

A291485


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


0



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

1,2


COMMENTS

Let b(n) be the smallest k such that sigma(k) is the nth 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

Table of n, a(n) for n=1..66.


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 A076386 A177801
Adjacent sequences: A291482 A291483 A291484 * A291486 A291487 A291488


KEYWORD

nonn,easy


AUTHOR

Altug Alkan, Aug 24 2017


STATUS

approved



