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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.)