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A069212
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a(n) = Sum_{k=1..n} 3^omega(k).
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6
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1, 4, 7, 10, 13, 22, 25, 28, 31, 40, 43, 52, 55, 64, 73, 76, 79, 88, 91, 100, 109, 118, 121, 130, 133, 142, 145, 154, 157, 184, 187, 190, 199, 208, 217, 226, 229, 238, 247, 256, 259, 286, 289, 298, 307, 316, 319, 328, 331, 340, 349, 358, 361, 370, 379, 388, 397
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OFFSET
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1,2
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COMMENTS
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More generally, if b is an integer =>3, Sum_{k=1..n} b^omega(k) ~ C(b)*n*log(n)^(b-1) where C(b)=1/(b-1)!*prod((1-1/p)^(b-1)*(1+(b-1)/p)).
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REFERENCES
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G. Tenenbaum and Jie Wu, Cours Spécialisés No. 2: "Théorie analytique et probabiliste des nombres", Collection SMF, Ordres moyens, p. 20.
G. Tenenbaum, Introduction to analytic and probabilistic number theory, 3rd ed., American Mathematical Soc. (2015). See page 59.
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LINKS
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FORMULA
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Asymptotic formula: a(n) ~ C*n*log(n)^2 with C = (1/2) * Product_{p} ((1-1/p)^2*(1+2/p)) where the product is over all the primes.
a(n) = Sum_{i=1..n} Sum_{j=1..n} mu(i*j)^2*floor(n/(i*j));
a(n) = Sum_{i=1..n} mu(i)^2*tau(i)*floor(n/i);
a(n) = Sum_{i=1..n} 2^Omega(i)*mu(i)^2*floor(n/i), where Omega = A001222. (End)
More precise asymptotics:
Let f(s) = Product_{primes p} (1 - 3/p^(2*s) + 2/p^(3*s)), then
a(n) ~ n * (f(1)*log(n)^2/2 + log(n)*((3*gamma - 1)*f(1) + f'(1)) + f(1)*(1 - 3*gamma + 3*gamma^2 - 3*sg1) + (3*gamma - 1)*f'(1) + f''(1)/2),
where f(1) = A065473 = Product_{primes p} (1 - 3/p^2 + 2/p^3) = 0.2867474284344787341078927127898384464343318440970569956414778593366522...,
f'(1) = f(1) * Sum_{primes p} 6*log(p) / (p^2 + p - 2) = 0.8023233847630974628467999132875783526536954420333140745016349208975965...,
f''(1) = f'(1)^2/f(1) + f(1) * Sum_{primes p} -6*p*(2*p+1) * log(p)^2 / (p^2 + p - 2)^2 = -0.255987592484328884627082229528266165335336670389046663124468278519...
and gamma is the Euler-Mascheroni constant A001620 and sg1 is the first Stieltjes constant (see A082633). (End)
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MATHEMATICA
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Accumulate @ Table[3^PrimeNu[n], {n, 1, 57}] (* Amiram Eldar, May 24 2020 *)
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PROG
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(Python)
from sympy.ntheory.factor_ import primenu
def A069212(n): return sum(3**primenu(m) for m in range(1, n+1)) # Chai Wah Wu, Sep 07 2023
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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