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A220018
Number of cyclotomic cosets of 3 mod 10^n.
2
4, 11, 29, 71, 129, 203, 293, 399, 521, 659, 813, 983, 1169, 1371, 1589, 1823, 2073, 2339, 2621, 2919, 3233, 3563, 3909, 4271, 4649, 5043, 5453, 5879, 6321, 6779, 7253, 7743, 8249, 8771, 9309, 9863, 10433, 11019, 11621, 12239, 12873, 13523, 14189
OFFSET
1,1
FORMULA
a(n) = A221855(n) - 1 for all n >= 2, a(1) = A221855(1).
Empirical G.f.: x*(8*x^4-13*x^3-8*x^2+x-4) / (x-1)^3. [Colin Barker, Feb 03 2013]
Conjecture: a(n) = 8*n^2-14*n-1 for n>2. a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>5. [Colin Barker, Apr 14 2013]
EXAMPLE
a(2) = 11 because there are 11 cyclotomic cosets of 3 mod 100:
O(0) = {0},
O(1) = {1, 3, 9, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 7, 21, 63, 89, 67},
O(2) = {2, 6, 18, 54, 62, 86, 58, 74, 22, 66, 98, 94, 82, 46, 38, 14, 42, 26, 78, 34},
O(4) = {4, 12, 36, 8, 24, 72, 16, 48, 44, 32, 96, 88, 64, 92, 76, 28, 84, 52, 56, 68},
O(5) = {5, 15, 45, 35},
O(10) = {10, 30, 90, 70},
O(11) = {11, 33, 99, 97, 91, 73, 19, 57, 71, 13, 39, 17, 51, 53, 59, 77, 31, 93, 79, 37},
O(20) = {20, 60, 80, 40},
O(25) = {25, 75},
O(50) = {50},
O(55) = {55, 65, 95, 85}.
MATHEMATICA
a[n_]:=DivisorSum[10^n, EulerPhi[#]/MultiplicativeOrder[3, #]&]; Array[a, 50] (* Ray Chandler, Jul 03 2023, after Jean-François Alcover *)
PROG
(PARI) A220018(n)=sumdiv(10^n, d, eulerphi(d)/znorder(Mod(3, d)))
CROSSREFS
Cf. A006694.
Sequence in context: A275012 A055418 A062432 * A131046 A109803 A262280
KEYWORD
base,nonn
AUTHOR
V. Raman, Jan 27 2013
EXTENSIONS
Edited by M. F. Hasler, Jan 28 2013
STATUS
approved