A293481 Numbers with last digit greater than or equal to 5 (in base 10).
5, 6, 7, 8, 9, 15, 16, 17, 18, 19, 25, 26, 27, 28, 29, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 65, 66, 67, 68, 69, 75, 76, 77, 78, 79, 85, 86, 87, 88, 89, 95, 96, 97, 98, 99, 105, 106, 107, 108, 109, 115, 116, 117, 118, 119, 125, 126, 127, 128, 129
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..500
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).
Programs
-
Magma
[n: n in [0..150] | n mod 10 ge 5];
-
Maple
select(n -> type(ceil(-n/5), odd), [$0..130]); # Peter Luschny, Oct 10 2017
-
Mathematica
LinearRecurrence[{1, 0, 0, 0, 1, -1}, {5, 6, 7, 8, 9, 15}, 70] (* Second program: *) Select[Range[129], Mod[#, 10] >= 5 &] (* Jean-François Alcover, Oct 10 2017 *) -
PARI
select(k -> (k\5) % 2, vector(130, k, k)) \\ Peter Luschny, Oct 10 2017
-
PARI
Vec(x*(5 + x + x^2 + x^3 + x^4 + x^5)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^100)) \\ Colin Barker, Oct 10 2017
-
Python
[k for k in range(130) if (k//5) % 2 == 1] # Peter Luschny, Oct 10 2017
-
Python
def A293481(n): return (n<<1)+3-(n-1)%5 # Chai Wah Wu, Oct 29 2024
-
Sage
[k for k in (0..130) if not 2.divides(k//5)] # Peter Luschny, Oct 10 2017
Formula
G.f.: x*(5 + x + x^2 + x^3 + x^4 + x^5)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-5) - a(n-6).
a(n) = A293292(n) + 5.
a(n) = 2n+3-((n-1) mod 5). - Chai Wah Wu, Oct 29 2024
Comments