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.

A272818 Numbers such that (sum + product) of all their prime factors equals (sum + product) of all exponents in their prime factorization.

Original entry on oeis.org

1, 4, 27, 72, 96, 108, 486, 800, 1280, 3125, 6272, 10976, 12500, 14336, 21600, 30375, 36000, 48600, 51840, 54675, 69120, 84375, 121500, 134456, 169344, 174960, 192000, 225000, 240000, 247808, 337500, 340736, 395136, 435456, 451584, 703125, 750141, 781250, 787320, 823543, 857304, 885735
Offset: 1

Views

Author

Giuseppe Coppoletta, May 08 2016

Keywords

Comments

For p prime, p^p satisfy the condition, hence A051674 (and also A048102) is a subsequence. Moreover, if p and q are primes and i and j are positive integers, if p^i * q^j verify the condition, then the same is true for p^j * q^i. So A122406 is also a subsequence. More generally, if a number is a term, then any permutation of the exponents in its prime factorization (i.e. any permutation of its prime signature) gives also a term. In addition, any number having no more than two distinct prime factors (apart their multiplicity) is a term iff it belongs also to A272858.

Examples

			885735 = 3^11 * 5 is included because (3+5) + 3*5 = (11+1) + 11*1.
2^10 * 3^6 * 19^2 is included because (2+3+19)+ 2*3*19 = (10+6+2)+ 10*6*2.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[10^6], Total@ First@ # + Times @@ First@ # == Total@ Last@ # + Times @@ Last@ # &@ Transpose@ FactorInteger@ # &] (* Michael De Vlieger, May 08 2016 *)
  • PARI
    spp(v) = vecsum(v) + prod(k=1, #v, v[k]);
    isok(n) = my(f = factor(n)); spp(f[,1]) == spp(f[,2]); \\ Michel Marcus, May 08 2016
  • Sage
    def d(n):
        v = factor(n)
        d1 = sum(w[0] for w in v) + prod(w[0] for w in v)
        d2 = sum(w[1] for w in v) + prod(w[1] for w in v)
        return d1 == d2
    [k for k in (1..10000) if d(k)]