A001208 a(n) = solution to the postage stamp problem with 3 denominations and n stamps.

3, 8, 15, 26, 35, 52, 69, 89, 112, 146, 172, 212, 259, 302, 354, 418, 476, 548, 633, 714, 805, 902, 1012, 1127, 1254, 1382, 1524, 1678, 1841, 2010, 2188, 2382, 2584, 2801, 3020, 3256, 3508, 3772, 4043, 4326, 4628, 4941, 5272, 5606, 5960, 6334, 6723, 7120

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• nice

Comments

Fred Lunnon [W. F. Lunnon] defines "solution" to be the smallest value not obtainable by the best set of stamps. The solutions given are one lower than this, that is, the sequence gives the largest number obtainable without a break using the best set of stamps.

References

• R. K. Guy, Unsolved Problems in Number Theory, C12.
• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• R. Alter and J. A. Barnett, A postage stamp problem, Amer. Math. Monthly, 87 (1980), 206-210.
• M. F. Challis, Two new techniques for computing extremal h-bases A_k, Comp. J. 36(2) (1993) 117-126.
• Erich Friedman, Postage stamp problem
• F. H. Kierstead, Jr.,, The Stamp Problem, J. Rec. Math., Vol. ?, Year ?, page 298. [Annotated and scanned copy]
• W. F. Lunnon, A postage stamp problem, Comput. J. 12 (1969) 377-380.
• Eric Weisstein's World of Mathematics, Postage stamp problem

Crossrefs

Postage stamp sequences: A001208, A001209, A001210, A001211, A001212, A001213, A001214, A001215, A001216, A005342, A005343, A005344, A014616, A053346, A053348, A075060, A084192, A084193.

Equals A195618 - 1.

A row or column of the array A196416 (possibly with 1 subtracted from it).

Programs

Maple

c2 :=array(0..8,[3,3,5,5,7,6,8,8,10]) ; c3 :=array(0..8,1..2,[[1,1],[1,1],[2,1],[2,1],[3,1],[2,2],[3,2],[3,2],[4,2]]); c4 :=array(0..8,1..3,[[0,0,0],[0,0,1],[1,0,1],[1,0,2],[2,0,2],[2,1,2],[3,1,2],[3,1,3],[4,1,3]]) ; for n from 23 to 100 do r := n mod 9 ; t := iquo(n,9) ; a2 := 6*t+c2[r] ; a3 := (2*t+c3[r,1])+(2*t+c3[r,2])*a2 ; printf("%a,",4*t+c4[r,1]+(2*t+c4[r,2])*a2+(3*t+c4[r,3])*a3) ; end: # R. J. Mathar, Apr 01 2006

Mathematica

ClearAll[c2, c3, c4, a]; Evaluate[ Array[c2, 9, 0]] = {3, 3, 5, 5, 7, 6, 8, 8, 10}; Evaluate[ Array[c3, {9, 2}, {0, 1}]] = {{1, 1}, {1, 1}, {2, 1}, {2,1}, {3, 1}, {2, 2}, {3, 2}, {3, 2}, {4, 2}}; Evaluate[ Array[c4, {9, 3}, {0, 1}]] = {{0, 0, 0}, {0, 0, 1}, {1, 0,1}, {1, 0, 2}, {2, 0, 2}, {2, 1, 2}, {3, 1, 2}, {3, 1, 3}, {4, 1,3}}; Evaluate[ Array[a, 19]] = {3, 8, 15, 26, 35, 52, 69, 89, 112, 146, 172, 212, 259, 302, 354, 418, 476, 548, 633}; a[n_] := (r = Mod[n, 9]; t = Quotient[n, 9]; a2 = 6t + c2[r]; a3 = (2t + c3[r, 1]) + (2t + c3[r, 2])*a2; 4t + c4[r, 1] + (2t + c4[r, 2])*a2 + (3t + c4[r, 3])*a3); Table[a[n], {n, 1, 48}] (* Jean-François Alcover, Dec 19 2011, after R. J. Mathar's Maple program *)

Extensions

Maple recursion program valid for n>=23 from Challis added by R. J. Mathar, Apr 01 2006

At least 64 terms are known, see Friedman link.

Entry improved by comments from John Seldon (johnseldon(AT)onetel.com), Sep 15 2004

More terms from Jean Gaumont (jeangaum87(AT)yahoo.com), Apr 16 2006