A248153 Start with a(0)=10, then a(n) = 7 times the n-th digit of the sequence.

10, 7, 0, 49, 0, 28, 63, 0, 14, 56, 42, 21, 0, 7, 28, 35, 42, 28, 14, 14, 7, 0, 49, 14, 56, 21, 35, 28, 14, 14, 56, 7, 28, 7, 28, 49, 0, 28, 63, 7, 28, 35, 42, 14, 7, 21, 35, 14, 56, 7, 28, 7, 28, 35, 42, 49, 14, 56, 49, 14, 56, 28, 63, 0, 14, 56, 42, 21, 49, 14, 56, 21, 35

Offset: 0

Author

M. F. Hasler, Oct 02 2014

Keywords

• nonn
• base

Comments

This sequence was inspired by E. Angelini's post to the SeqFan list, cf. links.

a(0)=10 is the smallest possible choice to ensure that the digit 0 appears anywhere in the sequence. a(0)=1 would lead to the same sequence with the terms 0 removed.

By construction, all terms a(n), n>0, are divisible by 7, and a(n)/7 yields the sequence of digits of the (concatenated) terms of this sequence.

It is easy to show that the distance between two 0's is strictly increasing from one occurrence to the next one. Thus, the asymptotic density of terms and/or digits 0 is zero, and the sequence can never "enter a loop".

Links

• E. Angelini, Brute force density: triples and cubes, SeqFan list, Oct 01 2014

Crossrefs

Cf. A248128, A248129, A248130, A248131.

Programs

PARI

a(n,s=10,m=7,d=[])={for(i=1,n,print1(s",");d=concat(d,if(s,digits(s)));s=m*d[1];d=vecextract(d,"^1"));s}

Python

def aupton(nn):
    alst, astr = [10], "X10"
    for n in range(1, nn+1):
        alst.append(7 * int(astr[n]))
        astr += str(alst[-1])
    return alst
print(aupton(72)) # Michael S. Branicky, Oct 07 2021