A293292 Numbers with last digit less than 5 (in base 10).
0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 70, 71, 72, 73, 74, 80, 81, 82, 83, 84, 90, 91, 92, 93, 94, 100, 101, 102, 103, 104, 110, 111, 112, 113, 114, 120, 121, 122, 123, 124, 130
Offset: 1
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).
Crossrefs
Programs
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Magma
[n: n in [0..130] | n mod 10 lt 5];
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Magma
[n: n in [0..130] | IsEven(Floor(n/5))];
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Magma
[n+5*Floor(n/5): n in [0..70]];
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Maple
select(k -> type(floor(k/5), even), [$0..130]); # Peter Luschny, Oct 05 2017
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Mathematica
Table[n + 5 Floor[n/5], {n, 0, 70}] Reap[For[k = 0, k <= 130, k++, If[Floor[k/5] == 2*Floor[k/10], Sow[k]]]][[2, 1]] (* or *) LinearRecurrence[{1, 0, 0, 0, 1, -1}, {0, 1, 2, 3, 4, 10}, 66] (* Jean-François Alcover, Oct 05 2017 *) -
PARI
concat(0, Vec(x^2*(1 + x + x^2 + x^3 + 6*x^4) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^70))) \\ Colin Barker, Oct 05 2017
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PARI
select(k->floor(k/5) == 2*floor(k/10), vector(1000, k, k)) \\ Colin Barker, Oct 05 2017
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Python
[k for k in range(131) if (k//5) % 2 == 0] # Peter Luschny, Oct 05 2017
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Python
def A293292(n): return (n-1<<1)-(n-1)%5 # Chai Wah Wu, Oct 29 2024
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Sage
[k for k in (0..130) if 2.divides(floor(k/5))] # Peter Luschny, Oct 05 2017
Formula
G.f.: x^2*(1 + x + x^2 + x^3 + 6*x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-5) - a(n-6).
a(n) = (n-1) + 5*floor((n-1)/5) = 10*floor((n-1)/5) + ((n-1) mod 5).
a(n) = 2n-2-((n-1) mod 5). - Chai Wah Wu, Oct 29 2024
Extensions
Definition by David A. Corneth, Oct 05 2017
Comments