%I #27 Jan 28 2023 22:08:16
%S 1,2,3,4,5,6,15,17,19,21,23,25,41,44,47,50,53,56,79,83,87,91,95,99,
%T 129,134,139,144,149,154,191,197,203,209,215,221,265,272,279,286,293,
%U 300,351,359,367,375,383,391,449,458,467,476,485,494,559,569,579,589,599
%N a(n) = Frobenius number for 7 successive numbers = F(n+1, n+2, n+3, n+4, n+5, n+6, n+7).
%H Harvey P. Dale, <a href="/A138987/b138987.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,2,-2,0,0,0,0,-1,1).
%F G.f.: x*(x^12-7*x^6-x^5-x^4-x^3-x^2-x-1) / ((x-1)^3*(x+1)^2*(x^2-x+1)^2*(x^2+x+1)^2). [_Colin Barker_, Dec 13 2012]
%e a(7) = 15 because 15 is the largest number k such that the equation 8*x_1 + 9*x_2 + 10*x_3 + 11*x_4 + 12*x_5 + 13*x_6 + 14*x_7 = k has no solution for any nonnegative x_i (in other words, for every k > 15 there exist one or more solutions).
%t Table[FrobeniusNumber[{n+1, n+2, n+3, n+4, n+5, n+6, n+7}], {n, 1, 100}]
%t Table[FrobeniusNumber[n+Range[7]],{n,100}] (* _Harvey P. Dale_, Dec 06 2021 *)
%Y Frobenius number for k successive numbers: A028387 (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), this sequence (k=7), A138988 (k=8).
%K nonn,easy
%O 1,2
%A _Artur Jasinski_, Apr 05 2008