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A362843
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Numbers that are equal to the sum of their digits raised to consecutive odd numbered powers (1,3,5,7,...).
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0
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 463, 3943, 371915027434113
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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1 = 1^1;
463 = 4^1 + 6^3 + 3^5;
3943 = 3^1 + 9^3 + 4^5 + 3^7.
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MATHEMATICA
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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 *)
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PROG
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(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)))
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CROSSREFS
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KEYWORD
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nonn,base,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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