A244144 Alternating sum of digits of n^n.

1, 4, -5, 3, -1, 5, 5, -5, 5, 1, -11, -10, 8, 4, 21, -38, 8, -2, 7, 1, 1, 0, 10, -5, 23, 26, 3, -7, 19, 23, -24, 23, 11, 56, 10, 36, 5, 37, 24, -32, 8, 15, -1, -33, -10, 20, 20, -35, 31, 23, -18, 24, -14, -34, 0, -1, 40, 16, 14, -21, 6, -27, -17, -5, -32, 11, 12, -41, 59, -23, -38, 52, -42, -29, -21, 12, 0, -1, -39, 1, -7, -19, -7, -25, -34

Offset: 1

Author

Anthony Sand, Jun 21 2014

Keywords

• sign
• base
• less

Comments

The alternating sum of the digits of n^n is the sum obtained by alternately adding and subtracting the digits of n^n from left to right. For example, 4^4 = 256, therefore the alternating sum = 2 - 5 + 6 = 3. 7^7 = 823543, alternating sum = 8 - 2 + 3 - 5 + 4 - 3 = 5.

Examples

			If the function f(x) alternately adds and subtracts the digits of x from left to right, then:
a(1) = f(1^1) = f(1) = 1.
a(2) = f(2^2) = f(4) = 4.
a(3) = f(3^3) = f(27) = 2 - 7 = -5.
a(4) = f(4^4) = f(256) = 2 - 5 + 6 = 3.
a(9) = f(9^9) = f(387420489) = 3 - 8 + 7 - 4 + 2 - 0 + 4 - 8 + 9 = 5.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A065796, A066588.

Programs

Maple

a:= n-> -(s->add(parse(s[i])*(-1)^i, i=1..length(s)))(""||(n^n)):
seq(a(n), n=1..80);  # Alois P. Heinz, Jun 21 2014

Mathematica

f[n_] := Block[ {d = IntegerDigits[ n], k = l = 1, s = 0}, l = Length[d]; While[ k <= l, s = s - (-1)^k*d[[k]]; k++ ]; Return[s]]; Table[ f[n^n], {n, 1, 100} ] (* Minor adaptation from program for A065796. *)