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a(n+1) = concatenation of a(n) and a(n)+1, with a(0)=1.

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a(n+1) = concatenation of a(n) followed byand a(n)+1.

The next term is too large to be included in the DATA section. - N. J. A. Sloane, Oct 20 2021

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a(n+1) is the concatenation of ) = a(n) andfollowed by a(n)+1; a(0)=1.

For every nonnegative integer n <= 8, a(n) is 2^n digits long, and each digit d in 1..n has exactly twice as many occurrences in a(n) as d+1 has, and each digit is evenly spaced throughout the sequence of digits in a(n). This makes this sequence useful as a model for spreading some integer number of items among some 2^n recipients if even spreading is a goal. For instance, to distribute five items among eight recipients, one could use a(3)=12131214, and distribute one item to each recipient marked with a 1 and one to each recipient marked with a 3, since 2^(3-1) + 2^(3-3) = 5. - Willis A. Hershey, Aug 25 2021

Cf. A001511, A018238, A082215.

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Sean A. Irvine: I'm not convinced about the value of this comment here, given that it only applies to n <= 8.

N. J. A. Sloane: Agreed.  Let's drop the comment.

Cf. A001511, A018238, A082215.

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Kevin Ryde: The distribution you're looking at may be the ruler A001511: equal spacings, halving frequencies (count low 0 bits).  The digits here are 2^n initial terms from A001511, though only until carries make a mess as David says :).

Kevin Ryde: Maybe the OFFSET=0 here (so a(0)=1) was conceived so a(n) has had n many times concat done to it (none yet at a(0)), or just for the the happy effect that a(n) has 2^n digits.

Kevin Ryde: Actually, there'd be a sum here using A001511 as coefficients rather than digits ...

Willis A. Hershey: A001511 follows the pattern I mentioned better, but in order to get the even spreading quality, you have to choose the first 2^n iterations of the sequence as your model, and since this sequence has 2^n kinda built in I thought it more relevant here

Willis A. Hershey: Also I think Jon's edits are very helpful

Jon E. Schoenfield: Okay, great!  Thanks!  :-)

Kevin Ryde: The sequence here is a bit obscure, and its pattern breaks down after initial terms, so I think you'd be better in A001511 or its friend A007814.  There's a lot of material in both those already.  Not sure if proportions are covered.  I confess I don't entirely follow the aim of the spreading!

As digits go, A018238 and A082215 do similar but slightly different.

For every nonnegative integer n <= 8, a(n) is 2^n digits long, and each digit in the sequenced fromin 1 to ..n has exactly twice as many occurrences as thein digita(n) as succeedingd+1 ithas, and each digit is evenly spaced throughout the sequence. of digits in a(n). This makes this sequence useful as a model for spreading some integer number of items among some 2^n recipients if even spreading is a goal. For instance, to distribute five items among eight recipients, one could use a(3)=12131214, and distribute one item to each recipient marked with a 1 and one to each recipient marked with a 3, since 2^(3-1) + 2^(3-3) = 5. - Willis A. Hershey, Aug 25 2021

Jon E. Schoenfield: I made a few changes to try to clarify things, e.g., that the even spacing "throughout the sequence" referred to the sequence of digits in a given single term a(n) -- not to the sequence of terms a(0), a(1), a(2), ....  Have I mangled the meaning anywhere (I hope not)?

For every nonnegative integer n from 0- <= 8 inclusive, a(n) is 2^n digits long, and the distribution of each digit in the sequence from 1 to n has exactly twice as many occurrences as the digit succeeding it, and each digit is evenly spaced throughout the sequence. This makes this sequence useful as a model for spreading some integer number of items among some 2^n recipients if even spreading is a goal. For instance, to distribute five items among eight recipients, one could use a(3)=12131214, and distribute one item to each recipient marked with a 1 and one to each recipient marked with a 3, since 2^(3-1) + 2^(3-3) = 5. - Willis A. Hershey, Aug 25 2021

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