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.

Showing 1-2 of 2 results.

A073395 Product of sums of prime factors of n: with and without repetition.

Original entry on oeis.org

0, 4, 9, 8, 25, 25, 49, 12, 18, 49, 121, 35, 169, 81, 64, 16, 289, 40, 361, 63, 100, 169, 529, 45, 50, 225, 27, 99, 841, 100, 961, 20, 196, 361, 144, 50, 1369, 441, 256, 77, 1681, 144, 1849, 195, 88, 625, 2209, 55, 98, 84, 400, 255, 2809, 55, 256, 117, 484
Offset: 1

Views

Author

Reinhard Zumkeller, Jul 30 2002

Keywords

Examples

			a(15) = (3+5)*(3+5) = 8*8 = 64 (n squarefree); a(16) = (2)*(2+2+2+2) = 2*8 = 16 (n prime-power); a(17) = (17)*(17) = 289 (n prime); a(18) = (2+3)*(2+3+3) = 5*8 = 40.
		

Crossrefs

Programs

  • Haskell
    a073395 n = a008472 n * a001414 n
    -- Reinhard Zumkeller, Oct 28 2015 (fixed), Mar 29 2012
    
  • Mathematica
    a[n_] := With[{fi = FactorInteger[n]}, Plus @@ fi[[All, 1]]*Plus @@ Apply[Times, fi, 1]]; a[1]=0; Table[a[n], {n, 1, 57}] (* Jean-François Alcover, Jul 19 2012 *)
  • Python
    from sympy import primefactors, factorint
    def a001414(n):
        f=factorint(n)
        return sum([i*f[i] for i in f])
    def a(n): return 0 if n==1 else sum(primefactors(n))*a001414(n)
    print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 23 2017

Formula

a(n) = A008472(n)*A001414(n).
a(m) is a square for all squarefree numbers m.
a(p^k) = k*p^2 for primes p.

A237798 Numbers k such that phi(k) equals the product of two numbers: sums of prime factors of k, with and without repetition.

Original entry on oeis.org

750, 5022, 8037, 18250, 60249, 105669, 1760490, 1965270, 4626489, 62546070, 202631245, 441416073, 286952692809
Offset: 1

Views

Author

Giovanni Resta, Feb 13 2014

Keywords

Comments

25446095286407049 and 27093686218810118911929 are terms.

Examples

			750 = 2*3*5^3 is a term since phi(750) = 200 = (2+3+5) * (2+3+5+5+5).
		

Crossrefs

Programs

  • Mathematica
    Select[Range[2, 10^5], (f=FactorInteger@#; EulerPhi[#] == Total[First /@ f]* Total[Times @@@ f]) &]
Showing 1-2 of 2 results.