# Recaman numbers

John Layman layman at calvin.math.vt.edu
Thu Oct 4 17:48:31 CEST 2001

```Following the recent flurry of notes on the Recaman numbers (A005132),
I finally couldn't resist the temptation to get involved.  I have completed
the calculation of R(n) to beyond n=2 billion and a calculation proceeding
as I write will push this to 15 billion.  The much sought-after R=1355 does
not occur by n=2 billion, the seven smallest values that have not occurred
by then being 1355, 1643, 1814, 2405, 2406, 2520, and 2991.

An interesting aspect of the Recamans that has not been previously noted is
the fact that the successive differences of the integer part of the ratio
R(n)/n is almost always negative, with only 36 occurrences where
FractPart(R(n+1)/(n+1))-FractPart(R(n)/n) is positive up to n=2*10^9, these
being at n=2,4,7,12,22,40,77,135,249,454,845,1521,2753,5046,9318,17224,31222,
57072,99742,181694,328256,589933,1034839,1788538,3225919,5784586,10212211,
18399785,32148795,58056876,101769230,173395920,302890749,511561221,904036925,
1610187039.  The ratio of successive numbers in this sequence is usually about
1.7 to 1.8.

Neil, my calculations use routines which I (pardon the expression!) "hacked", that allow 15 boolean,
or bit, values to be set or read per 2-byte INTEGER*2 Fortran data value.  This allows a 1-dim
array of 10^9 INTEGER*2 values to suffice for calculating R(n) up to n=15*10^9.  My server Fortran
permits array indexes only up to 2^30 or just over 1 billion.

```