%I #20 Dec 20 2025 17:25:45
%S 0,1,2,3,4,5,10,11,12,13,14,15,20,21,22,23,24,25,30,31,32,33,34,35,40,
%T 41,42,43,44,45,50,51,52,53,54,55,60,61,62,63,64,65,70,71,72,73,74,75,
%U 80,81,82,83,84,85,90,91,92,93,94,95,100,101,102,103,104,105
%N Numbers whose final digit is at most 5.
%C Numbers that are {0, 1, 2, 3, 4, 5} mod 10.
%C Union of A008592, A017281, A017293, A017305, A017317, and A017329.
%C Differs from A007092 for n >= 36. A007092(36) = 100, while a(36) = 60.
%C In cricket, balls bowled are measured in whole overs of 6 balls and a partial over of 0 to 5 balls.
%H Zhuorui He, <a href="/A391164/b391164.txt">Table of n, a(n) for n = 0..10000</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Over_(cricket)">Over (cricket)</a>.
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,1,-1).
%F a(n) = 10 * floor(n/6) + (n mod 6) = 10 * A152467(n) + A010875(n).
%F G.f.: x*(1 + x + x^2 + x^3 + x^4 + 5*x^5)/((-1 + x)^2*(1 + x)*(1 - x + x^2)*(1 + x + x^2)).
%e Bowling 63 balls results in 10.3 overs, so a(63) = 103.
%t a[n_] := Total[QuotientRemainder[n, 6] * {10, 1}]; Array[a, 66, 0]
%Y Cf. A010875, A017281, A017293, A017305, A017317, A017329, A152467.
%Y Cf. A008592 (at most 0), A197652 (at most 1), A391163 (at most 2), A390774 (at most 3), A293292 (at most 4), A272576 (at most 7), A001477 (trivial, at most 9).
%K nonn,easy,base
%O 0,3
%A _Jason Bard_, Dec 01 2025