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

1,2

Comments

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

Links

Table of n, a(n) for n=1..85.

Formula

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

Example

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

Mathematica

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

Prog

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

Crossrefs

Cf. A055017 (alternating digit sum), A055642 (number of digits).

Cf. A353907 (fixed points).

Cf. A113009 (normal sum instead of alternating).

Sequence in context: A225693 A040997 A256851 * A247149 A330350 A222210

Adjacent sequences: A353903 A353904 A353905 * A353907 A353908 A353909

Keyword

sign,easy,base

Author

Malo David, May 10 2022

Status

approved