A253253 a(n) = smallest divisor of the concatenation of n and n+1 that did not occur earlier.

1, 23, 2, 3, 4, 67, 6, 89, 5, 337, 8, 1213, 9, 283, 379, 7, 859, 17, 10, 43, 1061, 13, 14, 25, 421, 37, 11, 41, 293, 433, 12, 53, 1667, 15, 16, 3637, 21, 349, 20, 449, 19, 4243, 24, 35, 2273, 1549, 1187, 373, 18, 5051, 28, 51, 2677, 1091, 463, 5657, 2879, 27

Offset

1,2

Comments

Is this a permutation of the integers > 0?

Comment from N. J. A. Sloane, May 19 2017: (Start)

It should not be difficult to prove that every positive integer appears.

If not, let m be the smallest missing number. There is an n_0 such that for all n >= n_0, a(n) > m. The theorem will follow if we can find an N > n_0 such that

m divides the concatenation of N and N+1.

Let N have k digits and suppose that

10^(k-1) <= N <= 10^k - 2.

The concatenation of N and N+1 is N*(10^k+1)+1, so we want to find numbers k and N such that

N*(10^k+1) == -1 mod m.

Case (i). If gcd(m,10)=1, then by Euler's theorem, 10^phi(m) == 1 mod m, so we can take k to be a sufficiently large multiple of phi(m), and then take N to be a number of the form r*m-1 in the range 10^(k-1) <= N <= 10^k - 2.

Case (ii). If m = 2^r or 5^r, then for large k, 10^k+1 == 1 mod m, and we take N to be of the form m*s-1 in the range 10^(k-1) <= N <= 10^k - 2.

The other cases are left to the reader. (End)

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Éric Angelini, Divisors of the concatenation of n and n+1, SeqFan list, Jun 03 2015.

Prog

(Haskell)
import Data.List (insert); import Data.List.Ordered (minus)
a253253 n = a253253_list !! (n-1)
a253253_list = f a001704_list [] where
   f (x:xs) ds = y : f xs (insert y ds) where
                 y = head (a027750_row' x `minus` ds)

Crossrefs

Cf. A027750, A001704, A256285.

Sequence in context: A040521 A040524 A306374 * A225003 A040525 A040526

Adjacent sequences: A253250 A253251 A253252 * A253254 A253255 A253256

Keyword

nonn,base,look

Author

Eric Angelini and Reinhard Zumkeller, Jun 05 2015

Status

approved