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Smallest k>n such that n^3+1 divides k*n^2+1.
9

%I #17 May 04 2021 13:19:57

%S 1,3,11,31,69,131,223,351,521,739,1011,1343,1741,2211,2759,3391,4113,

%T 4931,5851,6879,8021,9283,10671,12191,13849,15651,17603,19711,21981,

%U 24419,27031,29823,32801,35971,39339,42911,46693,50691,54911,59359

%N Smallest k>n such that n^3+1 divides k*n^2+1.

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = n^3+n+1.

%F G.f.: (1 - x + 5*x^2 + x^3)/(1 - x)^4. - _Philippe Deléham_, Jun 06 2015

%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - _Wesley Ivan Hurt_, May 04 2021

%t sk[n_]:=Module[{k=n+1,n2=n^2,n3=n^3+1},While[!Divisible[k*n2+1,n3], k++]; k]; Array[sk,40] (* _Harvey P. Dale_, Jun 13 2013 *)

%o (PARI) for(n=1,50,s=n+1; while((s*n^2+1)%(n^3+1)>0,s++); print1(s,","))

%Y a(n+1) = A101220(n, n+1, 4).

%K easy,nonn

%O 0,2

%A _Benoit Cloitre_, May 31 2002

%E a(0) from _Philippe Deléham_, Jun 06 2015