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Numbers that are equal to the sum of their digits raised to consecutive odd numbered powers (1,3,5,7,...).
2

%I #29 Jul 08 2023 02:42:57

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

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

%C 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.

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

%e 1 = 1^1;

%e 463 = 4^1 + 6^3 + 3^5;

%e 3943 = 3^1 + 9^3 + 4^5 + 3^7.

%t 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 *)

%o (PARI) isok(k) = my(d=digits(k)); sum(i=1, #d, d[i]^(2*i-1)) == k; \\ _Michel Marcus_, May 06 2023

%o (Python)

%o from itertools import count, islice

%o def A362843_gen(startvalue=0): # generator of terms >= startvalue

%o return filter(lambda n:n==sum(int(d)**((i<<1)+1) for i,d in enumerate(str(n))),count(max(startvalue,0)))

%o A362843_list = list(islice(A362843_gen(),12)) # _Chai Wah Wu_, Jun 26 2023

%Y Cf. A032799, A154160, A208130.

%K nonn,base,more

%O 1,3

%A _Wolfe Padawer_, May 05 2023

%E a(13) from _Martin Ehrenstein_, Jul 07 2023