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A109733 Sequence is its own base-10 ASCII representation.

%I A109733 #29 Dec 31 2020 02:20:20
%S A109733 53,51,53,49,53,51,52,57,53,51,53,49,53,50,53,55,53,51,53,49,53,51,52,
%T A109733 57,53,51,53,48,53,51,53,53,53,51,53,49,53,51,52,57,53,51,53,49,53,50,
%U A109733 53,55,53,51,53,49,53,51,52,56,53,51,53,49,53,51,53,51,53,51,53,49,53
%N A109733 Sequence is its own base-10 ASCII representation.
%C A109733 Out of the digits 0,1,...,9, 5 is the only one whose ASCII representation, converted to base 10, begins with itself. So this sequence is the unique one with this property. - _N. J. A. Sloane_, Aug 25 2015
%C A109733 The octal version of this idea is simply 66,66,66,66,.., not interesting.
%H A109733 David W. Wilson, <a href="/A109733/b109733.txt">Table of n, a(n) for n = 1..10000</a>
%H A109733 <a href="/index/Fi#FIXEDPOINTS">Index entries for sequences that are fixed points of mappings</a>
%F A109733 I conjecture that a(n) = 53 when n is congruent to 1, 3 or 5 (mod 8) and a(n) = 51 when n is congruent to 2 (mod 8). - Jonathan Cross (jcross(AT)juggler.net), Oct 14 2005
%F A109733 a(n) = (if valuation(n/2^v + 1, 2) mod 5 = 3 then 56 else 57) - 2*((v-3) mod 5), where v = valuation(n,2), i.e., n = (2s+1)*2^v. (Translation of my PARI code from June 2011.) - _M. F. Hasler_, Feb 02 2016
%e A109733 We use the following table, giving digit d, ASCII equivalent in base 8, ASCII equivalent in base 10:
%e A109733   .0..1..2..3..4..5..6..7..8..9
%e A109733   60 61 62 63 64 65 66 67 70 71
%e A109733   48 49 50 51 52 53 54 55 56 57
%e A109733 We must start with 5 (see comment above), so the sequence grows like this:
%e A109733   5
%e A109733   53
%e A109733   53 51
%e A109733   53 51 53 49
%e A109733   53 51 53 49 53 51 52 57
%e A109733   ...
%o A109733 (PARI) a(n) = (valuation(1+n>>n=valuation(n,2),2)%5!=3)+56-(n-3)%5*2  \\ _M. F. Hasler_, Jun 20 2011
%Y A109733 See A109648 for another version.
%K A109733 easy,nonn,base
%O A109733 1,1
%A A109733 _N. J. A. Sloane_ and _Nadia Heninger_, Aug 10 2005
%E A109733 More terms from Jonathan Cross (jcross(AT)juggler.net), Oct 14 2005