A065085 Smallest prime having alternating bit sum (A065359) equal to -n, or 0 if no such prime exists.

3, 2, 43, 0, 683, 2731, 0, 43691, 174763, 0, 2796203, 44608171, 0, 715827881, 715827883, 0, 114532461227, 183251938027, 0, 2931494136491, 2932031007403, 0, 187647836990123, 748400914639531, 0, 11446649052900011, 45786596211600043, 0

Offset: 0

Author

Robert G. Wilson v, Nov 09 2001

Keywords

• nonn

Examples

			The smallest number having alternating bit sum -n is (2/3)(4^n-1), which for n=4 is 170 = (10101010)2. The least odd number with alternating bit sum -4 is (10)2 || ( (10101010)2 + (1)2 ) = 683, which is prime, so a(4) = 683. - _Washington Bomfim_, Jan 27 2011
		

Links

• Washington Bomfim, Table of n, a(n) for n = 0..117

Crossrefs

Cf. A065359, A020988, A065084.

Programs

Mathematica

f[n_] := Plus @@ (-(-1)^Range[Floor[Log2@n + 1]] Reverse@ IntegerDigits[n, 2]); a = Table[ f[ Prime[n]], {n, 1, 10^6} ]; b = Table[0, {12} ]; Do[ If[ a[[n]] < 1 && b[[ -a[[n]] + 1]] == 0, b[[ -a[[n]] + 1]] = Prime[n]], {n, 1, 10^6} ]; b

PARI

II()={i = (2/3)*(4^n-1) + 1 + 2^(2*n+1); if(isprime(i),return(1)); return(0)};
III()={w = 2^(2*n+3); for(j=1, n+1, i += w; w /= 4; i -= w; if(isprime(i),return(1))); return(0)};
IV()={i+=6; if(isprime(i), return(1)); w=4; for(j=1, n, i -= w; w*=4; i+=w; if(isprime(i), return(1)));return(0)};
V()={i += 2^(2*n+4) - 2^(2*n+2); if(isprime(i),return(1));w = i + 2^(2*n+5) - 2^(2*n+4); i = w - 2^(2*n+3) - 2^(2*n+1); if(isprime(i),return(1));w = 2^(2*n+1);for(j=1, n,i += w; w /= 4; i -= w;if(isprime(i),return(1) ));return(0)};
print("0 3");print("1 2");for(n=2,117, if(II(),print(n," ",i), if(III(),print(n," ",i), if(IV(),print(n," ",i), if(V(),print(n," ",i),if(n%3==0,print(n," 0"),print(n," not found."))))))) \\ Washington Bomfim, Feb 06 2011

Extensions

a(13)-a(27) from Washington Bomfim, Jan 27 2011