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A291485
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Numbers m such that sigma(x) = m*(m+1)/2 has at least one solution.
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1
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1, 2, 3, 5, 7, 8, 12, 13, 15, 18, 20, 24, 27, 30, 31, 32, 35, 38, 39, 47, 48, 51, 55, 56, 62, 63, 64, 79, 80, 84, 90, 92, 95, 96, 104, 111, 116, 119, 120, 128, 135, 140, 142, 143, 144, 147, 152, 155, 156, 159, 160, 167, 168, 170, 171, 175, 176, 182, 184, 188, 191, 192, 195, 203, 207, 208
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OFFSET
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1,2
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COMMENTS
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Let b(n) be the smallest k such that sigma(k) is the n-th triangular number, or 0 if no such k exists. For n >= 1, b(n) sequence is 1, 2, 5, 0, 8, 0, 12, 22, 0, 0, 0, 45, 36, 0, 54, 0, 0, 98, 0, 104, 0, 0, 0, 152, 0, 0, 160, 0, 0, 200, ...
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LINKS
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EXAMPLE
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15 is a term because sigma(54) = sigma(56) = sigma(87) = sigma(95) = A000217(15).
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MAPLE
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N:= 1000: # to get all terms <= N
Sigmas:= {seq(numtheory:-sigma(x), x=1..N*(N+1)/2)}:
select(t -> member(t*(t+1)/2, Sigmas), [$1..N]); # Robert Israel, Aug 25 2017
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MATHEMATICA
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invT[n_] := (Sqrt[8*n+1]-1)/2; Union@ Select[invT /@ DivisorSigma[1, Range[ 208*209/2]], IntegerQ[#] && # <= 208 &] (* Giovanni Resta, Aug 25 2017 *)
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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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