A069755 Frobenius number of the numerical semigroup generated by 3 consecutive triangular numbers.
17, 29, 89, 125, 251, 323, 539, 659, 989, 1169, 1637, 1889, 2519, 2855, 3671, 4103, 5129, 5669, 6929, 7589, 9107, 9899, 11699, 12635, 14741, 15833, 18269, 19529, 22319, 23759, 26927, 28559, 32129, 33965, 37961, 40013, 44459, 46739, 51659
Offset: 2
Examples
a(2)=17 because 17 is not a nonnegative linear combination of 3, 6 and 10 but all numbers greater than 17 are.
Links
- Harvey P. Dale, Table of n, a(n) for n = 2..1000
- R. Fröberg, C. Gottlieb and R. Häggkvist, On numerical semigroups, Semigroup Forum, 35 (1987), 63-83 (for definition of Frobenius number).
- Aureliano M. Robles-Pérez, José Carlos Rosales, The Frobenius number for sequences of triangular and tetrahedral numbers, arXiv:1706.04378 [math.NT], 2017.
Programs
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Mathematica
tri=Range[40]Range[2,41]/2; Table[t=CoefficientList[Series[1/(1-x^tri[[n]])/(1-x^tri[[n+1]])/(1-x^tri[[n+2]]), {x,0,n(n+1)(n+2)}], x]; Last[Position[t,0]-1][[1]], {n,2,33}] (* T. D. Noe, Nov 27 2006 *) Rest[FrobeniusNumber/@Partition[Accumulate[Range[50]],3,1]] (* Harvey P. Dale, Oct 04 2011 *)
Formula
Conjectures from Colin Barker, Nov 22 2012: (Start)
a(n) = (-14 + 6*(-1)^n + (3+9*(-1)^n)*n + 3*(5+(-1)^n)*n^2 + 6*n^3)/8.
G.f.: x^2*(17 + 12*x + 9*x^2 - 3*x^4 + x^6) / ((1 - x)^4*(1 + x)^3). (End)
Conjectures from Colin Barker, Mar 21 2017: (Start)
a(n) = (6*n^3 + 18*n^2 + 12*n - 8)/8 for n even.
a(n) = (6*n^3 + 12*n^2 - 6*n - 20)/8 for n odd. (End)
Extensions
Corrected by T. D. Noe, Nov 27 2006
Comments