A137323 a(n) = sum(d divides n, 2^(n/d-1) - 1 ), omitting d=1 and d=n.
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
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.
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..6645
Crossrefs
Cf. A138904.
Programs
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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 *)
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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
Comments