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A264750
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Number of sequences of 5 throws of an n-sided die (with faces numbered 1, 2, ..., n) in which the sum of the throws first reaches or exceeds n on the 5th throw.
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2
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5, 29, 99, 259, 574, 1134, 2058, 3498, 5643, 8723, 13013, 18837, 26572, 36652, 49572, 65892, 86241, 111321, 141911, 178871, 223146, 275770, 337870, 410670, 495495, 593775, 707049, 836969, 985304, 1153944, 1344904, 1560328, 1802493, 2073813, 2376843, 2714283
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OFFSET
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5,1
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COMMENTS
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Number of 5-tuples (t_1, ..., t_5) with 1 <= t_j <= n, Sum_{j <= 4} t_j < n and Sum_{j<=5} t_j >= n. - Robert Israel, Nov 25 2015
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LINKS
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FORMULA
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a(n) = (n - 4)*(n - 3)*(n - 2)*(n - 1)*(4*n + 5)/120.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>10.
G.f.: x^5*(5 - x) / (1 - x)^6. (End)
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EXAMPLE
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For n=5, the a(5) = 5 sequences (i.e., quintuples or 5-tuples) are {1,1,1,1,1}, {1,1,1,1,2}, {1,1,1,1,3}, {1,1,1,1,4} and {1,1,1,1,5}. (Each of the first four throws must be a 1; otherwise, the sum of the throws would reach or exceed 5 before the 5th throw.)
For n=6, each of the quintuples must have four throws whose sum is less than 6, followed by a fifth throw that brings the sum to at least 6, so the a(6) = 29 quintuples are the 5 quintuples {1,1,1,1,t_5} where t_5 is any value in 2..6 and the four sets of 6 quintuples {1,1,1,2,t_5}, {1,1,2,1,t_5}, {1,2,1,1,t_5} and {2,1,1,1,t_5} where t_5 is any value in 1..6. (End)
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MAPLE
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MATHEMATICA
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f[n_, k_] := Module[
{i, total = 0, part, perm},
part = IntegerPartitions[n, {k}];
perm = Flatten[Table[Permutations[part[[i]]], {i, 1, Length[part]}], 1];
For[i = 1, i <= Length[perm], i++, total += n + 1 - perm[[i, k]] ];
Return[total]; ]
And the sequences are obtained by:
h[k_] := Table[f[i, k], {i, k, number_of_terms_wanted}]
Table[(n - 4) (n - 3) (n - 2) (n - 1) (4 n + 5)/120, {n, 5, 40}] (* Bruno Berselli, Nov 24 2015 *)
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PROG
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(PARI) Vec(x^5*(5-x)/(1-x)^6 + O(x^100)) \\ Colin Barker, Nov 23 2015
(PARI) for(n=5, 40, print1((n-4)*(n-3)*(n-2)*(n-1)*(4*n+5)/120", ")); \\ Bruno Berselli, Nov 24 2015
(Magma) [(n-4)*(n-3)*(n-2)*(n-1)*(4*n+5)/120: n in [5..40]]; // Vincenzo Librandi, Nov 24 2015
(Sage) [(n-4)*(n-3)*(n-2)*(n-1)*(4*n+5)/120 for n in (5..40)] # Bruno Berselli, Nov 24 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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Formulae, b-file adapted to the new offset and definition rephrased by the Editors of the OEIS, Nov 26 2015
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STATUS
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approved
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