A047241 Numbers that are congruent to {1, 3} mod 6.
1, 3, 7, 9, 13, 15, 19, 21, 25, 27, 31, 33, 37, 39, 43, 45, 49, 51, 55, 57, 61, 63, 67, 69, 73, 75, 79, 81, 85, 87, 91, 93, 97, 99, 103, 105, 109, 111, 115, 117, 121, 123, 127, 129, 133, 135, 139, 141, 145, 147, 151, 153, 157, 159, 163, 165, 169, 171, 175, 177, 181, 183
Offset: 1
References
- L. Lovasz, J. Pelikan, K. Vesztergombi, Discrete Mathematics, Springer (2003); 14.4, p. 225.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
- L. Lovász, J. Pelikán and K. Vesztergombi, Discrete Mathematics, Elementary and Beyond, Springer (2003); 14.4, p. 225.
- Kival Ngaokrajang, Illustration of initial terms.
- Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
Crossrefs
Programs
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Haskell
a047241 n = a047241_list !! (n-1) a047241_list = 1 : 3 : map (+ 6) a047241_list -- Reinhard Zumkeller, Feb 19 2013
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Maple
seq(3*k-2-((k+1) mod 2), k=1..100); # Wesley Ivan Hurt, Sep 28 2013
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Mathematica
Table[{2, 4}, {30}] // Flatten // Prepend[#, 1]& // Accumulate (* Jean-François Alcover, Jun 10 2013 *) Select[Range[200], MemberQ[{1, 3}, Mod[#, 6]]&] (* or *) LinearRecurrence[{1, 1, -1}, {1, 3, 7}, 70] (* Harvey P. Dale, Oct 01 2013 *)
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PARI
a(n)=bitor(3*n-3,1) \\ Charles R Greathouse IV, Sep 28 2013
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Python
for n in range(1,10**5):print(3*n-2-((n+1)%2)) # Soumil Mandal, Apr 14 2016
Formula
From Paul Barry, Sep 04 2003: (Start)
O.g.f.: (1 + 2*x + 3*x^2)/((1 + x)*(1 - x)^2) = (1 + 2*x + 3*x^2)/((1 - x)*(1 - x^2)).
E.g.f.: (6*x + 1)*exp(x)/2 + exp(-x)/2;
a(n) = 3*n - 5/2 - (-1)^n/2. (End)
a(n) = 2*floor((n-1)/2) + 2*n - 1. - Gary Detlefs, Mar 18 2010
a(n) = 6*n - a(n-1) - 8 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 05 2010
a(n) = 3*n - 2 - ((n+1) mod 2). - Wesley Ivan Hurt, Jun 29 2013
a(1)=1, a(2)=3, a(3)=7; for n>3, a(n) = a(n-1) + a(n-2) - a(n-3). - Harvey P. Dale, Oct 01 2013
From Benedict W. J. Irwin, Apr 13 2016: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)) + log(3)/4. - Amiram Eldar, Dec 11 2021
Extensions
Formula corrected by Bruno Berselli, Jun 24 2010
Comments