A367906 Numbers k such that it is possible to choose a different binary index of each binary index of k.
1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 24, 26, 28, 32, 33, 34, 35, 36, 37, 38, 40, 41, 44, 48, 49, 50, 52, 56, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 76, 80, 81, 82, 84, 88, 96, 97, 98, 100, 104, 112, 128, 129, 130, 131, 132
Offset: 1
Examples
The set-system {{2,3},{1,2,3},{1,4}} with BII-number 352 has choices such as (2,1,4) that satisfy the axiom, so 352 is in the sequence. The terms together with the corresponding set-systems begin: 1: {{1}} 2: {{2}} 3: {{1},{2}} 4: {{1,2}} 5: {{1},{1,2}} 6: {{2},{1,2}} 8: {{3}} 9: {{1},{3}} 10: {{2},{3}} 11: {{1},{2},{3}} 12: {{1,2},{3}} 13: {{1},{1,2},{3}} 14: {{2},{1,2},{3}} 16: {{1,3}} 17: {{1},{1,3}}
Links
- John Tyler Rascoe, Table of n, a(n) for n = 1..10000
- Wikipedia, Axiom of choice.
Crossrefs
Programs
-
Mathematica
bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1]; Select[Range[100], Select[Tuples[bpe/@bpe[#]], UnsameQ@@#&]!={}&]
-
Python
from itertools import count, islice, product def bin_i(n): #binary indices return([(i+1) for i, x in enumerate(bin(n)[2:][::-1]) if x =='1']) def a_gen(): #generator of terms for n in count(1): for j in list(product(*[bin_i(k) for k in bin_i(n)])): if len(set(j)) == len(j): yield(n); break A367906_list = list(islice(a_gen(),100)) # John Tyler Rascoe, Dec 23 2023
Comments