|
%I #52 Sep 28 2023 11:44:37
%S 8424432925592889329288197322308900672459420460792433,
%T 17361015163508605989239159575667846308252873717727992,
%U 26297597401424322649190121829026791944046326974663551,35234179639340039309141084082385737579839780231599110
%N Numbers k such that gcd(k^17 + 9, (k+1)^17 + 9) > 1.
%C In other words, let f(n) = gcd(n^17 + 9, (n+1)^17 + 9). Then f(n) = 1 for all n <= 8424432925592889329288197322308900672459420460792432, but f(8424432925592889329288197322308900672459420460792433) > 1.
%C In fact f(8424432925592889329288197322308900672459420460792433) = 8936582237915716659950962253358945635793453256935559.
%H M. F. Hasler, <a href="/A010034/b010034.txt">Table of n, a(n) for n = 1..100</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Stan Wagon, <a href="https://web.archive.org/web/20150906000525/http://mathforum.org/wagon/spring96/p805.html">Macalester College Problem of the week # 805</a>, MacPOW archive on MathForum.org. Spring 1996.
%H Péter E. Frenkel, József Pelikán, <a href="https://arxiv.org/abs/1608.07936">On the greatest common divisor of the value of two polynomials</a>, Amer. Math. Monthly 124:5 (2017), 446-450. arXiv:1608.07936 [math.NT]
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F a(n) = 8424432925592889329288197322308900672459420460792433 + 8936582237915716659950962253358945635793453256935559*(n-1). - _Max Alekseyev_, Jul 26 2009
%F a(1) = A255859(17). - _M. F. Hasler_, Mar 17 2015
%t Table[8424432925592889329288197322308900672459420460792433+ 8936582237915716659950962253358945635793453256935559(n-1),{n,5}] (* or *) LinearRecurrence[{2,-1},{8424432925592889329288197322308900672459420460792433,17361015163508605989239159575667846308252873717727992},5] (* _Harvey P. Dale_, Jun 12 2014 *)
%o (PARI) A010034(n)=8936582237915716659950962253358945635793453256935559*n-512149312322827330662764931050044963334032796143126 \\ _M. F. Hasler_, Mar 17 2015
%o (PARI) \\ The values (a(1),p) can also be found using:
%o {p=polresultant(x^17+9,(x+1)^17+9);s=vector(2,i,Mod(-9,p)^(1/17));(u=s[2]/s[1])!=1&&until(setsearch(Set(s=concat(s,s[#s]*u)),s[#s]+1),)}
%o \\ Then the last element s[#s] equals Mod(a(1),p). - _M. F. Hasler_, Mar 26 2015
%Y Cf. A118119, A255859.
%K nonn,easy
%O 1,1
%A Ilan Vardi, _Stan Wagon_
%E More terms from _Max Alekseyev_, Jul 26 2009
| 