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A114617
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Numbers k such that k and k+1 are both refactorable numbers.
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15
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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
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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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
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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