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.

A039752 Ruth-Aaron numbers (2): sum of prime divisors of n = sum of prime divisors of n+1 (both taken with multiplicity).

Original entry on oeis.org

5, 8, 15, 77, 125, 714, 948, 1330, 1520, 1862, 2491, 3248, 4185, 4191, 5405, 5560, 5959, 6867, 8280, 8463, 10647, 12351, 14587, 16932, 17080, 18490, 20450, 24895, 26642, 26649, 28448, 28809, 33019, 37828, 37881, 41261, 42624, 43215, 44831, 44891, 47544, 49240
Offset: 1

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Comments

So called because 714 is Babe Ruth's lifetime home run record, Hank Aaron's 715th home run broke this record and 714 and 715 have the same sum of prime divisors, taken with multiplicity.
An infinite number of terms would follow from A175513 and the assumption of Schinzel's Hypothesis H. - Hans Havermann, Dec 15 2010
A 3109-digit term determined by Jens Kruse Andersen is currently the largest-known. - Hans Havermann, Dec 21 2010
The sum of this sequence's reciprocals is 0.42069... - Hans Havermann, Dec 21 2010
Both 417162 and 417163 are in the sequence. Hence these two numbers along with 417164 constitute a Ruth-Aaron "triple". The smallest member of the next triple is 6913943284. - Hans Havermann, Dec 01 2010, Dec 13 2010
The number of terms <= x is at most O(x (loglog x)^4 / (log x)^2) (Pomerance 1999/2002). - Tomohiro Yamada, Apr 22 2017

Examples

			7129199 (7*11^2*19*443, with 7129200 = 2^4*3*5^2*13*457) is in this sequence because 7+11+11+19+443 = 2+2+2+2+3+5+5+13+457.
		

References

  • John L. Drost, Ruth/Aaron Pairs, J. Recreational Math. 28 (No. 2), 120-122.
  • S. G. Krantz, Mathematical Apocrypha, MAA, 2002, see p. 26.
  • Dana Mackenzie, Homage to an itinerant master, Science 275, p. 759, 1997.

Crossrefs

Programs

  • Maple
    anzahl:=0: n:=4: nr:=0: g:=nops(ifactors(n)[2]):
    s[nr]:=sum(ifactors(n)[2,u][1]*ifactors(n)[2,u][2],u=1..g):
    for j from n+1 to 1000000 do nr:=(nr+1) mod 2: g:=nops(ifactors(j)[2]):
    s[nr]:=sum(ifactors(j)[2,u][1]*ifactors(j)[2,u][2],u=1..g):
    if (s[0]=s[1]) then anzahl):=anzahl+1: print(anzahl,j-1,j,s[0]): end if:
    end do:
    # Paul Weisenhorn, Jul 02 2009
  • Mathematica
    ppf[n_] := Plus @@ ((#[[1]] #[[2]]) & /@ FactorInteger[n]); Select[Range[50000], ppf[#] == ppf[#+1] &] (* Harvey P. Dale, Apr 27 2009 *)
  • PARI
    is_A039752(n)=A001414(n)==A001414(n+1) \\ M. F. Hasler, Mar 01 2014
    
  • Python
    from sympy import factorint
    def aupton(terms):
      alst, k, sopfrk, sopfrkp1 = [], 2, 2, 3
      while len(alst) < terms:
        if sopfrkp1 == sopfrk: alst.append(k)
        k += 1
        fkp1 = factorint(k+1)
        sopfrk, sopfrkp1 = sopfrkp1, sum(p*fkp1[p] for p in fkp1)
      return alst
    print(aupton(42)) # Michael S. Branicky, May 08 2021