cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A137323 a(n) = sum(d divides n, 2^(n/d-1) - 1 ), omitting d=1 and d=n.

Original entry on oeis.org

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

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Author

Max Sills, Apr 06 2008

Keywords

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).

Examples

			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.
		

References

  • John B. Fraleigh, A first course in abstract algebra, Pearson Education, 2003.

Crossrefs

Cf. A138904.

Programs

  • 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 *)
  • PARI
    a(n) = sumdiv(n, d, if ((d==1) || (d==n), 0, 2^((n/d -1)) -1)); \\ Michel Marcus, Aug 13 2013

Formula

a(n) = sum(d divides n, 2^(n/d-1) - 1 ), omitting d=1 and d=n.

Extensions

Corrected, extended, and new name by Michel Marcus, Aug 13 2013