A236067 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.

1, 0, 12, 4624, 3909511, 0, 13177388, 1033, 10, 0, 0, 0, 0, 0, 2758053616, 1053202, 7413245658, 419370838921, 52135640, 1347536041, 833904227332, 5117557126, 3606012949057, 5398293152472, 31301, 0, 15554976231978, 405287637330, 35751665247, 19705624111111

Offset

1,3

Comments

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.

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.

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.

For n = 21, 23, and 24, a(n) > 10^10.

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.

For n in {51, 52, 53, ..., 64, 65} and {67, 68, 69, ..., 73, 74}, a(n) > 10^11.

For n in {75, 76, 77} and {79, 80, 81, ..., 96, 97, 99}, a(n) > 5*10^11.

A few nonzero terms were added by math4pad.net @PascalCardin

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

Links

Chai Wah Wu, Table of n, a(n) for n = 1..500 (n = 1..100 from Hiroaki Yamanouchi)

John D. Cook, Monday morning math puzzle (2012)

Dean Morrow, Cycles of a family of digit functions

Example

12 is the smallest number such that 3^1 + 3^2 = 12 so a(3) = 12.
4624 is the smallest number such that 4^4 + 4^6 + 4^2 + 4^4 = 4624 so a(4) = 4624.
1033 is the smallest number such that 8^1 + 8^0 + 8^3 + 8^3 = 1033 so a(8) = 1033.

Prog

(PARI)
Min(n)=for(k=1, 10^3, if(n+k<=10^k, return(10^k)))
Max(n)=for(k=1, 10^3, if(k*n^9<=10^k-1, return(10^(k-1))))
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))))
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)
n=1; while(n<100, print1(a(n), ", "); n++) \\ PARI program more advanced than Python program \\ Derek Orr, Aug 01 2014
(Python)
def Min(n):
..for k in range(1, 10**3):
....if n+k <= 10**k:
......return 10**k
def Max(n):
..for k in range(1, 10**3):
....if k*(n**9) <= 10**k-1:
......return 10**(k-1)
def div10(n):
..for j in range(10**3):
....if n%10**j!=0:
......return j
def a(n):
..k = Min(n)
..while k <= Max(n):
....tot = 0
....for i in str(k):
......tot += n**(int(i))
....if tot == k:
......return k
....if tot < k:
......k += 1
....if tot > k-1:
......k = (1+k//10**div10(k))*10**div10(k)
n = 1
while n < 100:
..if a(n):
....print(a(n), end=', ')
..else:
....print(0, end=', ')
..n += 1
# Derek Orr, Aug 01 2014

Crossrefs

Cf. A139410 (for 4th term), A003321, A296138, A296139.

Sequence in context: A361106 A009094 A061701 * A134821 A229669 A013508

Adjacent sequences: A236064 A236065 A236066 * A236068 A236069 A236070

Keyword

nonn,base

Author

Derek Orr, Jan 19 2014

Extensions

More terms and edited extensively by Derek Orr, Aug 26 2014

a(21)-a(30) from Hiroaki Yamanouchi, Sep 27 2014

Status

approved