Le mar. 24 nov. 2020 à 07:15, Éric Angelini <eric.angelini@skynet.be> a écrit :
Hello Math-Fun Say we add the last odd digit of S to a(n): S=1,2,3,6,9,18,19,28,37,44,51,52,57,64,71,72, 79,88,97,104,105,110,111,112,113,116,117,124,... Say we have a hit when the concatenation of two or more initial terms of S reappear in S; will we ever find a hit here (the hit « 123 » was missed by a single unit with « 124 », in the above example;
As it is worded, you should count the appearance of concat(1,2) in 112 and later in 124... But I understand the concatenation must be equal to a term and not just appear in it. For this case I confirm Kris' result of a(2679) = 12369, and no other match below concat(1, 2, 3, 6, 9, 18, 19, 28, 37) = 1'236'918'192'837. (To see that it's enough to pass a(n) = 12'369'181'928, takes < 1 min with PARI code below.) The ratio a(n)/n seems to tend to some value between 4.6 and 4.7. So one would have to go up to ~ n = 2.5e13 to pass a(n) = 123'691'819'283'744 ~ 1.23...e14 and exclude a hit below a(n) = 1.23...e16. -Maximilian (PARI) lastodd(a)=until(!a\=10,bittest(a,0)&&return(L=a%10));L L=!a=0;V=vector(100,i,a+=lastodd(a)) j=c=L=!a=0;for(i=1,oo,c>(a+=lastodd(a))&&next; c==a&&print1("*** ",[i,a]" *** ");print1([c=eval(Str(c,V[j++]))])) the next possibilities would be
with 1236, 12369, 1236918,... but the margin of this post, etc. Best, É.