cp's OEIS Frontend

A195166 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.

%I A195166 #28 Oct 23 2022 01:01:21
%S A195166 1,2,6,12,30,24,60,120,252,240,504,16380,32760,65520
%N A195166 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 A195166       1 = 2^1  - 2^0. (n^1  - n^0)/1     : A000027
%C A195166       2 = 2^2  - 2^1. (n^2  - n^1)/2     : A000217
%C A195166       6 = 2^3  - 2^1. (n^3  - n^1)/6     : A000292
%C A195166      12 = 2^4  - 2^2. (n^4  - n^2)/12    : A002415
%C A195166      30 = 2^5  - 2^1. (n^5  - n^1)/30    : A033455
%C A195166      24 = 2^5  - 2^3. (n^5  - n^3)/24    : A006414
%C A195166      60 = 2^6  - 2^2. (n^6  - n^2)/60    : A213547
%C A195166     120 = 2^7  - 2^3. (n^7  - n^3)/120   : A114239
%C A195166     252 = 2^8  - 2^2. (n^8  - n^2)/252   :
%C A195166     240 = 2^8  - 2^4. (n^8  - n^4)/240   : A078876
%C A195166     504 = 2^9  - 2^3. (n^9  - n^3)/504   :
%C A195166   16380 = 2^14 - 2^2. (n^14 - n^2)/16380 :
%C A195166   32760 = 2^15 - 2^3. (n^15 - n^3)/32760 :
%C A195166   65520 = 2^16 - 2^4. (n^16 - n^4)/65520 :
%C A195166 Comment from _Tomohiro Yamada_, Oct 05 2022: (Start)
%C A195166 "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 A195166 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 A195166 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 A195166 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 A195166 Harry Ruderman, <a href="https://doi.org/10.2307/2319015">Problem E2468</a>, Amer. Math. Monthly 81 (1974), p. 405.
%H A195166 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 A195166 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 A195166 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 A195166 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 A195166 a(3) = 6 belongs to this sequence since (n^3 - n)/6 = C(n+1, 3) = A000292(n-1).
%K A195166 nonn,fini,full
%O A195166 1,2
%A A195166 _Michel Marcus_, Dec 21 2012