

A253253


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


2



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
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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^(k1) <= 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*m1 in the range 10^(k1) <= 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*s1 in the range 10^(k1) <= 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 !! (n1)
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



