cp's OEIS Frontend

A275666 Multisets of numbers such that the sum of reciprocals is 1 and each element e occurs at most lpf(e) - 1 times.

%I A275666 #37 Nov 19 2024 02:49:10
%S A275666 2,3,6,2,5,5,10,3,5,5,6,10,2,4,6,12,3,3,4,12,4,5,5,6,10,12,2,7,7,7,14,
%T A275666 3,6,7,7,7,14,4,6,7,7,7,12,14,5,5,7,7,7,10,14,2,3,10,15,2,4,10,12,15,
%U A275666 2,5,6,15,15,3,3,5,15,15,3,3,6,10,15,3,5,5,5,15,3,4
%N A275666 Multisets of numbers such that the sum of reciprocals is 1 and each element e occurs at most lpf(e) - 1 times.
%C A275666 lpf(n) gives the least prime factor of n (A020639(n)).
%C A275666 The multisets are ordered primarily by largest element, then secondly by length, then thirdly by smallest distinct element.
%C A275666 Sets are placed in nondecreasing order. The separation between two sets is where an element a(n) > a(n + 1).
%C A275666 Let M be the largest element of a multiset. No primes >= (M + 1)/2 are in that multiset.
%C A275666 This sequence may be used to find multisets that have the sum of reciprocals an integer, along with the following operations.
%C A275666 In a set we can replace a number by twice its double. For example, in (2, 3, 6), we can replace the 6 by two twelves, giving {2, 3, 12, 12}. The sum of reciprocals is still 1. However 12 occurs twice in the set, more than lpf(12) - 1 = 1.
%C A275666 {3, 4, 4, 6} isn't included as we could replace two fours by one 2 to get {2, 3, 6}. To exclude such possibilities, each element e is in a multiset at most lpf(e) - 1.
%C A275666 We can also take the union of two sets, for example the sets {2, 3, 6} and {2, 5, 5, 10} giving {2, 2, 3, 5, 5, 6, 10} which has sum of reciprocals 1 + 1 = 2.
%C A275666 We can put 1 in a set, so from {2, 3, 6} we can find {1, 2, 3, 6}.
%C A275666 Using any such operations, we get out of the sequence but more terms of A034708 can be found and A020473 may be investigated further.
%e A275666 {2, 3, 6}
%e A275666 {2, 5, 5, 10}
%e A275666 {3, 5, 5, 6, 10}
%e A275666 {2, 4, 6, 12}
%e A275666 {3, 3, 4, 12}
%e A275666 {4, 5, 5, 6, 10, 12}
%e A275666 {2, 7, 7, 7, 14}
%e A275666 {3, 6, 7, 7, 7, 14}
%e A275666 {4, 6, 7, 7, 7, 12, 14}
%e A275666 {5, 5, 7, 7, 7, 10, 14}
%e A275666 {2, 3, 10, 15}
%e A275666 {2, 4, 10, 12, 15}
%e A275666 {2, 5, 6, 15, 15}
%e A275666 {3, 3, 5, 15, 15}
%e A275666 {3, 3, 6, 10, 15}
%e A275666 {3, 5, 5, 5, 15}
%e A275666 ...
%e A275666 {3, 3, 4, 12} comes before {2, 7, 7, 7, 14} because the largest element of the first is less than the one from the second.
%e A275666 {2, 5, 5, 10} comes before {3, 5, 5, 6, 10} because they both have the largest element 10 but the latter has more elements.
%e A275666 {2, 4, 10, 12, 15} comes before {2, 5, 6, 15, 15} because they both have the largest element 15 and the same number of elements but the first smallest different element, 4 resp. 5, is less for the first.
%Y A275666 Cf. A020473, A020639, A034708, A091784.
%K A275666 nonn,base,tabf
%O A275666 1,1
%A A275666 _David A. Corneth_, Aug 23 2016