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A188169
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The number of divisors d of n of the form d == 1 (mod 8).
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12
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1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 2
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OFFSET
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1,9
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COMMENTS
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a(n) >= 1 as the divisor d=1 is always counted.
The largest terms up to n = 10^6 are each equal to 24. Those 8 terms are for n = 675675, 765765, 799425, 855855, 863379, 883575, 945945, or 987525. - Harvey P. Dale, May 31 2017
a(n) can be computed from the prime factorization of n. Let v(n) = (n1, n3, n5, n7) where n_r is the number of divisors of n in class r (mod 8) (we do not care about even remainders). Then if gcd(k, m) = 1 we have v(k) = (k1, k3, k5, k7) so a(k) = k1, v(m) = (m1, m3, m5, m7) so a(m) = k1.
We have a(k*m) = (km)_1 = k1*m1 + k2*m2 + k3*m3 + k4*m4. The other (km)_3..(km)_7 have a similar expression.
If p == 1 (mod 8) then a(p^e) = e + 1 otherwise floor(e/2) + 1. (End)
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LINKS
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FORMULA
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Sum_{k=1..n} a(k) = n*log(n)/8 + c*n + O(n^(1/3)*log(n)), where c = gamma(1,8) - (1 - gamma)/8 = A256781 - (1 - A001620)/8 = 0.735783... (Smith and Subbarao, 1981). - Amiram Eldar, Nov 25 2023
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MAPLE
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sigmamr := proc(n, m, r) local a, d ; a := 0 ; for d in numtheory[divisors](n) do if modp(d, m) = r then a := a+1 ; end if; end do: a; end proc:
A188169 := proc(n) sigmamr(n, 8, 1) ; end proc:
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MATHEMATICA
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Table[Count[Divisors[n], _?(Mod[#, 8]==1&)], {n, 100}] (* Harvey P. Dale, May 31 2017 *)
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PROG
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(PARI) a(n) = {my(d = divisors(n)); #select(x -> x%8 == 1, d)} \\ David A. Corneth, Apr 06 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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