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

0,1

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

Table of n, a(n) for n=0..72.

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

Prog

(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

Crossrefs

Cf. A248128, A248129, A248130, A248131.

Sequence in context: A185221 A349845 A358394 * A079166 A332980 A246662

Adjacent sequences: A248150 A248151 A248152 * A248154 A248155 A248156

Keyword

nonn,base

Author

M. F. Hasler, Oct 02 2014

Status

approved