

A114617


Numbers k such that k and k+1 are both refactorable numbers.


15



1, 8, 1520, 50624, 62000, 103040, 199808, 221840, 269360, 463760, 690560, 848240, 986048, 1252160, 1418480, 2169728, 2692880, 2792240, 3448448, 3721040, 3932288, 5574320, 5716880, 6066368, 6890624, 6922160, 8485568
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OFFSET

1,2


COMMENTS

It is not possible to have three consecutive refactorable numbers (see the link). The sequence is best viewed in base 12, with X for 10 and E for 11: 1, 8, X68, 25368, 2EX68, 4E768, 97768, X8468, 10EX68, 1X4468, 293768, 34XX68, 3E6768, 504768, 584X68, 887768, X9X468, E27X68, 11X3768, 12E5468, 1397768, 1X49X68, 1XE8468, 2046768, 2383768, 2399X68, 2X12768. After the first two terms all terms are 68, 368, 468, 668, 768, X68 mod 1000 (base 12).  Walter Kehowski, Jun 19 2006
No successive refactorables seem to be of the form odd, odd+1. If such a pair exist, they must be very large. The first pair of successive refactorables not divisible by 3 is (5*19)^41, (5*19)^4.  Walter Kehowski, Jun 25 2006
Zelinsky (2002, Theorem 59, p. 15) proved that all the terms above 1 are even.  Amiram Eldar, Feb 20 2021


LINKS



FORMULA

a(n) mod tau(a(n)) = 0 and (a(n)+1) mod tau(a(n)+1) = 0 where tau(n) is the number of divisors of n.  Walter Kehowski, Jun 19 2006


MAPLE

with(numtheory); RFC:=[]: for w to 1 do for k from 1 to 12^6 do n:=144*k+(6*12+8); if andmap(z> z mod tau(z) = 0, [n, n+1]) then RFC:=[op(RFC), n]; print(n); fi od od; # it is possible to remove the condition n = (6*12+8) mod 12^2 but you'll get the same sequence.  Walter Kehowski, Jun 19 2006


MATHEMATICA

Select[Join[{1, 8}, 144*Range[10^5] + 80], Mod[#, DivisorSigma[0, #]] == 0 && Mod[#+1, DivisorSigma[0, #+1]] == 0 & ](* JeanFrançois Alcover, Oct 25 2012, after Walter Kehowski *)


PROG

(PARI) isok(n) = !(n % numdiv(n)) && !((n+1) % numdiv(n+1)); \\ Michel Marcus, Dec 21 2018
(GAP) Filtered([1..10^6], n>n mod Tau(n)=0 and (n+1) mod Tau(n+1)=0 ); # Muniru A Asiru, Dec 21 2018


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



