A062762 Number of powerful numbers not exceeding 2^n.
1, 1, 2, 3, 5, 8, 11, 18, 26, 38, 55, 80, 116, 166, 240, 345, 497, 710, 1016, 1453, 2073, 2955, 4211, 5992, 8523, 12111, 17202, 24423, 34648, 49152, 69694, 98795, 140009, 198378, 281016, 398002, 563612, 797999, 1129737, 1599166, 2263457, 3203381
Offset: 0
Keywords
Examples
Below 128, the 18 powerful numbers {1,4,8,9,16,25,...,100,108,121,125,128} can be found, so a(7)=18.
Links
- Chai Wah Wu, Table of n, a(n) for n = 0..127 (terms 0..90 from Daniel Suteu)
Programs
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Mathematica
nn = 41; s = Union@ Flatten@ Table[a^2*b^3, {b, (2^nn)^(1/3)}, {a, Sqrt[(2^nn)/b^3]}]; Table[FirstPosition[s, 2^k][[1]], {k, 2, nn}] (* Michael De Vlieger, Oct 29 2023 *) -
PARI
a(n) = my(s=0,N=2^n); forsquarefree(k=1, sqrtnint(N, 3), s += sqrtint(N\k[1]^3)); s; \\ Daniel Suteu, Feb 18 2020
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Python
from math import isqrt from sympy import mobius, integer_nthroot def A062762(n): def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+1))) m = 1<
1: k2 = integer_nthroot(m//j**2,3)[0]+1 w = squarefreepi(k2-1) c += j*(w-l) l, j = w, isqrt(m//k2**3) return c-l # Chai Wah Wu, Sep 13 2024
Formula
a(n) = Sum_{k=0..n} A062761(k). - Daniel Suteu, Feb 18 2020
Extensions
a(19)-a(41) from Donovan Johnson, Oct 01 2009
Comments