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%I #17 Nov 12 2017 09:13:26
%S 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,
%T 4096,258,8262,0,16944,0,32902,1026,65536,78,133415,0,262144,4098,
%U 524948,0,1056864,0,2098182,16656,4194304,0,8423590,63,16777742
%N a(n) = sum(d divides n, 2^(n/d-1) - 1 ), omitting d=1 and d=n.
%C 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.
%C 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.
%C We discount numbers with l symmetry, because we know this only occurs once, at (2^l) - 1.
%C 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).
%C 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.
%C 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).
%D John B. Fraleigh, A first course in abstract algebra, Pearson Education, 2003.
%H Seiichi Manyama, <a href="/A137323/b137323.txt">Table of n, a(n) for n = 1..6645</a>
%F a(n) = sum(d divides n, 2^(n/d-1) - 1 ), omitting d=1 and d=n.
%e 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.
%e For p prime, a(p) = 0.
%t a[1] = 0; a[n_] := Sum[2^(n/d - 1) - 1, {d, Divisors[n][[2 ;; -2]]}];
%t Array[a, 50] (* _Jean-François Alcover_, Nov 12 2017 *)
%o (PARI) a(n) = sumdiv(n, d, if ((d==1) || (d==n), 0, 2^((n/d -1)) -1)); \\ _Michel Marcus_, Aug 13 2013
%Y Cf. A138904.
%K base,nonn
%O 1,6
%A _Max Sills_, Apr 06 2008
%E Corrected, extended, and new name by _Michel Marcus_, Aug 13 2013
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