A065796 Alternating sum of digits of n^2.

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, 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, 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

Offset

1,2

Comments

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

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

Links

Harry J. Smith, Table of n, a(n) for n=1..1000

Formula

a(n) = n^2 mod 10 - n^2 mod 100 div 10 + n^2 mod 1000 div 100 - ...

a(n) = A055017(n^2). - Robert Israel, Oct 24 2017

Example

a(18)=5 because 18^2 is 324 and 4-2+3=5

Maple

asd:= proc(n) local L, j;
  L:= convert(n, base, 10);
  add((-1)^(j+1)*L[j], j=1..nops(L))
end proc:
seq(asd(n^2), n=1..100); # Robert Israel, Oct 24 2017

Mathematica

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} ]
Table[Total[Times@@@Partition[Riffle[IntegerDigits[n^2], {1, -1}, {-2, 1, -2}], 2]], {n, 1, 100}] (* Vincenzo Librandi, Oct 24 2017 *)

Prog

(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) }
{ for (n=1, 1000, write("b065796.txt", n, " ", SumAD(n^2)) ) } \\ Harry J. Smith, Oct 30 2009

Crossrefs

Cf. A055017.

Sequence in context: A011513 A340350 A164818 * A070434 A011003 A344078

Adjacent sequences: A065793 A065794 A065795 * A065797 A065798 A065799

Keyword

base,easy,sign

Author

Benny Wegner (jaeger(AT)clan-efg.de), Dec 05 2001

Extensions

More terms from Robert G. Wilson v, Dec 06 2001

Status

approved