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A047390
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Numbers that are congruent to {0, 3, 5} mod 7.
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2
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0, 3, 5, 7, 10, 12, 14, 17, 19, 21, 24, 26, 28, 31, 33, 35, 38, 40, 42, 45, 47, 49, 52, 54, 56, 59, 61, 63, 66, 68, 70, 73, 75, 77, 80, 82, 84, 87, 89, 91, 94, 96, 98, 101, 103, 105, 108, 110, 112, 115, 117, 119, 122, 124, 126, 129, 131, 133, 136, 138, 140, 143
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OFFSET
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1,2
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COMMENTS
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Also numbers k such that k*(k+2)*(k+4) is divisible by 7. - Bruno Berselli, Dec 28 2017
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LINKS
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FORMULA
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a(n) = 2*n + floor(n/3) + (n^2 mod 3), with offset 0, a(0)=0. - Gary Detlefs, Mar 19 2010
G.f.: x^2*(3 + 2*x + 2*x^2)/((1 - x)^2*(1 + x + x^2)).
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = 7*n/3 - 2 - 2*sin(2*n*Pi/3)/(3*sqrt(3)).
a(3*k) = 7*k-2, a(3*k-1) = 7*k-4, a(3*k-2) = 7*k-7. (End)
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MAPLE
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seq(2*n+floor(n/3)+(n^2 mod 3), n=0..52); # Gary Detlefs, Mar 19 2010
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MATHEMATICA
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Select[Range[0, 150], MemberQ[{0, 3, 5}, Mod[#, 7]]&] (* Harvey P. Dale, Dec 07 2011 *)
CoefficientList[Series[x (3 + 2 x + 2 x^2)/((1 - x)^2 (1 + x + x^2)), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 02 2014 *)
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PROG
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(Magma) [n: n in [0..122] | n mod 7 in [0, 3, 5]]; // Bruno Berselli, Mar 29 2011
(Python)
import math
a = lambda n: 2*(n-1)+math.ceil((n-1)/3.0)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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