A331108 -id:A331108 - cp's OEIS frontend

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

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

Offset: 1

Amiram Eldar, Jan 09 2020

No more terms below 2.8*10^10.

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

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

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]

A376889 Numbers k such that A376888(k) = 2*k.

Original entry on oeis.org

6, 60, 90, 336, 5040, 87360, 764400, 11466000, 620568000, 9478560000, 14217840000, 22805874000

Offset: 1

Amiram Eldar, Oct 08 2024

a(12) > 7*10^10, if it exists.
28279283760000, 282792837600000 and 1583639890560000 are also terms.
k! is a term for k = 3 and 7, and for no other factorial of k < 10^4.

Cf. A376888.
Subsequence of A023196.
Similar sequences: A007357, A038182, A074849, A097464, A331108, A331111.

ff[q_, s_] := (q^(s + 1) - 1)/(q - 1); f[p_, e_] := Module[{k = e, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 0, AppendTo[s, {p^(m - 1)!, r}];]; m++]; Times @@ ff @@@ s]; fsigma[1] = 1; fsigma[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[10^6], fsigma[#] == 2*# &]

fdigits(n) = {my(k = n, m = 2, r, s = []); while([k, r] = divrem(k, m); k != 0 || r != 0, s = concat(s, r); m++); s;}
fsigma(n) = {my(f = factor(n), p = f[, 1], e = f[, 2], d); prod(i = 1, #p, prod(j = 1, #d=fdigits(e[i]), (p[i]^(j!*(d[j]+1)) - 1)/(p[i]^j! - 1)));}
is(k) = fsigma(k) == 2*k;

cp's OEIS Frontend

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

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