array[PrimeNu, 10^7], 2, 1]], Sign], w_ /; First@ w < 0] (* Michael De Vlieger, May 19 2017 *)
PROG
(PARI) alias("A", "A286287"); A=vector(19); apply(scan(N, s=1, t=omega(s))=for(k=s+1, N, t>(t=omega(k))||next; k-s>#A||A[k-s]||printf(" a(%d)=%d, ", k-s, s)||A[k-s]=s; s=k); done, [4e6]) \\ Then the search may be extended using scan(END, START). - M. F. Hasler, May 16 2017
CROSSREFS
KEYWORD
nonn,more
AUTHOR
N. J. A. Sloane, May 16 2017
EXTENSIONS
Edited by M. F. Hasler, May 16 2017
a(17)-a(24) from Giovanni Resta, Jul 16 2019
STATUS
approved