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A065796 Alternating sum of digits of n^2.

%I A065796 #13 Oct 24 2017 02:29:07
%S A065796 1,4,9,5,3,3,5,-2,-7,1,0,1,4,-2,5,3,3,5,-2,4,1,0,12,4,9,5,14,3,5,9,4,
%T A065796 1,0,1,4,-2,5,3,3,5,-2,4,12,11,1,4,9,5,3,3,5,9,15,12,0,1,4,-2,-6,3,3,
%U A065796 5,9,-7,1,0,1,4,-2,5,-8,-8,5,-2,4,1,11,-10,-7,-2,-6,3,3,-6,-2,-7,1,0,1,-7,-13,-6,3,3,-6,-2,4,1,0,1
%N A065796 Alternating sum of digits of n^2.
%C A065796 Conjecture; there are an infinite number of values which do not appear in this sequence (in the signed version, of course). The first example appears to be 2. _Sean A. Irvine_ has checked this up to 10^9. - _Robert G. Wilson v_, Dec 10 2001
%C A065796 a(n) == n^2 (mod 11).  In particular, values == 2, 6, 7, 8, 10 (mod 11) do not appear, and the conjecture is true. - _Robert Israel_, Oct 24 2017
%H A065796 Harry J. Smith, <a href="/A065796/b065796.txt">Table of n, a(n) for n=1..1000</a>
%F A065796 a(n) = n^2 mod 10 - n^2 mod 100 div 10 + n^2 mod 1000 div 100 - ...
%F A065796 a(n) = A055017(n^2). - _Robert Israel_, Oct 24 2017
%e A065796 a(18)=5 because 18^2 is 324 and 4-2+3=5
%p A065796 asd:= proc(n) local L,j;
%p A065796   L:= convert(n,base,10);
%p A065796   add((-1)^(j+1)*L[j],j=1..nops(L))
%p A065796 end proc:
%p A065796 seq(asd(n^2),n=1..100); # _Robert Israel_, Oct 24 2017
%t A065796 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^2], {n, 1, 100} ]
%t A065796 Table[Total[Times@@@Partition[Riffle[IntegerDigits[n^2], {1, -1}, {-2, 1, -2}], 2]], {n, 1, 100}] (* _Vincenzo Librandi_, Oct 24 2017 *)
%o A065796 (PARI) SumAD(x)= { local(a=1, s=0); while (x>9, s+=a*(x-10*(x\10)); x\=10; a=-a); return(s + a*x) }
%o A065796 { for (n=1, 1000, write("b065796.txt", n, " ", SumAD(n^2)) ) } \\ _Harry J. Smith_, Oct 30 2009
%Y A065796 Cf. A055017.
%K A065796 base,easy,sign
%O A065796 1,2
%A A065796 Benny Wegner (jaeger(AT)clan-efg.de), Dec 05 2001
%E A065796 More terms from _Robert G. Wilson v_, Dec 06 2001