A008854 Numbers that are congruent to {0, 1, 4} mod 5.
0, 1, 4, 5, 6, 9, 10, 11, 14, 15, 16, 19, 20, 21, 24, 25, 26, 29, 30, 31, 34, 35, 36, 39, 40, 41, 44, 45, 46, 49, 50, 51, 54, 55, 56, 59, 60, 61, 64, 65, 66, 69, 70, 71, 74, 75, 76, 79, 80, 81, 84, 85, 86, 89, 90, 91, 94, 95, 96, 99, 100, 101, 104, 105, 106, 109
Offset: 1
References
- L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, p. 459.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Craig Knecht, Triangle where every "triple" contains 1,2,2.
- Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
Programs
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Magma
[n : n in [0..150] | n mod 5 in [0, 1, 4]]; // Wesley Ivan Hurt, Jun 14 2016
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Maple
for n to 1000 do if n^3 - n mod 10 = 0 then print(n); fi; od;
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Mathematica
Select[Range[0, 150], MemberQ[{0, 1, 4}, Mod[#, 5]] &] (* or *) LinearRecurrence[{1, 0, 1, -1}, {0, 1, 4, 5}, 91] (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *) CoefficientList[Series[x (1 + 3 x + x^2) / ((1 + x + x^2) (x - 1)^2), {x, 0, 70}], x] (* Vincenzo Librandi, Jun 11 2013 *)
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PARI
concat(0, Vec(x^2*(1+3*x+x^2)/((1+x+x^2)*(x-1)^2) + O(x^100))) \\ Altug Alkan, Dec 03 2015
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PARI
a(n) = vecsum(divrem(5*n-7,3)); \\ Kevin Ryde, Aug 08 2022
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Python
def a(n): return sum(divmod(5*n-7, 3)) print([a(n) for n in range(1, 67)]) # Michael S. Branicky, Aug 08 2022 after Kevin Ryde
Formula
G.f.: x^2*(1+3*x+x^2) / ((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
a(n) = A047217(n+1)-1. - R. J. Mathar, Aug 04 2015
E.g.f: (5/3)*(x-1)*exp(x) + (2/3)*exp(-x/2)*cos(sqrt(3)*x/2) + (2/9)*exp(-x/2)*sin(sqrt(3)*x/2) + 1. - Robert Israel, Aug 04 2015
From Wesley Ivan Hurt, Jun 14 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = (15*n-15+6*cos(2*n*Pi/3)+2*sqrt(3)*sin(2*n*Pi/3))/9.
a(3k) = 5k-1, a(3k-1) = 5k-4, a(3k-2) = 5k-5. (End)
a(n) = 5*n/3 - 2*(n mod 3)/3 - 1. - Ammar Khatab, Aug 26 2020
Sum_{n>=2} (-1)^n/a(n) = 3*log(2)/5 - arccoth(3/sqrt(5))/sqrt(5). - Amiram Eldar, Dec 10 2021
From Peter Bala, Aug 04 2022: (Start)
a(n) = a(floor(n/2)) + a(1 + ceiling(n/2)) for n >= 4 with a(1) = 0, a(2) = 1 and a(3) = 4.
Comments