OFFSET
1,1
COMMENTS
This is an extension to Ruth-Aaron pairs. Sum of prime factors (inclusive of multiplicity) of pair of Consecutive positive integers are also consecutive.
The number of pairs less than 10^k (k=1,2,3,4,5,6,..) with this property are 4,7,8,19,55,149,...
Up to 10^13 there are only 5 sets of consecutive terms, namely, (2, 3), (3,4), (27574665988, 27574665989), (862179264458, 1862179264459) and (9600314395008, 9600314395009). - Giovanni Resta, Dec 24 2013
The sum of reciprocals of this sequence is approximately equal to 1.3077. - Abhiram R Devesh, Jun 14 2014
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..300
Giovanni Resta, eRAPs: the 446139 terms < 10^12
Carlos Rivera, Extension to Ruth Aaron pairs
EXAMPLE
For n=20: prime factors = 2,2,5; sum of prime factors = 9.
For n+1=21: prime factors = 3,7; sum of prime factors = 10.
MATHEMATICA
spd[n_]:=Total[Flatten[Table[#[[1]], #[[2]]]&/@FactorInteger[n]]]; Rest[ Position[ Partition[Array[spd, 70000], 2, 1], _?(#[[2]]-#[[1]]==1&), {1}, Heads->False]//Flatten] (* Harvey P. Dale, Sep 07 2016 *)
PROG
(Python)
## sumdivisors(n) is a function that would return the sum of prime
## divisors of n.
i=2
while i < 100000:
..sdi=sumdivisors(i)
..sdip=sumdivisors(i+1)
..if sdi==sdip-1:
....print i, i+1
..i=i+1
(PARI) sopfm(n)=my(f=factor(n)); sum(i=1, #f[, 1], f[i, 1]*f[i, 2])
for(n=1, 10^5, if(sopfm(n)==sopfm(n+1)-1, print1(n, ", "))) /* Ralf Stephan, Aug 12 2013 */
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Abhiram R Devesh, Aug 11 2013
EXTENSIONS
More terms from Ralf Stephan, Aug 12 2013
STATUS
approved