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

Author

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

Keywords

• base
• easy
• sign

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

Examples

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

Links

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

Crossrefs

Cf. A055017.

Programs

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 *)

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

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

Extensions

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