A065081 Alternating bit sum (A065359) for n-th prime p: replace 2^k with (-1)^k in binary expansion of p.

-1, 0, 2, 1, -1, 1, 2, 1, 2, 2, 1, 1, -1, -2, -1, 2, -1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, -1, 1, 2, 1, -1, -1, -2, 2, 1, 1, -2, -1, -1, -1, 1, -1, 1, 2, 1, 1, 1, -1, 1, -1, -1, 1, -1, 2, 2, 2, 1, 4, 2, 1, 2, 1, 2, 1, 2, 1, 4, 2, 4, 2, 2, 1, 4, 1, 2, 2, 1, 2, 1, -1, 1, -1, 1, 1, -1, 2, 1, 2, 1, 2, 2, 1, -1, 1, 2, 2, -1, -2, 1

Offset: 1

Author

Robert G. Wilson v, Nov 09 2001

Keywords

• base
• easy
• sign

Comments

Only 3d = 11b has an alternating sum of 0.

Examples

			The sixth prime is 13d = 1101b -> -(1)+(1)-(0)+(1) = 1 = a(6)
		

Links

• Harry J. Smith, Table of n, a(n) for n=1..1000
• William Paulsen, wpaulsen(AT)csm.astate.edu, Partitioning the [prime] maze

Crossrefs

Cf. A065359.

Programs

Mathematica

f[n_] := (d = Reverse[ IntegerDigits[n, 2]]; l = Length[d]; s = 0; k = 1; While[k < l + 1, s = s - (-1)^k*d[[k]]; k++ ]; s); Table[ Prime[ f[n]], {n, 1, 100} ]

PARI

baseE(x, b)=
{
  local(d, e=0, f=1);
  while (x>0, d=x-b*(x\b); x\=b; e+=d*f; f*=10);
  return(e)
}
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, p=prime(n);
  s=SumAD(baseE(p, 2)); write("b065081.txt", n, " ", s) )
} \\ Harry J. Smith, Oct 06 2009

PARI

f(p)=
{
  v=binary(p);
  L=#v; u=1; s=0;
  forstep(k=L,1,-1, if(v[k]==1,s+=u); u=-u;);
  return(s)
};
for(n=1,100,p=prime(n); an=f(p);print1(an,", ")) \\ Washington Bomfim, Jan 16 2011

Python

from sympy.ntheory import digits, prime
def A065081(n): return sum((0,1,-1,0)[i] for i in digits(prime(n),4)[1:]) # Chai Wah Wu, Jul 19 2024