A220021 Number of cyclotomic cosets of 11 mod 10^n.
10, 27, 65, 119, 189, 275, 377, 495, 629, 779, 945, 1127, 1325, 1539, 1769, 2015, 2277, 2555, 2849, 3159, 3485, 3827, 4185, 4559, 4949, 5355, 5777, 6215, 6669, 7139, 7625, 8127, 8645, 9179, 9729, 10295, 10877, 11475, 12089, 12719, 13365, 14027, 14705, 15399, 16109, 16835, 17577, 18335, 19109, 19899
Offset: 1
Examples
a(2) = 27 because there are 27 cyclotomic cosets of 11 mod 100:
{1, 11, 21, 31, 41, 51, 61, 71, 81, 91}
{3, 33, 63, 93, 23, 53, 83, 13, 43, 73}
{7, 77, 47, 17, 87, 57, 27, 97, 67, 37}
{9, 99, 89, 79, 69, 59, 49, 39, 29, 19}
{2, 22, 42, 62, 82}
{12, 32, 52, 72, 92}
{4, 44, 84, 24, 64}
{14, 54, 94, 34, 74}
{6, 66, 26, 86, 46}
{16, 76, 36, 96, 56}
{8, 88, 68, 48, 28}
{18, 98, 78, 58, 38}
{5, 55}
{15, 65}
{25, 75}
{35, 85}
{45, 95}
{0}
{10}
{20}
{30}
{40}
{50}
{60}
{70}
{80}
{90}
Links
- Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).
Programs
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Mathematica
a[n_] := DivisorSum[10^n, EulerPhi[#] / MultiplicativeOrder[11, #] &]; Array[a, 50] (* Jean-François Alcover, Dec 18 2015 *)
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PARI
for(n=1,50,print1(sumdiv(10^n, d, eulerphi(d)/znorder(Mod(11, d)))", "))
Formula
Conjecture: a(n) = 8*n^2-2*n-1 for n>1. a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>4. G.f.: x*(5*x^3-14*x^2+3*x-10) / (x-1)^3. - Colin Barker, Apr 13 2013
Comments