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%I #28 Oct 23 2022 01:01:21
%S 1,2,6,12,30,24,60,120,252,240,504,16380,32760,65520
%N Numbers expressible as 2^a - 2^b, with 0 <= b < a, such that n^a - n^b is divisible by 2^a - 2^b for all n.
%C 1 = 2^1 - 2^0. (n^1 - n^0)/1 : A000027
%C 2 = 2^2 - 2^1. (n^2 - n^1)/2 : A000217
%C 6 = 2^3 - 2^1. (n^3 - n^1)/6 : A000292
%C 12 = 2^4 - 2^2. (n^4 - n^2)/12 : A002415
%C 30 = 2^5 - 2^1. (n^5 - n^1)/30 : A033455
%C 24 = 2^5 - 2^3. (n^5 - n^3)/24 : A006414
%C 60 = 2^6 - 2^2. (n^6 - n^2)/60 : A213547
%C 120 = 2^7 - 2^3. (n^7 - n^3)/120 : A114239
%C 252 = 2^8 - 2^2. (n^8 - n^2)/252 :
%C 240 = 2^8 - 2^4. (n^8 - n^4)/240 : A078876
%C 504 = 2^9 - 2^3. (n^9 - n^3)/504 :
%C 16380 = 2^14 - 2^2. (n^14 - n^2)/16380 :
%C 32760 = 2^15 - 2^3. (n^15 - n^3)/32760 :
%C 65520 = 2^16 - 2^4. (n^16 - n^4)/65520 :
%C Comment from _Tomohiro Yamada_, Oct 05 2022: (Start)
%C "The Mod Set Stanford University" and Carl Pomerance independently noted that the completeness of this sequence follows from a result of Schinzel on primitive prime factors of sequences a^n - b^n in the remark to a problem of Harry Ruderman asking whether if 2^a - 2^b divides 3^a - 3^b, then 2^a - 2^b belongs to this sequence.
%C The Mod Set verified that if a > b, 2^a - 2^b divides 3^a - 3^b, but 2^a - 2^b does not belong to this sequence, then a - b > 1900. Ram Murty and Kumar Murty proved that there are only finitely many natural numbers a, b such that 2^a - 2^b divides 3^a - 3^b.
%C Qi Sun and Ming Zhi Zhang also showed that if a > b and n^a - n^b is divisible by 2^a - 2^b for all n, then 2^a - 2^b belongs to this sequence. (End)
%H M. Ram Murty and V. Kumar Murty, <a href="https://doi.org/10.4169/amer.math.monthly.118.07.644">On a Problem of Ruderman</a>, Amer. Math. Monthly 118 (2011), 644-650, available from <a href="https://www.mast.queensu.ca/~murty/monthly-ruderman.pdf">the first author's website</a>.
%H Harry Ruderman, <a href="https://doi.org/10.2307/2319015">Problem E2468</a>, Amer. Math. Monthly 81 (1974), p. 405.
%H A. Schinzel, <a href="https://doi.org/10.1017/S0305004100040561">On primitive prime factors of a^n - b^n</a>, Proc. Cambridge Phil. Soc. 58 (1962), 556-562.
%H Qi Sun and Ming Zhi Zhang, <a href="https://doi.org/10.1090/S0002-9939-1985-0770523-6">Pairs where 2^a-2^b divides n^a-n^b for all n</a>, Proc. Amer. Math. Soc. 93 (1985), 218-220.
%H The Mod Set Stanford University and Carl Pomerance, <a href="https://doi.org/10.2307/2318318">When 2^m - 2^n divides 3^m - 3^n, remarks to Problem E2468*</a>, Amer. Math. Monthly 84 (1977), 59-60.
%H W. Y. Velez, <a href="https://doi.org/10.2307/2318231">When 2^m - 2^n divides 3^m - 3^n, remarks to Problem E2468</a>, Amer. Math. Monthly 83 (1976), 288-289.
%e a(3) = 6 belongs to this sequence since (n^3 - n)/6 = C(n+1, 3) = A000292(n-1).
%K nonn,fini,full
%O 1,2
%A _Michel Marcus_, Dec 21 2012
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