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

1,2

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

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

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

Cf. A000203, A000217, A045746.

Sequence in context: A224858 A090704 A112055 * A228898 A352695 A076386

Adjacent sequences: A291482 A291483 A291484 * A291486 A291487 A291488

Keyword

nonn,easy

Author

Altug Alkan, Aug 24 2017

Status

approved