A245636 Number of terms of A245630 <= n.

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7

Offset

1,6

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Paul Erdős, Solution to Advanced Problem 4413, American Mathematical Monthly, 59 (1952) 259-261.

Formula

As n -> infinity, a(n)/sqrt(n) -> Product_{i=1..infinity} (1 - 1/prime(i))/(1 - (prime(i)*prime(i+1))^(-1/2)), see Erdős reference.

Example

The first two terms of A245630 are 1 and 6, so a(n) = 1 for 1 <= n <= 5 and a(6) = 2.

Maple

N:= 10^4: # to get a(1) to a(N)
PP:= [seq(ithprime(i)*ithprime(i+1), i=1.. numtheory[pi](floor(sqrt(N)))-1)]:
ext:= (x, p) -> seq(x*p^i, i=0..floor(log[p](N/x))):
S:= {1}: for i from 1 to nops(PP) do S:= map(ext, S, PP[i]) od:
E:= Array(1..N):
for s in S do E[s]:= 1 od:
A:= map(round, Statistics:-CumulativeSum(E)):
seq(A(i), i=1..N);

Mathematica

M = 10^4;
T = Table[Prime[n] Prime[n+1], {n, 1, PrimePi[Sqrt[M]]}];
T2 = Select[Join[T, T^2], # <= M&];
S = Join[{1}, T2 //. {a___, b_, c___, d_, e___} /; b*d <= M && FreeQ[{a, b, c, d, e}, b*d] :> Sort[{a, b, c, d, e, b*d}]];
ee = Table[0, {M}];
Scan[Set[ee[[#]], 1]&, S];
Accumulate[ee] (* Jean-François Alcover, Apr 17 2019 *)

Crossrefs

Cf. A006094, A028260, A245630.

Sequence in context: A309432 A230261 A242306 * A171622 A351846 A111858

Adjacent sequences: A245633 A245634 A245635 * A245637 A245638 A245639

Keyword

nonn

Author

Robert Israel, Jul 28 2014

Status

approved