A306681 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.

1, 3, 9, 27, 121, 313, 939, 27927, 12127121, 313121313, 939313939, 2792793927927, 121271212792712127121, 31312131312127121313121313, 939313939313121313939313939, 27927939279279393139392792793927927, 1212712127927121271212792793927927121271212792712127121

Offset

1,2

Comments

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.

Links

Table of n, a(n) for n=1..17.

Example

a(1) = 1;
a(2) = 3*1 = 3;
a(3) = 3*3 = 9;
a(4) = 3*9 = 27;
a(5) = 2/2 concatenated with 3*7 = 121;
a(6) = 3*1 concatenated with 2/2 concatenated with 3*1 = 313;
and so on.

Prog

(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)
terms(n) = my(x=1, i=0); while(i < n, print1(x, ", "); i++; x=replace_digits(x))
/* Print initial 17 terms as follows: */
terms(17) \\ Felix Fröhlich, Mar 05 2019

Crossrefs

Cf. A006370.

Sequence in context: A121746 A323927 A146151 * A254334 A375093 A028855

Adjacent sequences: A306678 A306679 A306680 * A306682 A306683 A306684

Keyword

nonn,base

Author

Ctibor O. Zizka, Mar 05 2019

Extensions

More terms from Felix Fröhlich, Mar 05 2019

Status

approved