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%I #19 Dec 08 2023 11:15:52
%S 3,2,43,0,683,2731,0,43691,174763,0,2796203,44608171,0,715827881,
%T 715827883,0,114532461227,183251938027,0,2931494136491,2932031007403,
%U 0,187647836990123,748400914639531,0,11446649052900011,45786596211600043,0
%N Smallest prime having alternating bit sum (A065359) equal to -n, or 0 if no such prime exists.
%H Washington Bomfim, <a href="/A065085/b065085.txt">Table of n, a(n) for n = 0..117</a>
%e 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
%t 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
%o (PARI)II()={i = (2/3)*(4^n-1) + 1 + 2^(2*n+1); if(isprime(i),return(1)); return(0)};
%o III()={w = 2^(2*n+3); for(j=1, n+1, i += w; w /= 4; i -= w; if(isprime(i),return(1))); return(0)};
%o 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)};
%o 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)};
%o 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
%Y Cf. A065359, A020988, A065084.
%K nonn
%O 0,1
%A _Robert G. Wilson v_, Nov 09 2001
%E a(13)-a(27) from _Washington Bomfim_, Jan 27 2011
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