cp's OEIS Frontend

Continued fraction expansion of tanh(1/5). - Benoit Cloitre, Dec 17 2002

n such that 5 divides the numerator of B(2n) where B(2n) = the 2n-th Bernoulli number. - Benoit Cloitre, Jan 01 2004

5 times odd numbers. - Omar E. Pol, May 02 2008

5th transversal numbers (or 5-transversal numbers): Numbers of the 5th column of positive numbers in the square array of nonnegative and polygonal numbers A139600. Also, numbers of the 5th column in the square array A057145. - Omar E. Pol, May 02 2008

Successive sums: 5, 20, 45, 80, 125, ... (see A033429). - Philippe Deléham, Dec 08 2011

3^a(n) + 1 is divisible by 61. - Vincenzo Librandi, Feb 05 2013

If the initial 5 is changed to 1, giving 1,15,25,35,45,..., these are values of m such that A323288(m)/m reaches a new record high value. - N. J. A. Sloane, Jan 23 2019

Take the decimal expansion of n, say n = d_1 d_2 ... d_k. We can choose to map it to any number that can be obtained by the following process. Take any substring d_i, d_{i+1}..., d_j that does not begin with 0. If the number represented by this substring is odd, replace it with twice the number. If it is even either halve it or double it.

The substring may increase or decrease in length. We do not pad it with zeros if it decreases in length.

For example, if n = 20129, then by acting on single-digit substrings we get 10129, 40129, 20229, 20119, 20149, 201218. Acting on 2-digit substrings we get in addition 2069 (halve the 12!), 20249, 20158. From 3-digit substrings we also get 40229, 20258; from 4-digit substrings we get 40249; and from 5-digit substrings we get 40258.

Eric Angelini asks what is the smallest number of steps needed to reach n if we start at 1 and repeatedly apply this process? We can reach 2 in 1 step, 4 in 2 steps, 13 in five steps, and so on.

Lars Blomberg has shown, by considering just the final digit of the numbers in the trajectory, that no number ending in 0 or 5 can be reached from 1. All other numbers can be reached (cf. A323454) - see proof below.

Update, Jan 15 2019: Lorenzo Angelini has found that 3 can be reached from 1 in 11 steps: 1, 2, 4, 8, 16, 112, 56, 28, 14, 12, 6, 3. No shorter path is possible.

From N. J. A. Sloane, Jan 16 2019: (Start)

Theorem: If k > 1 does not end in 0 or 5 then it can be reached from 1.

Proof: Suppose not, and let k be the smallest such number. Note that the allowed operations are invertible: if a -> b then also b -> a. So that means that

*** all the descendants of k must be bigger than k ***

(if there was a descendant < k, then it would also be unreachable from 1, which is a contradiction to k being the smallest).

All digits of k must be odd (if there were an even digit > 0, halve it and get a smaller number; if there is a zero digit, say we see a0, then we halve a0 and get a smaller number).

If all the digits of k are 1, do 111...1 -> 111...2 -> 55..56, a smaller number.

If there is a digit 3, 7, or 9, we know we can get that single digit down to 1 (see A323454), again a contradiction.

But all the digits can't be 5. QED (End)

Smallest element in row n of irregular triangle in A323460.

Theorem: Let the decimal expansion of n be d_1 d_2 ... d_k. (i) If there is a substring d_r ... d_s which starts with d_r = 1 and ends with an even digit d_s = e, take that string which starts with the leftmost 1 and ends with the rightmost even digit, and halve it. (ii) Otherwise, if there is an even digit e, take the substring from d_1 to the rightmost such e and halve it. (iii) Otherwise, all d_i are odd, and a(n) = n.