A275120 - OEIS

%I #45 Jul 21 2016 16:28:01

%S 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,

%T 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,

%U 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

%N 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.

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

%H Charles R Greathouse IV, <a href="/A275120/b275120.txt">Table of n, a(n) for n = 1..10000</a>

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

%e lcm({}) = lcm({1}) = 1, so a(1) = 2.

%e lcm({1, 2}) = 2, so a(2) = 1.

%e lcm({1, 2, 3}) = 6, so a(3) = 1.

%e lcm({1, 2, 3, 4}) = 12, so a(4) = 1.

%e lcm({1, ..., 5}) = lcm({1, ..., 6}) = 60, so a(5) = 2.

%e lcm({1, ..., 7}) = 420, so a(6) = 1.

%e lcm({1, ..., 8}) = 840, so a(7) = 1.

%e lcm({1, ..., 9}) = lcm({1, ..., 10}) = 2520, so a(8) = 2.

%e lcm({1, ..., 11}) = lcm({1, ..., 12}) = 27720, so a(9) = 2.

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

%o (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

%Y Frequency of given numbers using A003418.

%Y Apart from the first term, same as A057820.

%K nonn

%O 1,1

%A _Tyler Skywalker_, Jul 18 2016