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

%I A114617 #31 Feb 16 2025 08:32:59
%S A114617 1,8,1520,50624,62000,103040,199808,221840,269360,463760,690560,
%T A114617 848240,986048,1252160,1418480,2169728,2692880,2792240,3448448,
%U A114617 3721040,3932288,5574320,5716880,6066368,6890624,6922160,8485568
%N A114617 Numbers k such that k and k+1 are both refactorable numbers.
%C A114617 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
%C A114617 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
%C A114617 Zelinsky (2002, Theorem 59, p. 15) proved that all the terms above 1 are even. - _Amiram Eldar_, Feb 20 2021
%H A114617 Jud McCranie, <a href="/A114617/b114617.txt">Table of n, a(n) for n = 1..94342</a> First 1000 terms by Donovan Johnson.
%H A114617 Joshua Zelinsky, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL5/Zelinsky/zelinsky9.html">Tau Numbers: A Partial Proof of a Conjecture and Other Results</a>, Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.8.
%H A114617 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RefactorableNumber.html">Refactorable Number</a>.
%F A114617 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
%p A114617 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
%t A114617 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_ *)
%o A114617 (PARI) isok(n) = !(n % numdiv(n)) && !((n+1) % numdiv(n+1)); \\ _Michel Marcus_, Dec 21 2018
%o A114617 (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
%Y A114617 Cf. A033950, A036898.
%K A114617 nonn
%O A114617 1,2
%A A114617 _Eric W. Weisstein_, Dec 16 2005