A125087 a(1)=1. For n>1, a(n) = the number of terms (from among the first n-1 terms of the sequence) that are equal to at least one exponent in the prime factorization of n.

1, 1, 2, 1, 3, 3, 3, 3, 1, 4, 4, 5, 4, 4, 4, 5, 4, 5, 4, 5, 4, 4, 4, 8, 1, 5, 4, 6, 5, 5, 5, 8, 5, 5, 5, 1, 6, 6, 6, 10, 6, 6, 6, 7, 7, 6, 6, 17, 1, 8, 7, 8, 7, 11, 7, 11, 7, 7, 7, 8, 7, 7, 8, 9, 7, 7, 7, 8, 7, 7, 7, 5, 7, 7, 8, 8, 7, 7, 7, 18, 11, 7, 7, 8, 7, 7, 7, 11, 7, 8, 7, 8, 7, 7, 7, 19, 7, 8, 8, 1, 8

Offset: 1

Author

Leroy Quet, Nov 19 2006

Keywords

• nonn

Examples

			12 = 2^2 * 3^1, So the exponents in the prime-factorization of 12 are 1 and 2. There are 4 terms that equal 1 among the first 11 terms of the sequence. (Those terms are a(1), a(2), a(4), a(9)). There is one term (a(3)) that equals 2. So a(12) = 4+1 = 5.
14 = 2^1 * 7^1. So the only distinct exponent in the prime-factorization of 14 is 1. There are 4 terms from among the first 13 terms of the sequence that equal 1. So a(14) = 4.
		

Links

• Katarzyna Matylla, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A125088.

Programs

Mathematica

f[l_List] :=Append[l, Length @Select[l, MemberQ[Last /@ FactorInteger[Length[l] + 1], # ] &]];Nest[f, {1}, 100] (* Ray Chandler, Nov 21 2006 *)

Maxima

/* program from Katarzyna Matylla, Feb 19 2008: Variable max = how much terms we need. Generating 1000 terms took a few hours: */
"|"(a,b):= if b=0 then true else if a=0 then false else if (floor(b/a))=(b/a) then true else false;
max:1000;
infix("|");
deg(n,p):=if p<2 then 0 else block(d:0, loop, if ((p^(d+1))|n) then d:d+1 else go(end), go(loop), end, d);
f:makelist(0,i,1,max);
dg:makelist(0, i, 1, max);
f[1]:1;
for n:2 through max do block(for i:2 through n do dg[i]:deg(n,i),k:0, loop1, k:k+1, p:2, loop, if (f[k]=dg[p]) then block(f[n]:f[n]+1, p:n+1), if (p<=n) then p:next_prime(p), if (p<=n) then go(loop), if k<(n-1) then go(loop1));

Extensions

Extended by Ray Chandler, Nov 21 2006