A275120 List the least common multiples of {1, 2, ..., k} for k = 0, 1, ...; this sequence gives the length of the n-th block of consecutive equal numbers.

2, 1, 1, 1, 2, 1, 1, 2, 2, 3, 1, 2, 4, 2, 2, 2, 2, 1, 5, 4, 2, 4, 2, 4, 6, 2, 3, 3, 4, 2, 6, 2, 2, 6, 8, 4, 2, 4, 2, 4, 8, 4, 2, 1, 3, 6, 2, 10, 2, 6, 6, 4, 2, 4, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4, 6, 2, 2, 8, 5, 1, 6, 6, 2, 6, 4, 2, 6, 4, 14, 4, 2, 4, 14, 6, 6, 4, 2, 4, 6, 2, 6, 6, 6, 4, 6

Offset: 1

Author

Tyler Skywalker, Jul 18 2016

Keywords

• nonn

Comments

a(n) is the count of how many consecutive terms in A003418 are equal.

Examples

			lcm({}) = lcm({1}) = 1, so a(1) = 2.
lcm({1, 2}) = 2, so a(2) = 1.
lcm({1, 2, 3}) = 6, so a(3) = 1.
lcm({1, 2, 3, 4}) = 12, so a(4) = 1.
lcm({1, ..., 5}) = lcm({1, ..., 6}) = 60, so a(5) = 2.
lcm({1, ..., 7}) = 420, so a(6) = 1.
lcm({1, ..., 8}) = 840, so a(7) = 1.
lcm({1, ..., 9}) = lcm({1, ..., 10}) = 2520, so a(8) = 2.
lcm({1, ..., 11}) = lcm({1, ..., 12}) = 27720, so a(9) = 2.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Frequency of given numbers using A003418.

Apart from the first term, same as A057820.

Programs

Mathematica

{2}~Join~Rest@ Most@ Map[Length, Split@ Table[LCM @@ Range@ n, {n, 396}]] (* Michael De Vlieger, Jul 18 2016 *)

PARI

do(lim)=my(v=List()); for(e=2,logint(lim\=1,2), forprime(p=2,sqrtnint(lim,e), listput(v,p^e))); v=Set(concat(Vec(v), primes([2,lim]))); concat(2, vector(#v-1,i,v[i+1]-v[i])) \\ Charles R Greathouse IV, Jul 18 2016

Formula

a(n) = A057820(n), n>1.