A181470 Numbers n such that 97 is the largest prime factor of n^2 - 1.

96, 98, 193, 195, 290, 389, 484, 581, 583, 775, 872, 874, 969, 971, 1066, 1163, 1165, 1359, 1456, 1551, 1553, 1648, 1747, 1844, 1939, 2036, 2133, 2135, 2232, 2521, 2715, 2911, 3008, 3103, 3299, 3394, 3396, 3590, 3976, 4267, 4269, 4463, 4558, 4946, 5045

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(25) = 99913980938200001; primepi(97) = 25.

Links

• Lucas A. Brown, Table of n, a(n) for n = 1..2734 (2679 terms from Artur Jasinski, 55 missing terms from Lucas A. Brown, complete subject to the effective abc conjecture c < rad(abc)^2)
• From _David A. Corneth_, Oct 03 2022: (Start)
• To verify full I listed all 97-smooth numbers k that are a multiple of 97 below (inclusive) A175607(25) + 2. I then checked if k+2 is 97-smooth. If so, k+1 is a term. Then similarily I checked if k-2 is 97-smooth. If so, k-1 is a term.
• Doing so found the 2734 terms from the b-file. All candidates have been checked completing the proof of full. (End)

Crossrefs

Cf. A175607, A181447-A181469, A181568.

Programs

Magma

[ n: n in [2..300000] | m eq 97 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 97 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 == 97, AppendTo[rr, n]]]; n++]; rr
(* or *)
Select[Range[300000], FactorInteger[#^2 - 1][[-1, 1]] == 97 &]

PARI

is(n)=n=n^2-1;forprime(p=2,89,n/=p^valuation(n,p));n>1 && 97^valuation(n,97)==n \\ Charles R Greathouse IV, Jul 01 2013