A369818 The sixth term of the greedy B_n set of natural numbers.
5, 20, 71, 153, 366, 668, 1153, 1822, 3119, 4448, 6348, 8559, 11565, 14976, 21023, 26220, 33066, 40306, 49601, 59354, 76031, 89248, 106008, 122909, 143989, 165196, 200759, 227660, 261030, 293736, 333825, 373110, 438191, 485952, 544356, 600523, 668573, 734072, 841679, 918988, 1012578, 1101374, 1208065, 1309426, 1474943, 1592000, 1732656
Offset: 1
Keywords
Examples
a(2) = 20, as all 21 nonincreasing sums from {0,1,3,7,12,20}, namely 0+0 < 0+1 < 1+1 < 0+3 < 1+3 < 3+3 < 0+7 < 1+7 < 3+7 < 0+12 < 1+12 < 7+7 < 3+12 < 7+12 < 0+20 < 1+20 < 3+20 < 12+12 < 7+20 < 12+20 < 20+20, are distinct, and all other 6-element sets of nonnegative integers with this property are lexicographically after {0,1,3,7,12,20}.
Links
- M. B. Nathanson, The third positive element in the greedy B_h-set, arXiv:2310.14426 [math.NT], 2023.
- M. B. Nathanson and Kevin O'Bryant, The fourth positive element in the greedy B_h-set, arXiv:2311.14021 [math.NT], 2023.
- Kevin O'Bryant, B_h-sets and Rigidity, arXiv:2312.10910 [math.NT], 2023.
Programs
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Python
from itertools import count, combinations_with_replacement def A369818(n): alist = [0,1,n+1,n*(n+1)+1,(n+3>>1)*n**2+(3*n+2>>1)] aset = set(sum(d) for d in combinations_with_replacement(alist,n)) blist = [] for i in range(n): blist.append(set(sum(d) for d in combinations_with_replacement(alist,i))) for k in count(max(alist[-1]+1,(n**3>>1)*(1+(n>>2)))): for i in range(n): if any((n-i)*k+d in aset for d in blist[i]): break else: return k # Chai Wah Wu, Feb 28 2024
Formula
Conjectured that a(6n+i) is a quartic polynomial sequence with lead term (1/3)n^4 for each i in {1,2,3,5,6,10} in arxiv:2312.10910.
Proved that (1/8)*n^4 + (1/2)*n^3 <= a(n) <= 0.406671*n^4 + O(n^3) in arxiv:2312.10910.
Comments