login
a(n) = Frobenius number for 6 successive numbers = F(n+1, n+2, n+3, n+4, n+5, n+6).
18

%I #24 Jan 28 2023 22:08:12

%S 1,2,3,4,5,13,15,17,19,21,35,38,41,44,47,67,71,75,79,83,109,114,119,

%T 124,129,161,167,173,179,185,223,230,237,244,251,295,303,311,319,327,

%U 377,386,395,404,413,469,479,489,499,509,571,582,593,604,615,683,695,707

%N a(n) = Frobenius number for 6 successive numbers = F(n+1, n+2, n+3, n+4, n+5, n+6).

%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,2,-2,0,0,0,-1,1).

%F G.f.: x*(x^10-6*x^5-x^4-x^3-x^2-x-1) / ((x-1)^3*(x^4+x^3+x^2+x+1)^2). [_Colin Barker_, Dec 13 2012]

%e a(6)=13 because 13 is the largest number k such that the equation 7*x_1 + 8*x_2 + 9*x_3 + 10*x_4 + 11*x_5 + 12*x_6 = k has no solution for any nonnegative x_i (in other words, for every k > 13 there exist one or more solutions).

%t Table[FrobeniusNumber[{n + 1, n + 2, n + 3, n + 4, n + 5, n + 6}], {n, 1, 100}]

%t Table[FrobeniusNumber[Range[n,n+5]],{n,2,100}] (* _Harvey P. Dale_, Dec 22 2018 *)

%Y Frobenius number for k successive numbers: A028387 (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), this sequence (k=6), A138987 (k=7), A138988 (k=8).

%K nonn,easy

%O 1,2

%A _Artur Jasinski_, Apr 05 2008