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 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. 3
 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 (list; graph; refs; listen; history; text; internal format)
 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(qs, for(j=1, 9, dd=digits(q+j); ss=sum(m=1, #dd, n^dd[m]); if(q+jss, 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: A288967 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

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Last modified December 3 18:19 EST 2021. Contains 349467 sequences. (Running on oeis4.)