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.

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, 1, 3, 3, 3, 0, 1, 3, 1, 2, 1, 1, 1, 3, 1, 0, 1, 3

Offset: 2

Author

Derek Orr, Jun 16 2014

Keywords

• nonn
• base

Comments

a(10*n) = 0 for all n > 0.

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.

a(53), a(71) and a(77) are conjectured to be infinite. See A244187.

Examples

			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.
		

Crossrefs

Cf. A243912, A243911, A244187.

Programs

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)))
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)))
n=2;while(n<100,print1(hup(n),", ");n++)

Python

def a(n,p):
  lst = []
  for c in range(10**p+1):
    m = n**c%10**p
    if m in lst:
      return [c,c-lst.index(m)]
    else:
      lst.append(m)
def cou(n):
  if n % 10 == 0:
    return 0
  ww = []
  p = 2
  aa = a(n,p)[0]
  ww.extend(range(aa))
  while p < 100:
    newlst = ww
    w = []
    for i in newlst:
      m = n**i%10**p
      if len(str(m))==p and m%int('1'*p)==0:
        w.append(i)
    if w:
      ww = []
      for k in w:
        j = k
        while j <= a(n,p+1)[0]:
          ww.append(j)
          j += a(n,p)[1]
      ww.sort()
      p += 1
    else:
      return p-1
n = 2
while n < 100:
  if cou(n):
    print(cou(n),end=', ')
  else:
    print(0,end=', ')
  n += 1

Extensions

Programs corrected and improved by Derek Orr, Aug 18 2014