A113009 -id:A113009 - cp's OEIS frontend

A353906 a(n) is the {alternating sum of the digits of n} raised to the power {number of digits of n}.

1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 4, 9, 16, 25, 36, 49, 64, 4, 1, 0, 1, 4, 9, 16, 25, 36, 49, 9, 4, 1, 0, 1, 4, 9, 16, 25, 36, 16, 9, 4, 1, 0, 1, 4, 9, 16, 25, 25, 16, 9, 4, 1, 0, 1, 4, 9, 16, 36, 25, 16, 9, 4, 1, 0, 1, 4, 9, 49, 36, 25, 16, 9, 4, 1, 0, 1, 4, 64, 49, 36, 25, 16, 9

Offset: 1

Malo David, May 10 2022

The first negative term is a(120) = -1.

Note that it does not matter whether the alternating sum starts from the first or from the last digit of n. - Jianing Song, Jun 12 2022

			For n=489, a(489) = (9-8+4)^3 = 125.

Cf. A055017 (alternating digit sum), A055642 (number of digits).
Cf. A353907 (fixed points).
Cf. A113009 (normal sum instead of alternating).

a[n_] := Total[-(-1)^Range[m = Length[d = IntegerDigits[n]]] * d]^m; Array[a, 100] (* Amiram Eldar, May 11 2022 *)

a(n) = my(d=digits(n)); sum(k=1, #d, (-1)^(k+1)*d[k])^#d; \\ Michel Marcus, May 10 2022

def a(n):
   counter = 0
   S = 0
   q = n
   while q:
      q, c = q//10, q % 10
      S += (-1)** counter * c
      counter += 1
   return S ** counter

def A353906(n): return sum((-1 if i % 2 else 1)*int(j) for i, j in enumerate(str(n)[::-1]))**len(str(n)) # Chai Wah Wu, May 11 2022

a(n) = A055017(n)^A055642(n).

cp's OEIS Frontend

A353906 a(n) is the {alternating sum of the digits of n} raised to the power {number of digits of n}.

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