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A137323
a(n) = sum(d divides n, 2^(n/d-1) - 1 ), omitting d=1 and d=n.
1
0, 0, 0, 1, 0, 4, 0, 8, 3, 16, 0, 42, 0, 64, 18, 135, 0, 290, 0, 534, 66, 1024, 0, 2216, 15, 4096, 258, 8262, 0, 16944, 0, 32902, 1026, 65536, 78, 133415, 0, 262144, 4098, 524948, 0, 1056864, 0, 2098182, 16656, 4194304, 0, 8423590, 63, 16777742
OFFSET
1,6
COMMENTS
Previous name was: Characteristic value of numbers used to compute number of binary expansions of a certain length that have a given number of rotational symmetries.
Sequence is useful in counting binary expansions of length l with r rotational symmetries (we only consider r>1), where r has to be a proper divisor not equal to 1.
We discount numbers with l symmetry, because we know this only occurs once, at (2^l) - 1.
For example, consider binary expansions of length 8. We know that for any number the possible symmetries are the proper divisors of 8 not equal to 1; (2, 4).
So if we would like to find the number of expansions of length 8 that have 2 rotational symmetries, it is [2^(8/2 -1) -1] - a(8/2) = 7 - 1 = 6.
In general it appears that the formula for r rotational-symmetric numbers of expansion length l is the following: [2^(l/r -1) -1] - a(l/r).
REFERENCES
John B. Fraleigh, A first course in abstract algebra, Pearson Education, 2003.
LINKS
FORMULA
a(n) = sum(d divides n, 2^(n/d-1) - 1 ), omitting d=1 and d=n.
EXAMPLE
a(6) = 4 because 6 has (2,3) as proper divisors not equal to one. Plugging these values into the formula we get [2^(6/2 -1)-1] + [2^(6/3 -1) -1] = 3 + 1 = 4.
For p prime, a(p) = 0.
MATHEMATICA
a[1] = 0; a[n_] := Sum[2^(n/d - 1) - 1, {d, Divisors[n][[2 ;; -2]]}];
Array[a, 50] (* Jean-François Alcover, Nov 12 2017 *)
PROG
(PARI) a(n) = sumdiv(n, d, if ((d==1) || (d==n), 0, 2^((n/d -1)) -1)); \\ Michel Marcus, Aug 13 2013
CROSSREFS
Cf. A138904.
Sequence in context: A020763 A229911 A244336 * A021075 A232705 A245174
KEYWORD
base,nonn
AUTHOR
Max Sills, Apr 06 2008
EXTENSIONS
Corrected, extended, and new name by Michel Marcus, Aug 13 2013
STATUS
approved