A291485 Numbers m such that sigma(x) = m*(m+1)/2 has at least one solution.

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

Author

Altug Alkan, Aug 24 2017

Keywords

• nonn
• easy

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, ...

Examples

			15 is a term because sigma(54) = sigma(56) = sigma(87) = sigma(95) = A000217(15).
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A000217, A045746.

Programs

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 *)