A131717 Natural numbers A000027 with 6n+4 and 6n+5 terms swapped.
1, 2, 3, 5, 4, 6, 7, 8, 9, 11, 10, 12, 13, 14, 15, 17, 16, 18, 19, 20, 21, 23, 22, 24, 25, 26, 27, 29, 28, 30, 31, 32, 33, 35, 34, 36, 37, 38, 39, 41, 40, 42, 43, 44, 45, 47, 46, 48, 49, 50, 51, 53, 52, 54, 55, 56, 57, 59, 58, 60, 61, 62, 63, 65, 64, 66, 67, 68, 69, 71, 70, 72
Offset: 1
Links
- Frieder Mittmann, Table of n, a(n) for n = 1..1002
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,1,-1).
Crossrefs
Cf. A131042.
Programs
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Maple
seq(seq(6*i+s,s=[1,2,3,5,4,6]),i=0..100); # Robert Israel, Nov 11 2014
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Mathematica
Drop[CoefficientList[Series[x (2x^5 - x^4 + 2x^3 + x^2 + x + 1)/((x - 1)^2 (x + 1) (x^2 - x + 1) (x^2 + x + 1)), {x, 0, 100}], x], 1] (* Indranil Ghosh, Apr 18 2017 *) Table[Sum[(7 #1 - 13 #2 + 17 #3 - 3 #4 + 2 #5 + 2 #6)/30 & @@ Mod[k + Range[0, 5], 6], {k, 0, n}], {n, 0, 71}] (* Michael De Vlieger, Apr 22 2017 *)
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PARI
Vec(x*(2*x^5-x^4+2*x^3+x^2+x+1)/((x-1)^2*(x+1)*(x^2-x+1)*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Nov 11 2014
Formula
a(n) = A008585(n/3) if n is congruent to 0 mod 3. - Frieder Mittmann, Nov 11 2014
a(n) = A007310((n-1)/3) if n is congruent to 1 mod 3. - Frieder Mittmann, Nov 11 2014
a(n) = A047235((n-2)/3) if n is congruent to 2 mod 3. - Frieder Mittmann, Nov 11 2014
G.f.: x*(2*x^5-x^4+2*x^3+x^2+x+1) / ((x-1)^2*(x+1)*(x^2-x+1)*(x^2+x+1)). - Colin Barker, Nov 11 2014
a(n) = (24*floor(n/6)-3*(n^2-3*n-2)-9*floor(n/3)*(3*floor(n/3)-2*n+3)+(-1)^floor(n/3)*(3*n^2-5*n-6+3*floor(n/3)*(9*floor(n/3)-6*n+5)))/4. - Luce ETIENNE, Apr 18 2017
Comments