A362843 Numbers that are equal to the sum of their digits raised to consecutive odd numbered powers (1,3,5,7,...).

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 463, 3943, 371915027434113

Offset: 1

Author

Wolfe Padawer, May 05 2023

Keywords

• nonn
• base
• more

Comments

Unlike A032799 and A208130, this sequence is not easily proven to be finite. With m >= 1, 10^(m - 1) exceeds 9^1 + 9^2 + ... + 9^m when m is approximately 22.97, meaning it is impossible for an integer with 23 or more digits to be equal to the sum of its digits raised to the consecutive powers. However, 10^(m - 1) will never exceed 9^1 + 9^3 + ... + 9^(2m - 1) over m >= 1. It appears that 10^(m - 1) will never exceed 9^1 + 9^(1 + x) + 9^(1 + 2x) ... 9^(mx - x + 1) over m >= 1 when x >= A154160, approximately 1.04795. For A032799, x = 1, and for this sequence, x = 2. This means this sequence could theoretically be infinite, although it is currently unknown whether it is.

a(14) > 10^24 if it exists. The expected number of k-digit terms can be heuristically estimated as about 10^(-0.15*k), which suggests that the sequence is likely finite. - Max Alekseyev, May 17 2025

Examples

			1 = 1^1;
463 = 4^1 + 6^3 + 3^5;
3943 = 3^1 + 9^3 + 4^5 + 3^7.
		

Crossrefs

Cf. A032799, A154160, A208130.

Programs

Mathematica

kmax=10^6; a={}; For[k=0, k<=kmax, k++,If[Sum[Part[IntegerDigits[k],i]^(2i-1),{i,IntegerLength[k]}]==k, AppendTo[a,k]]]; a (* Stefano Spezia, May 06 2023 *)

PARI

isok(k) = my(d=digits(k)); sum(i=1, #d, d[i]^(2*i-1)) == k; \\ Michel Marcus, May 06 2023

Python

from itertools import count, islice
def A362843_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n:n==sum(int(d)**((i<<1)+1) for i,d in enumerate(str(n))),count(max(startvalue,0)))
A362843_list = list(islice(A362843_gen(),12)) # Chai Wah Wu, Jun 26 2023

Extensions

a(13) from Martin Ehrenstein, Jul 07 2023