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A338434
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Sum of the numbers less than or equal to sqrt(n) whose square does not divide n.
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6
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0, 0, 0, 0, 2, 2, 2, 0, 2, 5, 5, 3, 5, 5, 5, 3, 9, 6, 9, 7, 9, 9, 9, 7, 9, 14, 11, 12, 14, 14, 14, 8, 14, 14, 14, 9, 20, 20, 20, 18, 20, 20, 20, 18, 17, 20, 20, 14, 20, 22, 27, 25, 27, 24, 27, 25, 27, 27, 27, 25, 27, 27, 24, 21, 35, 35, 35, 33, 35, 35, 35, 24, 35, 35, 30, 33, 35, 35, 35, 29, 32, 44, 44, 42, 44, 44, 44, 42, 44, 41, 44, 42, 44, 44, 44, 38, 44, 37, 41, 37
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OFFSET
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1,5
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COMMENTS
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Equal to sum of the numbers less than sqrt(n) whose square does not divide n. - Michel Marcus and Chai Wah Wu, Feb 07 2021
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LINKS
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FORMULA
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a(n) = Sum_{k=1..floor(sqrt(n))} k * (ceiling(n/k^2) - floor(n/k^2)).
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EXAMPLE
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a(9) = 2; floor(sqrt(n)) = 3 and 2^2 does not divide 9, so a(9) = 2.
a(10) = 5; floor(sqrt(n)) = 3 and the squares of 2 and 3 do not divide 10, so a(10) = 2 + 3 = 5.
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MATHEMATICA
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Table[Sum[k (Ceiling[n/k^2] - Floor[n/k^2]), {k, Sqrt[n]}], {n, 100}]
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PROG
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(PARI) a(n) = sum(k=1, sqrtint(n), if (n % k^2, k)); \\ Michel Marcus, Jan 31 2021
(Python)
from sympy import divisor_sigma, integer_nthroot
from sympy.ntheory.factor_ import core
m = integer_nthroot(n, 2)[0]
return m*(m+1)//2-divisor_sigma(integer_nthroot(n//core(n, 2), 2)[0]) # Chai Wah Wu, Jan 31 2021
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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