cp's OEIS Frontend

a(n) points to the center of each palindromic row/subrange of A233270, and to the lower position nearest to the center, if the length of range is even.

For all n, A218602(a(n)) = a(n) + (1-A000035(A213709(n-1))).

The sequence tells how many positions to the left of center of each row/subrange (of irregular tables like A233270, central point given by A233268) the sequences A233271 and A218616 cross each other (please see the linked graph).

a(n) tells in what number we end in n steps, when we start climbing up the infinite trunk of the "binary beanstalk" from its root (zero). The name "beanstalk" is due to Antti Karttunen.

There are many finite sequences such as 0,1,2; 0,1,3,4,7,9; etc. obeying the same condition (see A218254) and as the length increases, so (necessarily) does the similarity to this infinite sequence.

These are iterates of A233272: a(0)=0, and for n>0, a(n) = A233272(a(n-1)). The difference from A216431 stems from the fact that it uses A023416 to count the 0-bits in the binary expansion of n, while this sequence uses A080791, which results a slightly different start for the iteration, and a much better alignment with sequences related to "infinite trunk of binary beanstalk", A179016.

Apart from term a(2)=2, it seems that each term a(n) >= A179016(n). Please see their ratio plotted with Plot2, and also their differences: A233270.

Conjecture: A179016(a(n))= 2^n for all n apart from n=2. This is true if all powers of 2 except 2 itself occur in A179016 as in that case they must occur at positions given by this sequence.

This is easy to prove: It suffices to note that after 3 no integer of form (2^k)+1 can occur in A005187, thus for all k >= 2, A213725((2^k)+1) = 1 or equally: A213714((2^k)+1) = 0. - Antti Karttunen, Jun 12 2013

This permutation can be used to map between the sequences A179016 and A218616. E.g. A179016(n) = A218616(a(n)) and vice versa: A218616(n) = A179016(a(n)).

This can be viewed as an irregular table: after the initial zero on row 0, start each row n with term x = (2^(n+1))-2 and subtract repeatedly the number of runs in binary representation of x to get successive x's, until the number that has already been listed (which is always (2^n)-2) is encountered, which is not listed second time, but instead, the current row is finished [and thus containing only terms of equal binary length, A000523(n) on row n]. The next row then starts with (2^(n+2))-2, with the same process repeated.

The terms for row n are computed as A227451(n), A226062(A227451(n)), A226062(A226062(A227451(n))), etc. until a term that is a fixed point of A226062 is reached (A037481(n)), which will be the last term of row n.

Row n has A002061(n) = 1,1,3,7,13,21,... terms.

Can be viewed also as an irregular table: after the initial zero on row 0, start each row n with (n!)-1 and subtract repeatedly the sum of factorial expansion digits (A034968) to get successive terms, until the number that has already been listed [which is always (n-1)!-1] is encountered, which is not listed second time, but instead, the current row is finished and the next row starts with ((n+1)!-1), with the same process repeated.

Contains the terms in the infinite trunk of factorial beanstalk (A219666) listed in partially reversed manner: after the initial zero each subsequence lists A219661(n) successive terms from A219666, descending from (n!)-1 downwards.

This can be viewed as an irregular table: after the initial zero on row 0, start each row n with k = F(n+2)-1 and subtract repeatedly the number of "1-fibits" (number of terms in Zeckendorf expansion of k) from k to get successive terms, until the number that has already been listed (which is always F(n+1)-1) is encountered, which is not listed second time, but instead, the current row is finished and the next row starts with (F(n+3))-1, with the same process repeated. Here F(n) = the n-th Fibonacci number, A000045(n).