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 A244144 Alternating sum of digits of n^n. 3

%I #12 Jun 21 2014 16:43:56

%S 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,

%T -7,19,23,-24,23,11,56,10,36,5,37,24,-32,8,15,-1,-33,-10,20,20,-35,31,

%U 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

%N Alternating sum of digits of n^n.

%C 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.

%H Alois P. Heinz, <a href="/A244144/b244144.txt">Table of n, a(n) for n = 1..5000</a>

%e If the function f(x) alternately adds and subtracts the digits of x from left to right, then:

%e a(1) = f(1^1) = f(1) = 1.

%e a(2) = f(2^2) = f(4) = 4.

%e a(3) = f(3^3) = f(27) = 2 - 7 = -5.

%e a(4) = f(4^4) = f(256) = 2 - 5 + 6 = 3.

%e a(9) = f(9^9) = f(387420489) = 3 - 8 + 7 - 4 + 2 - 0 + 4 - 8 + 9 = 5.

%p a:= n-> -(s->add(parse(s[i])*(-1)^i, i=1..length(s)))(""||(n^n)):

%p seq(a(n), n=1..80); # _Alois P. Heinz_, Jun 21 2014

%t f[n_] := Block[ {d = Reverse[ 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.

%Y Cf. A065796, A066588.

%K sign,base,less

%O 1,2

%A _Anthony Sand_, Jun 21 2014

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