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A291485
Numbers m such that sigma(x) = m*(m+1)/2 has at least one solution.
1
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
OFFSET
1,2
COMMENTS
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, ...
LINKS
EXAMPLE
15 is a term because sigma(54) = sigma(56) = sigma(87) = sigma(95) = A000217(15).
MAPLE
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
MATHEMATICA
invT[n_] := (Sqrt[8*n+1]-1)/2; Union@ Select[invT /@ DivisorSigma[1, Range[ 208*209/2]], IntegerQ[#] && # <= 208 &] (* Giovanni Resta, Aug 25 2017 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Altug Alkan, Aug 24 2017
STATUS
approved