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Let S_n denote the list of decimal numbers 0 to n, written backwards (allowing leading zeros) and arranged in lexicographic order; a(n) = position where backwards-n appears, starting indexing at 0.
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%I #35 Nov 05 2023 09:04:08

%S 0,1,2,3,4,5,6,7,8,9,1,3,5,7,9,11,13,15,17,19,2,5,8,11,14,17,20,23,26,

%T 29,3,7,11,15,19,23,27,31,35,39,4,9,14,19,24,29,34,39,44,49,5,11,17,

%U 23,29,35,41,47,53,59,6,13,20,27,34,41,48,55,62,69,7,15,23,31,39,47,55,63,71,79,8,17,26,35,44,53,62,71,80,89,9,19,29,39,49,59,69,79,89,99,1,12

%N Let S_n denote the list of decimal numbers 0 to n, written backwards (allowing leading zeros) and arranged in lexicographic order; a(n) = position where backwards-n appears, starting indexing at 0.

%H Alois P. Heinz, <a href="/A264600/b264600.txt">Table of n, a(n) for n = 0..20000</a> (first 10000 terms from Chai Wah Wu)

%F a(0) = 0, a(10n+m) = a(n) + m*(n+1) for m in {0,...,9}. - _Alois P. Heinz_, Nov 20 2015

%e S_0 = [0], so a(0)=0,

%e ...

%e S_9 = [0,1,2,3,4,5,6,7,8,9], so a(9) = 9,

%e S_10 = [0,01,1,2,3,4,5,6,7,8,9], so a(10) = 1,

%e S_11 = [0,01,1,11,2,3,4,5,6,7,8,9], so a(11) = 3,

%e ...

%e S_20 = [0,01,02,1,11,2,21,3,31,4,41,5,51,6,61,7,71,8,81,9,91], so a(20) = 2, and so on

%p a:= proc(n) option remember; `if`(n=0, 0,

%p irem(n, 10, 'r')*(r+1)+a(r))

%p end:

%p seq(a(n), n=0..101); # _Alois P. Heinz_, Nov 20 2015

%t A264600[0]=0;A264600[n_]:=A264600[n]=Mod[n,10](Floor[n/10]+1)+A264600[Floor[n/10]];Array[A264600,100,0] (* _Paolo Xausa_, Nov 04 2023, after _Alois P. Heinz_ *)

%o (Python)

%o def A264600(n):

%o return sorted(str(i)[::-1] for i in range(n+1)).index(str(n)[::-1]) # _Chai Wah Wu_, Nov 20 2015

%Y Decimal analog of A264596.

%Y Has same beginning as A061486 but is ultimately different: see A264668.

%K nonn,base

%O 0,3

%A _N. J. A. Sloane_, Nov 20 2015

%E More terms from _Alois P. Heinz_, Nov 20 2015