OFFSET
1,1
COMMENTS
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.
In fact f(8424432925592889329288197322308900672459420460792433) = 8936582237915716659950962253358945635793453256935559.
LINKS
M. F. Hasler, Table of n, a(n) for n = 1..100
Tanya Khovanova, Recursive Sequences
Stan Wagon, Macalester College Problem of the week # 805, MacPOW archive on MathForum.org. Spring 1996.
Péter E. Frenkel, József Pelikán, On the greatest common divisor of the value of two polynomials, Amer. Math. Monthly 124:5 (2017), 446-450. arXiv:1608.07936 [math.NT]
Index entries for linear recurrences with constant coefficients, signature (2,-1).
FORMULA
a(n) = 8424432925592889329288197322308900672459420460792433 + 8936582237915716659950962253358945635793453256935559*(n-1). - Max Alekseyev, Jul 26 2009
a(1) = A255859(17). - M. F. Hasler, Mar 17 2015
MATHEMATICA
Table[8424432925592889329288197322308900672459420460792433+ 8936582237915716659950962253358945635793453256935559(n-1), {n, 5}] (* or *) LinearRecurrence[{2, -1}, {8424432925592889329288197322308900672459420460792433, 17361015163508605989239159575667846308252873717727992}, 5] (* Harvey P. Dale, Jun 12 2014 *)
PROG
(PARI) A010034(n)=8936582237915716659950962253358945635793453256935559*n-512149312322827330662764931050044963334032796143126 \\ M. F. Hasler, Mar 17 2015
(PARI) \\ The values (a(1), p) can also be found using:
{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), )}
\\ Then the last element s[#s] equals Mod(a(1), p). - M. F. Hasler, Mar 26 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Ilan Vardi, Stan Wagon
EXTENSIONS
More terms from Max Alekseyev, Jul 26 2009
STATUS
approved