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A243977 a(n) is the largest run of identical digits that n^k can end with for some k, or 0 if there is no limit to such runs.

%I A243977 #23 May 22 2025 10:21:38
%S A243977 3,1,2,1,1,1,3,1,0,2,3,3,2,1,1,5,1,2,0,1,3,1,1,1,1,1,3,1,0,3,1,4,2,1,
%T A243977 1,3,3,3,0,1,3,1,2,1,1,1,3,1,0,1,3
%N A243977 a(n) is the largest run of identical digits that n^k can end with for some k, or 0 if there is no limit to such runs.
%C A243977 a(10*n) = 0 for all n > 0.
%C A243977 a(54) > 30 or 0. The sequence continues for 54 < n < 100 {2, 2, 1, 1, 3, 2, 0, 1, 3, 1, 2, 1, 2, 1, 1, 1, 0, ?, 3, 3, 1, 1, 1, ?, 3, 4, 0, 1, 1, 1, 2, 1, 1, 1, 3, 1, 0, 2, 3, 1, 2, 1, 1, 4, 3, 2} where the question marks represent a(71) and a(77). a(71) > 44 or 0 and a(77) > 16 or 0.
%C A243977 a(53), a(71) and a(77) are conjectured to be infinite. See A244187.
%e A243977 For a(1), 2^k ends in 1 identical digit when k = 1, 2 identical digits when k = 18, and 3 identical digits when k = 39. 2^k doesn't end in 4 identical digits for any k. Thus a(1) = 3.
%o A243977 (PARI) a(n,p)=lst=[];for(c=0,10^p,m=n^c%10^p;if(vecsearch(vecsort(lst),m),for(i=1,#lst,if(vecextract(lst,2^(i-1),)==[m],return([c,c-i+1]))));if(!vecsearch(vecsort(lst),m),lst=concat(lst,m)))
%o A243977 hup(n)=if(n%10==0,return(0));ww=[];p=2;for(ii=1,a(n,p)[1],ww=concat(ww,ii));while(p<100,v=ww;w=[];for(q=1,#v,h=digits(n^v[q]%10^p);if(#h==p&&(vecmin(h)==vecmax(h)),w=concat(w,v[q])));if(w,ww=[];for(k=1,#w,j=w[k];while(j<=a(n,p+1)[1],ww=concat(ww,j);j+=a(n,p)[2]));ww=vecsort(ww,,8);p++);if(!w,return(p-1)))
%o A243977 n=2;while(n<100,print1(hup(n),", ");n++)
%o A243977 (Python)
%o A243977 def a(n,p):
%o A243977   lst = []
%o A243977   for c in range(10**p+1):
%o A243977     m = n**c%10**p
%o A243977     if m in lst:
%o A243977       return [c,c-lst.index(m)]
%o A243977     else:
%o A243977       lst.append(m)
%o A243977 def cou(n):
%o A243977   if n % 10 == 0:
%o A243977     return 0
%o A243977   ww = []
%o A243977   p = 2
%o A243977   aa = a(n,p)[0]
%o A243977   ww.extend(range(aa))
%o A243977   while p < 100:
%o A243977     newlst = ww
%o A243977     w = []
%o A243977     for i in newlst:
%o A243977       m = n**i%10**p
%o A243977       if len(str(m))==p and m%int('1'*p)==0:
%o A243977         w.append(i)
%o A243977     if w:
%o A243977       ww = []
%o A243977       for k in w:
%o A243977         j = k
%o A243977         while j <= a(n,p+1)[0]:
%o A243977           ww.append(j)
%o A243977           j += a(n,p)[1]
%o A243977       ww.sort()
%o A243977       p += 1
%o A243977     else:
%o A243977       return p-1
%o A243977 n = 2
%o A243977 while n < 100:
%o A243977   if cou(n):
%o A243977     print(cou(n),end=', ')
%o A243977   else:
%o A243977     print(0,end=', ')
%o A243977   n += 1
%Y A243977 Cf. A243912, A243911, A244187.
%K A243977 nonn,base
%O A243977 2,1
%A A243977 _Derek Orr_, Jun 16 2014
%E A243977 Programs corrected and improved by _Derek Orr_, Aug 18 2014