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%I #85 Dec 08 2017 15:50:22
%S 1,0,12,4624,3909511,0,13177388,1033,10,0,0,0,0,0,2758053616,1053202,
%T 7413245658,419370838921,52135640,1347536041,833904227332,5117557126,
%U 3606012949057,5398293152472,31301,0,15554976231978,405287637330,35751665247,19705624111111
%N a(n) is the least number m such that m = n^d_1 + n^d_2 + ... + n^d_k where d_k represents the k-th digit in the decimal expansion of m, or 0 if no such number exists.
%C The 0's in the sequence are definite. There exists both a maximum and a minimum number that a(n) can be based on n. They are given in the programs below as Max(n) and Min(n), respectively.
%C It is known that a(22) = 5117557126, a(25) = 31301, a(29) = 35751665247, a(32) = 2112, a(33) = 1224103, a(37) = 111, a(40) = 102531321, a(48) = 25236435456, a(50) = 101, a(66) = 2524232305, a(78) = 453362316342, a(98) = 100, and a(100) = 20102.
%C There are an infinite number of nonzero entries. First, note if a(n) is nonzero, a(n) >= n. Further, a(9) = 10, a(98) = 100, a(997) = 1000, ..., a(10^k-k) = 10^k for all k >= 0.
%C For n = 21, 23, and 24, a(n) > 10^10.
%C For n in {26, 27, 28, 30, 31, 34, 35, 36, 38, 39, 41, 42, 43, 44, 45, 46, 47, 49}, a(n) > 5*10^10.
%C For n in {51, 52, 53, ..., 64, 65} and {67, 68, 69, ..., 73, 74}, a(n) > 10^11.
%C For n in {75, 76, 77} and {79, 80, 81, ..., 96, 97, 99}, a(n) > 5*10^11.
%C A few nonzero terms were added by math4pad.net @PascalCardin
%C a(1000) = 1000000000000002002017, a(10000) = 0, a(1000000) = 1000002000010, a(10000000) = 200000020000011. It looks like a(10^k) in decimal consists of mostly the digits 0, 1 and 2. - _Chai Wah Wu_, Dec 07 2017
%H Chai Wah Wu, <a href="/A236067/b236067.txt">Table of n, a(n) for n = 1..500</a> (n = 1..100 from Hiroaki Yamanouchi)
%H John D. Cook, <a href="http://www.johndcook.com/blog/2012/02/20/monday-morning-math-puzzle/">Monday morning math puzzle</a> (2012)
%H Dean Morrow, <a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.64.3408&rep=rep1&type=pdf">Cycles of a family of digit functions</a>
%e 12 is the smallest number such that 3^1 + 3^2 = 12 so a(3) = 12.
%e 4624 is the smallest number such that 4^4 + 4^6 + 4^2 + 4^4 = 4624 so a(4) = 4624.
%e 1033 is the smallest number such that 8^1 + 8^0 + 8^3 + 8^3 = 1033 so a(8) = 1033.
%o (PARI)
%o Min(n)=for(k=1,10^3,if(n+k<=10^k,return(10^k)))
%o Max(n)=for(k=1,10^3,if(k*n^9<=10^k-1,return(10^(k-1))))
%o side(n,q)=v=digits(q);for(i=1,10,qq=digits((floor(q/10^i)+1)*10^i);st=sum(j=1,#qq,n^qq[j]);if(q+10^i>st,return((floor(q/10^i)+1)*10^(i-1))))
%o a(n)=k=Min(n);while(k<=Max(n),q=10*k;d=digits(q);s=sum(i=1,#d,n^d[i]);if(q<s,k=side(n,q));if(q>s,for(j=1,9,dd=digits(q+j);ss=sum(m=1,#dd,n^dd[m]);if(q+j<ss,k++;break);if(q+j==ss,return(q+j)));if(q+9>ss,k++));if(q==s,return(q)));return(0)
%o n=1;while(n<100,print1(a(n),", ");n++) \\ PARI program more advanced than Python program \\ _Derek Orr_, Aug 01 2014
%o (Python)
%o def Min(n):
%o ..for k in range(1,10**3):
%o ....if n+k <= 10**k:
%o ......return 10**k
%o def Max(n):
%o ..for k in range(1,10**3):
%o ....if k*(n**9) <= 10**k-1:
%o ......return 10**(k-1)
%o def div10(n):
%o ..for j in range(10**3):
%o ....if n%10**j!=0:
%o ......return j
%o def a(n):
%o ..k = Min(n)
%o ..while k <= Max(n):
%o ....tot = 0
%o ....for i in str(k):
%o ......tot += n**(int(i))
%o ....if tot == k:
%o ......return k
%o ....if tot < k:
%o ......k += 1
%o ....if tot > k-1:
%o ......k = (1+k//10**div10(k))*10**div10(k)
%o n = 1
%o while n < 100:
%o ..if a(n):
%o ....print(a(n),end=', ')
%o ..else:
%o ....print(0,end=', ')
%o ..n += 1
%o # _Derek Orr_, Aug 01 2014
%Y Cf. A139410 (for 4th term), A003321, A296138, A296139.
%K nonn,base
%O 1,3
%A _Derek Orr_, Jan 19 2014
%E More terms and edited extensively by _Derek Orr_, Aug 26 2014
%E a(21)-a(30) from _Hiroaki Yamanouchi_, Sep 27 2014
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