|
|
A339081
|
|
Initial prime in the least binary Ormiston n-tuple: n consecutive primes whose binary representations are anagrams of each other.
|
|
0
|
|
|
2, 11, 103, 167, 941, 6287, 6287, 150287, 866087, 4813583, 53376151, 80522263, 564779279, 1300664983, 1786616407, 1971072527, 4149916763, 133076127097, 515655598279, 4572291787807, 4572291787807, 4572291787807, 9039081952627, 189984035976239
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Equivalently, a(n) is the start of the least run of n consecutive primes with the same length of binary representation (A070939) and the same binary weight (A000120).
|
|
LINKS
|
Andy Edwards, Ormiston Pairs, Australian Mathematics Teacher, Vol. 58, No. 2 (2002), pp. 12-13.
|
|
EXAMPLE
|
a(1) = 2 since 2 is the least prime number and its binary representation, 10, is not an anagram of the binary representation of the next prime, 3, whose binary representation is 11.
a(2) = 11 since 11 and 13 are the least pair of consecutive primes whose binary representations, 1011 and 1101, are anagrams of each other.
a(3) = 103 since 103, 107 and 109 are the least triple of consecutive primes whose binary representations, 1100111, 1101011 and 1101101, are anagrams of each other.
|
|
MATHEMATICA
|
s[n_] := Sort[IntegerDigits[n, 2]]; orm[mx_] := Module[{p1 = p2 = 2, c = 1, m = 0, seq, s1, s2}, s1 = s[p1]; seq = Table[0, {mx}]; seq[[1]] = p1; While[c < mx, p2 = NextPrime[p2]; If[(s2 = s[p2]) == s1, c++; If[seq[[c]] == 0, seq[[c]] = p1], c = 1; p1 = p2; s1 = s2]]; seq]; orm[10]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base,more
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|