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Numbers of pairs (i, j), i, j > 1, such that i^j <= n.
4

%I #30 Jan 26 2021 04:00:23

%S 0,0,0,1,1,1,1,2,3,3,3,3,3,3,3,5,5,5,5,5,5,5,5,5,6,6,7,7,7,7,7,8,8,8,

%T 8,9,9,9,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,10,10,10,

%U 10,10,13,13,13,13,13,13,13,13,13,13,13,13,13,13,13,13,13,15,15,15,15,15,15

%N Numbers of pairs (i, j), i, j > 1, such that i^j <= n.

%C These numbers are related to the divergent series r sum(n^(1/k)) = n^(1/2) + n^(1/3) + ... + n^(1/r) for abs(n) > 0 and r=sqrt(n). Notice some numbers are missing, such as 4, 11, 12, 14.

%C Gaps (i.e., a(n) - a(n-1) > 1) occur for values of n > 1 in A117453. a(n) - a(n-1) = number of factors of j > 1, for the j in the pair (i,j) with the smallest value of i. Where n = A117453(x), a(n) = a(n-1) + A175066(x). For example: n = 64, a(64) = 13, a(63) = 10, 13 - 10 = 3; 64 = 2^6, 6 has three factors (2,3,6), corresponding to the three perfect powers for 64 (2^6, 4^3, 8^2). Also, A117453(3) = 64 and A175066(3) = 3. - _Doug Bell_, Jun 23 2015

%H Robert Israel, <a href="/A089361/b089361.txt">Table of n, a(n) for n = 1..10000</a>

%F a(1) = 0; for n > 1, if n not in A001597, a(n) = a(n-1), otherwise a(n) = a(n-1) + number of factors of j > 1 (A000005(j) - 1), for the j in the positive integer pair (i,j) where i^j = n with the smallest value of i. - _Doug Bell_, Jun 23 2015

%F a(n) = Sum(j = 2 .. floor(log_2(n)), floor(n^(1/j) - 1)). - _Robert Israel_, Jun 24 2015

%e There are 5 perfect powers greater than 1 that are less than or equal to 16: 2^2, 2^3, 2^4, 3^2, 4^2, ergo the first 5 in the table.

%p N:= 1000; # to get a(1) to a(N)

%p B:= Vector(N);

%p for i from 2 to floor(sqrt(N)) do

%p for j from 2 while i^j <= N do

%p B[i^j]:= B[i^j]+1

%p od

%p od:

%p convert(map(round,Statistics:-CumulativeSum(B)),list); # _Robert Israel_, Jun 24 2015

%o (PARI) plessn(n,m) = { for(k=1,n, s=0; r = floor(sqrt(k)); for(x=m,r, for(y=2,r, p = floor(x^y); if(p<=k,s++) ) ); print1(s",") ) }

%o (PARI) A = vector(100); for (p = 2, 6, i = 2; while (i^p <= 100, A[i^p]++; i++)); for (n = 2, 100, A[n] += A[n - 1]); \\ _David Wasserman_

%Y Cf. A001597, A117453, A175066. - _Doug Bell_, Jun 23 2015

%K easy,nonn

%O 1,8

%A _Cino Hilliard_, Dec 27 2003