login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A290952
Multi de Bruijn Sequences: Number of ways to arrange 2^(n+1) binary digits in a circle so that every length n binary string occurs exactly twice.
1
2, 5, 82, 52496, 44079843328, 62177039921456290463744, 247422994777239366039696433386055989663945981952, 7835921708100840781377057397856335571660942358870727003819788990112934851947892015462438777389056
OFFSET
1,1
COMMENTS
Let m,q,n be positive integers. A cyclic multi de Bruijn sequence is a cyclic sequence over a q-ary alphabet in which every q-ary word of length n occurs exactly m times. Each such sequence has length m*q^n. Tesler (2017) shows the number of cyclic multi de Bruijn sequences is 1/(m*q^n) Sum_{r|m} phi(m/r) * ((r*q)! / (r!^q))^(q^(n-1)), where phi() is the Euler totient function A000010. Case m=1 is de Bruijn sequences; see A016031 for binary de Bruijn sequences (m=1, q=2, n>=1). Case m=2, q=2, n>=1 is a(n).
LINKS
Glenn Tesler, Multi de Bruijn sequences, Journal of Combinatorics, Vol. 8, No. 3 (2017), 439-474 Preprint.
FORMULA
a(n) = (6^(2^(n-1)) + 2^(2^(n-1))) / 2^(n+1).
EXAMPLE
For n=1, the a(1)=2 solutions are (0011) and (0101); each has two 0's and two 1's. Cyclic sequences have multiple representations via circular shifts: (0011)=(1001)=(1100)=(0110) all count as the same cyclic sequence, as do (0101)=(1010).
For n=2, the a(2)=5 solutions are (00010111), (00011011), (00011101), (00100111), and (00110011); each has two occurrences of each of 00, 01, 10, and 11. Note that in a cyclic sequence, occurrences may wrap around: in (00010111), there is one 10 in the middle, and another 10 that wraps around from the end to the start. Or, use a different rotation of the sequence: (00010111)=(10001011) shows both occurrences of 10 without wrapping.
For n=3 and 4, see the links section.
MATHEMATICA
Table[(6^(2^(n - 1)) + 2^(2^(n - 1)))/2^(n + 1), {n, 8}] (* Michael De Vlieger, Aug 15 2017 *)
PROG
(PARI) a(n) = (6^(2^(n-1)) + 2^(2^(n-1))) / 2^(n+1) \\ Felix Fröhlich, Aug 15 2017
CROSSREFS
Cf. A016031.
Sequence in context: A123978 A303482 A360503 * A201113 A309089 A120798
KEYWORD
nonn,easy
AUTHOR
Glenn Tesler, Aug 14 2017
STATUS
approved