A331111 Dual-Zeckendorf-infinitary perfect numbers: numbers k such that A331110(k) = 2*k.

6, 60, 90, 655200, 28828800, 238140000, 10478160000

Offset: 1

Author

Amiram Eldar, Jan 09 2020

Keywords

• nonn
• more

Comments

No more terms below 2.8*10^10.

Examples

			6 is a term since A331110(6) = 12 = 2*6.
		

Crossrefs

Cf. A007357, A038182, A074849, A097464, A331108, A331110.

Programs

Mathematica

fibTerms[n_] := Module[{k = Ceiling[Log[GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; fr];
dualZeck[n_] := Module[{v = fibTerms[n]}, nv = Length[v]; i = 1; While[i <= nv - 2, If[v[[i]] == 1 && v[[i + 1]] == 0 && v[[i + 2]] == 0, v[[i]] = 0; v[[i + 1]] = 1; v[[i + 2]] = 1; If[i > 2, i -= 3]]; i++]; i = Position[v, _?(# > 0 &)]; If[i == {}, {}, v[[i[[1, 1]] ;; -1]]]];
f[p_, e_] := p^Fibonacci[1 + Position[Reverse @ dualZeck[e], _?(# == 1 &)]];
dzsigma[1] = 1; dzsigma[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); seqQ[n_] := dzsigma[n] == 2n; Select[Range[10^6], seqQ]