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

1,3

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.

If a(13) exists, it exceeds 1.3*10^13.

Links

Table of n, a(n) for n=1..13.

Example

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

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 *)

Prog

(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

Crossrefs

Cf. A032799, A154160, A208130.

Sequence in context: A335151 A065110 A307887 * A344822 A004893 A069282

Adjacent sequences: A362840 A362841 A362842 * A362844 A362845 A362846

Keyword

nonn,base,more

Author

Wolfe Padawer, May 05 2023

Extensions

a(13) from Martin Ehrenstein, Jul 07 2023

Status

approved