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.

1, 2, 6, 12, 30, 24, 60, 120, 252, 240, 504, 16380, 32760, 65520

Offset

1,2

Comments

1 = 2^1 - 2^0. (n^1 - n^0)/1 : A000027

2 = 2^2 - 2^1. (n^2 - n^1)/2 : A000217

6 = 2^3 - 2^1. (n^3 - n^1)/6 : A000292

12 = 2^4 - 2^2. (n^4 - n^2)/12 : A002415

30 = 2^5 - 2^1. (n^5 - n^1)/30 : A033455

24 = 2^5 - 2^3. (n^5 - n^3)/24 : A006414

60 = 2^6 - 2^2. (n^6 - n^2)/60 : A213547

120 = 2^7 - 2^3. (n^7 - n^3)/120 : A114239

252 = 2^8 - 2^2. (n^8 - n^2)/252 :

240 = 2^8 - 2^4. (n^8 - n^4)/240 : A078876

504 = 2^9 - 2^3. (n^9 - n^3)/504 :

16380 = 2^14 - 2^2. (n^14 - n^2)/16380 :

32760 = 2^15 - 2^3. (n^15 - n^3)/32760 :

65520 = 2^16 - 2^4. (n^16 - n^4)/65520 :

Comment from Tomohiro Yamada, Oct 05 2022: (Start)

"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.

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.

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)

Links

Table of n, a(n) for n=1..14.

M. Ram Murty and V. Kumar Murty, On a Problem of Ruderman, Amer. Math. Monthly 118 (2011), 644-650, available from the first author's website.

Harry Ruderman, Problem E2468, Amer. Math. Monthly 81 (1974), p. 405.

A. Schinzel, On primitive prime factors of a^n - b^n, Proc. Cambridge Phil. Soc. 58 (1962), 556-562.

Qi Sun and Ming Zhi Zhang, Pairs where 2^a-2^b divides n^a-n^b for all n, Proc. Amer. Math. Soc. 93 (1985), 218-220.

The Mod Set Stanford University and Carl Pomerance, When 2^m - 2^n divides 3^m - 3^n, remarks to Problem E2468*, Amer. Math. Monthly 84 (1977), 59-60.

W. Y. Velez, When 2^m - 2^n divides 3^m - 3^n, remarks to Problem E2468, Amer. Math. Monthly 83 (1976), 288-289.

Example

a(3) = 6 belongs to this sequence since (n^3 - n)/6 = C(n+1, 3) = A000292(n-1).

Crossrefs

Sequence in context: A093387 A324408 A229487 * A225646 A225627 A358070

Adjacent sequences: A195163 A195164 A195165 * A195167 A195168 A195169

Keyword

nonn,fini,full

Author

Michel Marcus, Dec 21 2012

Status

approved