A138984 a(n) = Frobenius number for 4 successive numbers = F(n+1, n+2, n+3, n+4).

1, 2, 3, 9, 11, 13, 23, 26, 29, 43, 47, 51, 69, 74, 79, 101, 107, 113, 139, 146, 153, 183, 191, 199, 233, 242, 251, 289, 299, 309, 351, 362, 373, 419, 431, 443, 493, 506, 519, 573, 587, 601, 659, 674, 689, 751, 767, 783, 849, 866, 883, 953, 971, 989, 1063, 1082

Offset: 1

Author

Artur Jasinski, Apr 05 2008

Keywords

• nonn
• easy

Examples

			a(4) = 9 because 9 is the largest number k such that the equation 5*x_1 + 6*x_2 + 7*x_3 + 9*x_4 = k has no solution for any nonnegative x_i (in other words, for every k > 9 there exist one or more solutions).
		

Links

• Index entries for linear recurrences with constant coefficients, signature (1,0,2,-2,0,-1,1).

Crossrefs

Frobenius number for k successive numbers: A028387 (k=2), A079326 (k=3), this sequence (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).

Programs

Mathematica

Table[FrobeniusNumber[{n + 1, n + 2, n + 3, n + 4}], {n, 1, 100}]
Table[n + Floor[(n-1)/3]*(n+1), {n,56}] (* Giorgos Kalogeropoulos, Apr 06 2025 *)

Formula

G.f.: x*(x^6-4*x^3-x^2-x-1) / ((x-1)^3*(x^2+x+1)^2). [Colin Barker, Dec 13 2012]

a(n) = n + (n+1)*floor((n-1)/3). - Giorgos Kalogeropoulos, Apr 06 2025