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A257010
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Number of sequences of positive integers with length 3 and alternant equal to n.
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4
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0, 2, 2, 4, 3, 6, 2, 9, 4, 6, 5, 11, 4, 9, 6, 10, 5, 14, 2, 16, 7, 6, 9, 16, 6, 11, 8, 17, 5, 14, 4, 20, 10, 8, 9, 22, 2, 17, 10, 16, 11, 14, 6, 18, 13, 12, 5, 28, 6, 19, 9, 15, 13, 16, 8, 24, 6, 12, 11, 32, 6, 15, 16, 16, 9, 19, 8, 30, 8, 14, 9, 30, 8, 15, 12, 21, 16, 22
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OFFSET
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3,2
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COMMENTS
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See A257009 for the definition of the alternant of a sequence. The number of sequences of length 1 with given alternant value n is 1, while the number of sequences of length 2 with given alternant value n is d(n), the number of divisors of n (see A000005).
There are infinitely many sequences of length 3 and alternant equal to 2. It is for this reason that the offset is 3.
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LINKS
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FORMULA
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a(n) = Sum_{b=1..n-1} (Dbm (b,b^2+nb+1)-2), where Dbm(b,m) is the number of positive divisors of m that are congruent to 1 modulo b. - Barry R. Smith, Jan 24 2016
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EXAMPLE
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For n=6, the a(6) = 4 sequences with alternant 6 are (1,1,3), (1,3,2), (2,3,1), (3,1,1)
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MAPLE
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Dbm:= proc(b, m) nops(select(t -> (t-1) mod b = 0, numtheory:-divisors(m))) end proc:
seq(add(Dbm(b, b^2+n*b+1)-2, b=1..n-1), n=3..100); # Robert Israel, Jan 24 2016
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MATHEMATICA
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Length3Q[x_, y_] :=
Module[{l = ContinuedFraction[(x[[2]] + 2*x[[1]] + y)/(2*x[[1]])]},
If[OddQ[Length[l]], Return[Length[l] == 3],
If[Last[l] == 1, Return[Length[l] - 1 == 3], Return[Length[l] + 1 == 3]]]];
Table[Length[
Select[Flatten[
Select[
Table[{a, k}, {k,
Select[Range[Ceiling[-Sqrt[n^2 - 4]], Floor[Sqrt[n^2 - 4]]],
Mod[# - n^2 + 4, 2] == 0 &]}, {a,
Select[Divisors[(n^2 - 4 - k^2)/4], # > (Sqrt[n^2 - 4] - k)/2 &]}],
UnsameQ[#, {}] &], 1], Length3Q[#, n] &]], {n, 3, 60}]
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PROG
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(PARI) a(n)={sum(b=1, n-1, sumdiv(b^2+n*b+1, d, (d-1)%b==0) - 2)} \\ Andrew Howroyd, May 01 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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