A332584 - OEIS

%I #52 Jan 02 2024 14:45:49

%S 2,82,1888,6842,6,50,20,10,1320,28,208,32,66,148,1008,60,192,124536,

%T 282,46,128,32,28,86,40,33198,36,42,346,738,1532,246,70,68,102,306,56,

%U 20226,78316,10778,328,2432,738,2783191412956,48,746,8350,398,70,150,2300,21378

%N a(n) = minimal value of n+k (with k >= 1) such that the concatenation of the decimal digits of n,n+1,...,n+k is divisible by n+k+1, or -1 if no such n+k exists.

%C Certainly a(n) must be even, since no odd number can be divisible by an even number.

%C The values of k = a(n)-n are given in the companion sequence A332580, which also has an extended table of values.

%C A heuristic argument suggests that n+k should always exist.

%H J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, <a href="http://arxiv.org/abs/2004.14000">Three Cousins of Recaman's Sequence</a>, arXiv:2004:14000 [math.NT], April 2020.

%F a(n) = n + A332580(n) (trivially from the definitions).

%e a(1) = 2 as '1' || '2' = '12', which is divisible by 3 (where || denotes decimal concatenation).

%e a(7) = 20 as '7' || '8' || '9' || '10' || '11' || '12' || ... || '20' = 7891011121314151617181920, which is divisible by 21.

%e a(8) = 10 as '8' || '9' || '10' = 8910, which is divisible by 11.

%e a(2) = 82: the concatenation 2 || 3 || ... || 82 is

%e   23456789101112131415161718192021222324252627282930313233343536373839\

%e   40414243444546474849505152535455565758596061626364656667686970717273747\

%e   576777879808182, which is divisible by 83.

%p grow := proc(n,M) # searches out to a limit of M, returns [n,n+k] or [n,-1] if no k was found

%p local R,i;

%p R:=n;

%p for i from n+1 to M do

%p R:=R*10^length(i)+i;

%p if (i mod 2) = 0 then

%p if (R mod (i+1)) = 0 then return([n, i]); fi;

%p fi;

%p od:

%p [n, -1];

%p end;

%p for n from 1 to 100 do lprint(grow(n,20000)); od;

%o (PARI) apply( {A332584(n,L=10^#Str(n),c=n)= until((c=c*L+n)%(n+1)==0, n++<L||L*=10);n}, [1..17]) \\ _M. F. Hasler_, Feb 20 2020

%o (Python)

%o def A332584(n):

%o     r, m = n, n + 1

%o     while True:

%o         r = r*10**(len(str(m))) + m

%o         if m % 2 == 0 and r % (m+1) == 0:

%o             return m

%o         m += 1 # _Chai Wah Wu_, Jun 12 2020

%Y Cf. A061836 (multiplication instead of concatenation), A332580, A332585.

%K nonn,base

%O 1,1

%A _Scott R. Shannon_ and _N. J. A. Sloane_, Feb 16 2020

%E a(44) onwards (using A332580) added by _Andrew Howroyd_, Jan 02 2024