A258272 The smallest amount which cannot be made with fewer than n British coins.

1, 3, 8, 18, 38, 88, 188, 388, 588, 788, 988, 1188, 1388, 1588, 1788, 1988, 2188, 2388, 2588, 2788, 2988, 3188, 3388, 3588, 3788, 3988, 4188, 4388, 4588, 4788, 4988, 5188, 5388, 5588, 5788, 5988, 6188, 6388, 6588, 6788, 6988, 7188, 7388, 7588, 7788, 7988, 8188

Offset

1,2

Comments

a(n) is the smallest amount (in pence) that cannot be made with fewer than n coins.

The coins included are those in common circulation in the UK: 1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

Links

Matthew Scroggs, Table of n, a(n) for n = 1..5005

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

Formula

From Georg Fischer, Jul 22 2022: (Start)

a(n) = 200*n - 1222 for n >= 7.

O.g.f.: (1 + x + 3*x^2 + 5*x^3 + 10*x^4 + 30*x^5 + 50*x^6 + 100*x^7)/(x - 1)^2. (End)

Example

The smallest value that requires 4 coins is 18p (10p, 5p, 2p and 1p). Therefore a(4)=18.

Mathematica

CoefficientList[Series[(1 + x + 3*x^2 + 5*x^3 + 10*x^4 + 30*x^5 + 50*x^6 + 100*x^7)/(x - 1)^2, {x, 0, 64}], x] (* Georg Fischer, Jul 22 2022 *)

Prog

(Python) #
coins = [1, 2, 5, 10, 20, 50, 100, 200]
need = [0]
while True:
....next = len(need)
....n_need = next
....for coin in coins:
........if coin>next:
............break
........n_need = min(n_need, 1+need[next-coin])
....need.append(n_need)
....if n_need == len(seq):
........print(next)
(PARI) Vec((1 + x + 3*x^2 + 5*x^3 + 10*x^4 + 30*x^5 + 50*x^6 + 100*x^7)/(x - 1)^2 + O(x^50)) \\ Felix Fröhlich, Jul 23 2022

Crossrefs

Cf. A258274.

Sequence in context: A131051 A051633 A172265 * A117727 A117713 A128552

Adjacent sequences: A258269 A258270 A258271 * A258273 A258274 A258275

Keyword

nonn

Author

Matthew Scroggs, May 25 2015

Extensions

More terms from Felix Fröhlich, Jul 23 2022

Status

approved