A114617 Numbers k such that k and k+1 are both refactorable numbers.

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

Offset: 1

Author

Eric W. Weisstein, Dec 16 2005

Keywords

• nonn

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)^4-1, (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

• Jud McCranie, Table of n, a(n) for n = 1..94342 First 1000 terms by Donovan Johnson.
• Joshua Zelinsky, Tau Numbers: A Partial Proof of a Conjecture and Other Results, Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.8.
• Eric Weisstein's World of Mathematics, Refactorable Number.

Crossrefs

Cf. A033950, A036898.

Programs

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

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 & ](* Jean-François Alcover, Oct 25 2012, after Walter Kehowski *)

PARI

isok(n) = !(n % numdiv(n)) && !((n+1) % numdiv(n+1)); \\ Michel Marcus, Dec 21 2018

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