A065816 -id:A065816 - cp's OEIS frontend

22, 55, 77, 99, 132, 187, 286, 1056, 1463, 1474, 1606, 1837, 2277, 2981, 4785, 4851, 5313, 5588, 5929, 7227, 8272, 8415, 8492, 8954, 11517, 12573, 12628, 13156, 14883, 15972, 17688, 22066, 23936, 24915, 25850, 27522, 34045, 36289, 36806, 38489, 40744, 43450, 46794, 48092

Offset: 1

Anthony Sand, Jun 23 2014

The result of alternately adding and subtracting the digits of n sometimes differs in sign when the procedure goes from left to right or right to left. For example, if n = 1234, 1 - 2 + 3 - 4 = -2, whereas 4 - 3 + 2 - 1 = +2. However, if the sum is zero when adding and subtracting from left to right, it will also be zero when adding and subtracting from right to left.
n such that A000312(n) is in A135499. - Robert Israel, Jul 13 2014
All terms are multiples of 11. This follows from the divisibility rule for 11. - Jens Kruse Andersen, Jul 13 2014
Number of terms less than 10^k: 0, 4, 7, 24, 55, 135, ..., . - Robert G. Wilson v, Jul 18 2014
Numbers for which the alternating sum of the digits of n^n are == 0 (Mod 10): 12, 22, 23, 35, 45, 46, 47, 55, 57, 77, 99, 117, 126, 132, 151, ..., . Obviously the members of A244212 are included here. - Robert G. Wilson v, Jul 20 2014

			22^22 = 341427877364219557396646723584, therefore the alternating sum = 4 - 8 + 5 - 3 + 2 - 7 + 6 - 4 + 6 - 6 + 9 - 3 + 7 - 5 + 5 - 9 + 1 - 2 + 4 - 6 + 3 - 7 + 7 - 8 + 7 - 2 + 4 - 1 + 4 - 3 = 0.

Jens Kruse Andersen and Robert G. Wilson v, Table of n, a(n) for n = 1..139 (a(48) to a(63) from Jens Kruse Andersen).

Cf. A000312, A065816, A135499, A244144.

filter:= proc(n) local x,j;
   x:= convert(n^n,base,10);
   evalb(add((-1)^j*x[j],j=1..nops(x)) = 0)
end proc;
select(filter, 11 * [$1..1000]); # Robert Israel, Jul 13 2014

fQ[n_] := Block[{id = IntegerDigits[ n^n]}, Sum[ id[[i]]*(-1)^i, {i, Length@ id}] == 0]; k = 11; lst = {}; While[k < 100001, If[ fQ@ k, AppendTo[ lst, k]; Print@ k]; k+= 11]; lst (* Robert G. Wilson v, Jul 13 2014 *)

isok(n) = d = digits(n^n) ; sum(i=1, #d, d[i]*(-1)^i) == 0; \\ Michel Marcus, Jun 25 2014

s = 0; m = 1; for digit[n,i=1..j] of n, s = s + digit[i] * m; m = -m; next i; if s = 0, print n;

a(9)-a(24) from Michel Marcus, Jun 23 2014

a(25)-a(44) from Robert G. Wilson v, Jul 13 2014

cp's OEIS Frontend

A244212 Numbers n for which the alternating sum of the digits of n^n is 0.

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