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A078181
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a(n) = Sum_{d|n, d==1(mod 3)} d.
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16
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1, 1, 1, 5, 1, 1, 8, 5, 1, 11, 1, 5, 14, 8, 1, 21, 1, 1, 20, 15, 8, 23, 1, 5, 26, 14, 1, 40, 1, 11, 32, 21, 1, 35, 8, 5, 38, 20, 14, 55, 1, 8, 44, 27, 1, 47, 1, 21, 57, 36, 1, 70, 1, 1, 56, 40, 20, 59, 1, 15, 62, 32, 8, 85, 14, 23, 68, 39, 1, 88, 1, 5, 74, 38, 26, 100, 8, 14, 80, 71, 1
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OFFSET
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1,4
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LINKS
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Seiichi Manyama, Table of n, a(n) for n = 1..10000
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FORMULA
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G.f.: Sum_{n>=0} (3*n+1)*x^(3*n+1)/(1-x^(3*n+1)).
G.f.: -q*P'/P where P = Product_{n>=0} (1 - q^(3*n+1)). - Joerg Arndt, Aug 03 2011
Conjecture. If a(n)=n+1 then n==1 (mod 3). (Is this easy to settle? It has been verified for n=1,2,3,...,2000.) - John W. Layman, Apr 03 2006
The conjecture is false. The first and only counterexample below 10^8 is a(6800) = 6801 and 6800 == 2 (mod 3). - Lambert Herrgesell (zero815(AT)googlemail.com), May 06 2008
Equals A051731 * [1, 0, 0, 4, 0, 0, 7, 0, 0, 10, ...]. - Gary W. Adamson, Nov 06 2007
A272027(n/3) + a(n) + A078182(n) = A000203(n). - R. J. Mathar, May 25 2020
G.f.: Sum_{n >= 1} x^n*(1 + 2*x^(3*n))/(1 - x^(3*n))^2. - Peter Bala, Dec 19 2021
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MAPLE
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A078181 := proc(n)
a := 0 ;
for d in numtheory[divisors](n) do
if modp(d, 3) =1 then
a :=a+d ;
end if;
end do:
a;
end proc: # R. J. Mathar, May 11 2016
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MATHEMATICA
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a[n_] := Plus @@ Select[Divisors[n], Mod[#, 3] == 1 &]; Array[a, 100] (* Giovanni Resta, May 11 2016 *)
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CROSSREFS
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Cf. A001817, A001822, A035382, A051731, A272716.
Cf. Sum_{d|n, d==1 mod k} d: A000593 (k=2), this sequence (k=3), A050449 (k=4), A284097 (k=5), A284098 (k=6), A284099 (k=7), A284100 (k=8).
Sequence in context: A299458 A300096 A294296 * A054110 A132048 A141398
Adjacent sequences: A078178 A078179 A078180 * A078182 A078183 A078184
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KEYWORD
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easy,nonn
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AUTHOR
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Vladeta Jovovic, Nov 21 2002
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STATUS
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approved
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