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A188430
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a(n) is the maximum of the largest elements of all n-full sets, or 0 if no such set exists.
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3
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1, 0, 2, 0, 0, 3, 4, 0, 0, 4, 5, 6, 7, 7, 8, 6, 7, 8, 9, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37, 37, 38, 38
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OFFSET
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1,3
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COMMENTS
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Let A be a set of positive integers. We say that A is n-full if (sum A)=[n] for a positive integer n, where (sum A) is the set of all positive integers which are a sum of distinct elements of A and [n]={1,2,...,n}. The number a(n) denotes the maximum of the set {max A: (sum A)=[n]}, or 0 if there is no n-full set.
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LINKS
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Table of n, a(n) for n=1..76.
Mohammad Saleh Dinparvar, Python program (github)
L. Naranjani and M. Mirzavaziri, Full Subsets of N, Journal of Integer Sequences, 14 (2011), Article 11.5.3.
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
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FORMULA
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a(n) = ceiling(n/2) for n >= 20.
From Colin Barker, May 11 2020: (Start)
G.f.: x*(1 - x + x^2 - x^3 - 2*x^4 + 5*x^5 + x^6 - 7*x^7 - x^8 + 8*x^9 + x^10 - 3*x^11 - x^13 - 2*x^15 + 3*x^17 - x^21) / ((1 - x)^2*(1 + x)).
a(n) = a(n-1) + a(n-2) - a(n-3) for n>22.
(End)
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MATHEMATICA
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LinearRecurrence[{1, 1, -1}, {1, 0, 2, 0, 0, 3, 4, 0, 0, 4, 5, 6, 7, 7, 8, 6, 7, 8, 9, 10, 11, 11}, 80] (* Harvey P. Dale, Jul 24 2021 *)
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PROG
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(Haskell)
a188430 n = a188430_list !! (n-1)
a188430_list = [1, 0, 2, 0, 0, 3, 4, 0, 0, 4, 5, 6, 7, 7, 8, 6, 7, 8, 9] ++
(drop 19 a008619_list)
-- Reinhard Zumkeller, Aug 06 2015
(PARI) Vec(x*(1 - x + x^2 - x^3 - 2*x^4 + 5*x^5 + x^6 - 7*x^7 - x^8 + 8*x^9 + x^10 - 3*x^11 - x^13 - 2*x^15 + 3*x^17 - x^21) / ((1 - x)^2*(1 + x)) + O(x^80)) \\ Colin Barker, May 11 2020
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CROSSREFS
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Cf. A188429, A188431.
Cf. A008619.
Sequence in context: A174956 A124182 A188429 * A013585 A261319 A230414
Adjacent sequences: A188427 A188428 A188429 * A188431 A188432 A188433
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KEYWORD
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nonn,easy
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AUTHOR
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Madjid Mirzavaziri, Mar 31 2011
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STATUS
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approved
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