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A047513 Numbers that are congruent to {0, 1, 2, 4, 6, 7} mod 8. 4
0, 1, 2, 4, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 20, 22, 23, 24, 25, 26, 28, 30, 31, 32, 33, 34, 36, 38, 39, 40, 41, 42, 44, 46, 47, 48, 49, 50, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 70, 71, 72, 73, 74, 76, 78, 79, 80, 81, 82, 84, 86, 87, 88 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
The product of any two terms belongs to the sequence and therefore also a(n)^2, a(n)^3, a(n)^4, etc. - Bruno Berselli, Nov 28 2012
Nonnegative m such that floor(k*(m/4)^2) = k*floor((m/4)^2), where k = 2 or 3. - Bruno Berselli, Dec 03 2015
LINKS
FORMULA
From Chai Wah Wu, May 29 2016: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - a(n-6), n>6.
G.f.: x^2*(x^2 + 1)^2/((x - 1)^2*(x^2 - x + 1)*(x^2 + x + 1)). (End)
From Wesley Ivan Hurt, Jun 15 2016: (Start)
a(n) = (24*n-24-2*sqrt(3)*(cos((1-4*n)*Pi/6)-3*cos((2*n+1)*Pi/6)))/18.
a(6k) = 8k-1, a(6k-1) = 8k-2, a(6k-2) = 8k-4, a(6k-3) = 8k-6, a(6k-4) = 8k-7, a(6k-5) = 8k-8. (End)
Sum_{n>=2} (-1)^n/a(n) = (4-sqrt(2))*log(2)/8 + sqrt(2)*log(sqrt(2)+2)/4. - Amiram Eldar, Dec 27 2021
MAPLE
A047513:=n->(24*n-24-2*sqrt(3)*(cos((1-4*n)*Pi/6)-3*cos((2*n+1)*Pi/6)))/18: seq(A047513(n), n=1..100); # Wesley Ivan Hurt, Jun 15 2016
MATHEMATICA
Select[Range[0, 75], Function[k, Mod[#, 8] == k] /@ Nor[3, 5] &] (* or *) Select[Range[0, 75], Function[k, Floor[k (#/4)^2] == k Floor[(#/4)^2]] /@ Or[2, 3] &] (* Michael De Vlieger, Dec 03 2015 *)
Select[Range[0, 100], MemberQ[{0, 1, 2, 4, 6, 7}, Mod[#, 8]]&] (* Harvey P. Dale, Apr 26 2016 *)
LinearRecurrence[{2, -2, 2, -2, 2, -1}, {0, 1, 2, 4, 6, 7}, 50] (* G. C. Greubel, May 29 2016 *)
PROG
(Magma) [n : n in [0..100] | n mod 8 in [0, 1, 2, 4, 6, 7]]; // Wesley Ivan Hurt, May 29 2016
CROSSREFS
Sequence in context: A153351 A153343 A136499 * A039068 A260399 A229785
KEYWORD
nonn,easy
AUTHOR
STATUS
approved

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Last modified May 16 04:39 EDT 2024. Contains 372549 sequences. (Running on oeis4.)