cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A331108 Zeckendorf-infinitary perfect numbers: numbers k such that A331107(k) = 2*k.

Original entry on oeis.org

6, 60, 90, 3024, 133056, 1330560, 6879600, 28828800, 302702400, 698544000, 11763214848
Offset: 1

Views

Author

Amiram Eldar, Jan 09 2020

Keywords

Comments

No more terms below 4*10^10.

Examples

			6 is a term since A331107(6) = 12 = 2*6.

Crossrefs

Cf. A007357, A038182, A074849, A097464, A331107.

Programs

  • Mathematica
    fb[n_] := Block[{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--]; Fibonacci[1 + Position[Reverse@fr, ?(# == 1 &)]]]; f[p, e_] := p^fb[e]; zsigma[1] = 1; zsigma[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); zPerfectQ[n_] := zsigma[n] == 2 n; Select[Range[10^4], zPerfectQ] (* after Robert G. Wilson v at A014417 *)

A376888 The sum of divisors of n that are products of factors of the form p^(k!) with multiplicities not larger than their multiplicity in n, where p is a prime and k >= 1, when the factorization of n is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.

Original entry on oeis.org

1, 3, 4, 5, 6, 12, 8, 15, 10, 18, 12, 20, 14, 24, 24, 21, 18, 30, 20, 30, 32, 36, 24, 60, 26, 42, 40, 40, 30, 72, 32, 63, 48, 54, 48, 50, 38, 60, 56, 90, 42, 96, 44, 60, 60, 72, 48, 84, 50, 78, 72, 70, 54, 120, 72, 120, 80, 90, 60, 120, 62, 96, 80, 65, 84, 144
Offset: 1

Views

Author

Amiram Eldar, Oct 08 2024

Keywords

Comments

See A376885 for details about this factorization.
First differs from A188999 at n = 16.
The number of these divisors is given by A376887(n).

Examples

			For n = 12 = 2^2 * 3^1, the representation of 2 in factorial base is 10, i.e., 2 = 2!, so 12 = (2^(2!))^1 * (3^(1!))^1 and a(12) is the sum of the 4 divisors 1 + 3 + 4 + 12 = 20.

Links

Crossrefs

Cf. A188999, A376885, A376886, A376887.
Similar sequences: A049417, A049418, A074847, A097863, A331107, A331110.

Programs

  • Mathematica
    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]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    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;}
a(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)));}

Formula

Multiplicative: if e = Sum_{k>=1} d_k * k! (factorial base representation), then a(p^e) = Product_{k>=1} (p^(k!*{d_k+1}) - 1)/(p^(k!) - 1).

A331110 The sum of dual-Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the dual Zeckendorf expansion (A104326) of each s(i) contains only terms that are in the dual Zeckendorf expansion of r(i).

Original entry on oeis.org

1, 3, 4, 5, 6, 12, 8, 15, 10, 18, 12, 20, 14, 24, 24, 27, 18, 30, 20, 30, 32, 36, 24, 60, 26, 42, 40, 40, 30, 72, 32, 45, 48, 54, 48, 50, 38, 60, 56, 90, 42, 96, 44, 60, 60, 72, 48, 108, 50, 78, 72, 70, 54, 120, 72, 120, 80, 90, 60, 120, 62, 96, 80, 135, 84, 144
Offset: 1

Views

Author

Amiram Eldar, Jan 09 2020

Keywords

Comments

First differs from A188999 at n = 32.

Examples

			a(32) = 45 since 32 = 2^5 and the dual Zeckendorf expansion of 5 is 110, i.e., its dual Zeckendorf representation is a set with 2 terms: {2, 3}. There are 4 possible exponents of 2: 0, 2, 3 and 5, corresponding to the subsets {}, {2}, {3} and {2, 3}. Thus 32 has 4 dual-Zeckendorf-infinitary divisors: 2^0 = 1, 2^2 = 4, 2^3 = 8, and 2^5 = 32, and their sum is 1 + 4 + 8 + 32 = 45.

Links

Crossrefs

The number of dual-Zeckendorf-infinitary divisors of n is in A331109.
Cf. A049417, A049418, A074847, A097863, A104326, A188999, A331107.

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 &)]];
a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100]

Formula

Multiplicative with a(p^e) = Product_{i} (p^s(i) + 1), where s(i) are the terms in the dual Zeckendorf representation of e (A104326).
Showing 1-3 of 3 results.

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