A184908 Let S_n be the set of the integers having alternating bit sum equal to -n. There are a(n) primes among the smallest 3n+5 odd numbers of S_n.

1, 7, 5, 0, 7, 5, 0, 7, 10, 0, 6, 3, 0, 5, 9, 0, 7, 7, 0, 7, 9, 0, 5, 6, 0, 6, 7, 0, 10, 7, 0, 4, 5, 0, 11, 7, 0, 10, 9, 0, 9, 4, 0, 4, 8, 0, 2, 6, 0, 9, 5, 0, 10, 9, 0, 8, 6, 0, 4, 3, 0, 4, 11, 0, 9, 3, 0, 5, 8, 0, 6, 3, 0, 11, 7, 0, 6, 8, 0, 5, 6

Offset: 0

Author

Washington Bomfim, Jan 27 2011

Keywords

• nonn
• base

Examples

			The smallest 3n+5 = 8 odd numbers of the set S_1 of the integers having alternating bit sum -1 are 11, 35, 41, 47, 59, 107, 131, and 137, so a(1)=7.
		

Links

• W. Bomfim, A method to find the first 3n+5 odd integers with alternating bit sum -n

Crossrefs

Cf. A065359, A065085, A184907, A020988.

Programs

Maple

A065359 := proc(n) local dgs ; dgs := convert(n,base,2) ; add( -op(i,dgs)*(-1)^i,i=1..nops(dgs)) ; end proc:
S := proc(n) local ads,k; ads := {} ; for k from 1 by 2 do if A065359(k) = -n then ads := ads union {k} ; end if; if nops(ads) = 3*n+5 then return ads; end if; end do: end proc:
A184908 := proc(n) local ads,a,p; a := 0 ; for p in S(n) do if isprime(p) then a := a+1 ; end if; end do: a ; end proc:
for n from 0 do print(A184908(n)); end do: # slow! R. J. Mathar, Feb 11 2011

PARI

II()={i = (2/3)*(4^n-1) + 1 + 2^(2*n+1); if(isprime(i),an++)};
III()={w = 2^(2*n+3); for(j=1, n+1, i += w; w /= 4; i -= w; if(isprime(i), an++ ))};
IV()={i+=6; if(isprime(i), an++ ); w=4; for(j=1, n, i -= w; w *= 4; i += w; if(isprime(i), an++))};
V()={i += 2^(2*n+4) - 2^(2*n+2); if(isprime(i),an++ );w = i + 2^(2*n+5) - 2^(2*n+4); i = w - 2^(2*n+3) - 2^(2*n+1); if(isprime(i),an++ );w = 2^(2*n+1);for(j=1, n,i += w; w /= 4; i -= w;if(isprime(i),an++ ))};
print1("1, 7, ");for(n=2,80, an=0; II(); III(); IV(); V(); print1(an,", ")) \\ Washington Bomfim, Feb 06 2011