%I #29 Jan 28 2023 22:08:25
%S 1,2,3,4,11,13,15,17,29,32,35,38,55,59,63,67,89,94,99,104,131,137,143,
%T 149,181,188,195,202,239,247,255,263,305,314,323,332,379,389,399,409,
%U 461,472,483,494,551,563,575,587,649,662,675,688,755,769,783,797,869
%N a(n) = Frobenius number for 5 successive numbers = F(n+1, n+2, n+3, n+4, n+5).
%H Harvey P. Dale, <a href="/A138985/b138985.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,2,-2,0,0,-1,1).
%F G.f.: x*(x^8 - 5*x^4 - x^3 - x^2 - x - 1) / ((x-1)^3*(x+1)^2*(x^2+1)^2). - _Colin Barker_, Dec 13 2012
%e a(5)=11 because 11 is the largest number k such that the equation 6*x_1 + 7*x_2 + 8*x_3 + 9*x_4 + 10*x_5 = k has no solution for any nonnegative x_i (in other words, for every k > 11 there exist one or more solutions).
%t Table[FrobeniusNumber[{n + 1, n + 2, n + 3, n + 4}], {n, 1, 100}]
%t Table[(Floor[(n-1)/4]+1)*(n+1)-1,{n,57}] (* _Zak Seidov_, Jan 10 2015 *)
%t FrobeniusNumber/@Partition[Range[2,70],5,1] (* or *) LinearRecurrence[ {1,0,0,2,-2,0,0,-1,1},{1,2,3,4,11,13,15,17,29},70] (* _Harvey P. Dale_, Oct 07 2016 *)
%o (PARI)for (n=1,57,print1((floor((n-1)/4)+1)*(n+1)-1 ","))\\ _Zak Seidov_, Jan 10 2015
%Y Frobenius number for k successive numbers: A028387 (k=2), A079326 (k=3), A138984 (k=4), this sequence (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).
%K nonn,easy
%O 1,2
%A _Artur Jasinski_, Apr 05 2008