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A258272 The smallest amount which cannot be made with fewer than n British coins. 2
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 (list; graph; refs; listen; history; text; internal format)
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: A051633 A131051 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

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Last modified November 27 13:03 EST 2022. Contains 358405 sequences. (Running on oeis4.)