1,2

As with A332580 a heuristic argument, based on the divergent sum of reciprocals which approximates the probability that the concatenation of 1,2,...,n will divide the concatenation of n+1,n+2,...,n+k suggests that k should always exist. However an examination of the prime factors of the concatenation of 1,2,...,n shows that most of these numbers contain one or more very large primes, suggesting the values of k will likely become extremely large as n increases.

The author thanks Joseph Myers for suggestions for finding the larger terms of this sequence.

J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], April 2020.

a(2) = 4 as '1'||'2' = 12 and '3'||'4'||'5'||'6' = 3456, which is divisible by 12 (where '||' denotes decimal concatenation).

a(3) = 20 as '1'||'2'||'3' = 123 and '4'||'5'||...||'22'||'23' = 4567891011121314151617181920212223, which is divisible by 123.

a:= proc(n) local i, t, m; t, m:= parse(cat($1..n)), 0;

for i from n+1 do m:= parse(cat(m, i)) mod t;

if m=0 then break fi od; i-n

end:

seq(a(n), n=1..6); # Alois P. Heinz, Feb 29 2020

(PARI) a(n) = {my(k=1, small="", big = n+1); for (j=1, n, small=concat(small, Str(j))); small = eval(small); while (big % small, k++; big = eval(concat(Str(big), Str(n+k)))); k; } \\ Michel Marcus, Feb 29 2020

(Python)

def A332867(n):

m, k = int(''.join(str(d) for d in range(1, n+1))), 1

i = n+k

i2, l = i % m, len(str(i))

t = 10**l

t2, r = t % m, i % m

while r != 0:

k += 1

i += 1

i2 = (i2+1) % m

if i >= t:

l += 1

t *= 10

t2 = (10*t2) % m

r = (r*t2 + i2) % m

return k # Chai Wah Wu, May 20 2020

Cf. A332580, A332830, A007908.

Sequence in context: A012841 A074195 A376049 * A217369 A146342 A103896

Adjacent sequences: A332864 A332865 A332866 * A332868 A332869 A332870

nonn,base,more,hard

Scott R. Shannon, Feb 27 2020

a(11) from Chai Wah Wu, Feb 20 2026

a(12) from Chai Wah Wu, Mar 18 2026

approved