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A325203 a(n) is 10^n represented in bijective base-9 numeration.

%I A325203 #38 Feb 16 2025 08:33:58
%S A325203 1,11,121,1331,14641,162151,1783661,19731371,228145181,2519596991,
%T A325203 27726678111,315994569221,3477151372431,38358665196741,
%U A325203 432956427275251,4763631711137761,53499948822526471,588621548147792281,6585837139636825191,73555318547116177211
%N A325203 a(n) is 10^n represented in bijective base-9 numeration.
%C A325203 Differs from A055479 first at n = 7: a(7) = 19731371 < 20731371 = A055479(7).
%C A325203 Also: the (10^n)-th zeroless number. - _M. F. Hasler_, Jan 13 2020
%H A325203 Alois P. Heinz, <a href="/A325203/b325203.txt">Table of n, a(n) for n = 0..954</a>
%H A325203 R. R. Forslund, <a href="http://www.emis.de/journals/SWJPAM/Vol1_1995/rrf01.ps">A logical alternative to the existing positional number system</a>, Southwest Journal of Pure and Applied Mathematics, Vol. 1, 1995, 27-29.
%H A325203 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Zerofree.html">Zerofree</a>
%H A325203 Wikipedia, <a href="https://en.wikipedia.org/wiki/Bijective_numeration">Bijective numeration</a>
%F A325203 a(n) = A052382(10^n) = A052382(A011557(n)).
%e A325203 a(1) = 11_bij9   =                 1*9^1 + 1*9^0 =           9+1 =   10.
%e A325203 a(2) = 121_bij9  =         1*9^2 + 2*9^1 + 1*9^0 =       81+18+1 =  100.
%e A325203 a(3) = 1331_bij9 = 1*9^3 + 3*9^2 + 3*9^1 + 1*9^0 =  729+243+27+1 = 1000.
%e A325203 a(7) = 19731371_bij9 = 9*(9*(9*(9*(9*(9*(9*1+9)+7)+3)+1)+3)+7)+1 = 10^7.
%p A325203 b:= proc(n) local d, l, m; m:= n; l:= "";
%p A325203       while m>0 do d:= irem(m, 9, 'm');
%p A325203         if d=0 then d:=9; m:= m-1 fi; l:= d, l
%p A325203       od; parse(cat(l))
%p A325203     end:
%p A325203 a:= n-> b(10^n):
%p A325203 seq(a(n), n=0..23);
%o A325203 (PARI) A325203(n)=A052382(10^n) \\ _M. F. Hasler_, Jan 13 2020
%Y A325203 Cf. A011557, A052382 (zeroless numbers), A055479, A309908.
%K A325203 nonn,base
%O A325203 0,2
%A A325203 _Alois P. Heinz_, Sep 05 2019