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%I #31 Mar 26 2019 19:36:38
%S 1,3,9,27,121,313,939,27927,12127121,313121313,939313939,
%T 2792793927927,121271212792712127121,31312131312127121313121313,
%U 939313939313121313939313939,27927939279279393139392792793927927,1212712127927121271212792793927927121271212792712127121
%N a(1) = 1, for n >= 2 let the digits of a(n-1) be d_1, ..., d_i in base 10. Starting from d_1 do the following procedure: if d_r is divisible by 2, then d_r_new = d_r / 2, otherwise d_r_new = 3*d_r. a(n) = concatenation of d_r_new for r = 1 to i.
%C Let a(1) has r digits x_1, ..., x_r. The count of the digits 1, ..., 9 in a(1) is y_1, ..., y_9; y_i >= 0. For the digits 1, ..., 9 there exist constants c_1, ..., c_9; c_j irrational. Let k = (c_1*y_1 + ... + c_9*y_9). Then the number of digits of a(n) is approximately equal to k * (1.290554)^n and the sum of the digits of a(n) is approximately equal to 3.719315 * k * (1.290554)^n. This sequence has a(1) = 1, thus c_1*y_1 = 0.670173 * 1 and the number of digits of a(n) is approximately equal to 0.670173 * (1.290554)^n. The sum of digits of a(n) is approximately equal to 2.492585 * (1.290554)^n. This is based on empirical observations.
%e a(1) = 1;
%e a(2) = 3*1 = 3;
%e a(3) = 3*3 = 9;
%e a(4) = 3*9 = 27;
%e a(5) = 2/2 concatenated with 3*7 = 121;
%e a(6) = 3*1 concatenated with 2/2 concatenated with 3*1 = 313;
%e and so on.
%o (PARI) replace_digits(n) = my(d=digits(n), e=[]); for(k=1, #d, if(d[k]%2==0, e=concat(e, d[k]/2), my(ee=digits(3*d[k])); for(r=1, #ee, e=concat(e, ee[r])))); subst(Pol(e), x, 10)
%o terms(n) = my(x=1, i=0); while(i < n, print1(x, ", "); i++; x=replace_digits(x))
%o /* Print initial 17 terms as follows: */
%o terms(17) \\ _Felix Fröhlich_, Mar 05 2019
%Y Cf. A006370.
%K nonn,base
%O 1,2
%A _Ctibor O. Zizka_, Mar 05 2019
%E More terms from _Felix Fröhlich_, Mar 05 2019
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