A181469 Numbers n such that 89 is the largest prime factor of n^2-1.

88, 90, 177, 179, 266, 355, 533, 535, 622, 624, 711, 713, 800, 802, 889, 1067, 1156, 1158, 1247, 1334, 1423, 1425, 1601, 1781, 1959, 2224, 2313, 2402, 2404, 2491, 2493, 2582, 2669, 2849, 3025, 3381, 3383, 3739, 3826, 4093, 4095, 4184, 4271, 4451, 4716

Offset: 1

Author

Artur Jasinski, Oct 21 2010

Keywords

• nonn
• fini
• full

Comments

Sequence is finite, for proof see A175607.

Search for terms can be restricted to the range from 2 to A175607(24) = 332110803172167361; primepi(89) = 24.

Links

• Lucas A. Brown, Table of n, a(n) for n = 1..2371 (terms 1..432 and 434..2371 from _Artur Jasinski_, term 433 from _Lucas A. Brown_, Oct 01 2022. This is the complete list of terms assuming an abc conjecture c < rad(abc)^2).
• From _David A. Corneth_, Oct 02 2022: (Start)
• To verify full I listed all 89-smooth numbers k that are a multiple of 89 below (inclusive) A175607(24) + 2. I then checked if k+2 is 89-smooth. If so, k+1 is a term. Then similarily I checked if k-2 is 89-smooth. If so, k-1 is a term.
• Doing so found the 2371 terms from the b-file. All candidates have been checked completing the proof of full. (End)

Crossrefs

Cf. A175607, A181447-A181468, A181470, A181568.

Programs

Magma

[ n: n in [2..300000] | m eq 89 where m is D[#D] where D is PrimeDivisors(n^2-1) ]; // Klaus Brockhaus, Feb 21 2011

Magma

p:=(97*89*83*79*73*71)^5 *(67*61*59*53*47*43*41)^6 *(37*31*29)^7 *(23*19*17)^8 *13^9 *11^10 *7^13 *5^15 *3^23 *2^36; [ n: n in [2..50000000] | p mod (n^2-1) eq 0 and (D[#D] eq 89 where D is PrimeDivisors(n^2-1)) ]; // Klaus Brockhaus, Feb 21 2011

Mathematica

jj = 2^36*3^23*5^15*7^13*11^10*13^9*17^8*19^8*23^8*29^7*31^7*37^7*41^6 *43^6*47^6*53^6*59^6*61^6*67^6*71^5*73^5*79^5*83^5*89^5*97^5; rr = {}; n = 2; While[n < 3222617400, If[GCD[jj, n^2 - 1] == n^2 - 1, k = FactorInteger[n^2 - 1]; kk = Last[k][[1]]; If[kk == 89, AppendTo[rr, n]]]; n++ ]; rr (* Artur Jasinski *)
Select[Range[300000], FactorInteger[#^2-1][[-1, 1]]==89&]

PARI

is(n)=n=n^2-1; forprime(p=2, 83, n/=p^valuation(n, p)); n>1 && 89^valuation(n, 89)==n \\ Charles R Greathouse IV, Jul 01 2013