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

2,1

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.

Links

Table of n, a(n) for n=2..52.

Example

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.

Prog

(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

Crossrefs

Cf. A243912, A243911, A244187.

Sequence in context: A086249 A176784 A176511 * A328569 A016569 A072801

Adjacent sequences: A243974 A243975 A243976 * A243978 A243979 A243980

Keyword

nonn,base

Author

Derek Orr, Jun 16 2014

Extensions

Programs corrected and improved by Derek Orr, Aug 18 2014

Status

approved