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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.
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%I #10 Mar 03 2015 00:54:44

%S 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,

%T 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,

%U 3,0,5,8,0,6,3,0,11,7,0,6,8,0,5,6

%N 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.

%H W. Bomfim, <a href="/w/images/c/cd/A184908.txt">A method to find the first 3n+5 odd integers with alternating bit sum -n</a>

%e 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.

%p A065359 := proc(n) local dgs ; dgs := convert(n,base,2) ; add( -op(i,dgs)*(-1)^i,i=1..nops(dgs)) ; end proc:

%p 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:

%p 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:

%p for n from 0 do print(A184908(n)); end do: # slow! _R. J. Mathar_, Feb 11 2011

%o (PARI)II()={i = (2/3)*(4^n-1) + 1 + 2^(2*n+1); if(isprime(i),an++)};

%o III()={w = 2^(2*n+3); for(j=1, n+1, i += w; w /= 4; i -= w; if(isprime(i), an++ ))};

%o IV()={i+=6; if(isprime(i), an++ ); w=4; for(j=1, n, i -= w; w *= 4; i += w; if(isprime(i), an++))};

%o 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++ ))};

%o print1("1, 7, ");for(n=2,80, an=0; II(); III(); IV(); V(); print1(an,", ")) \\ _Washington Bomfim_, Feb 06 2011

%Y Cf. A065359, A065085, A184907, A020988.

%K nonn,base

%O 0,2

%A _Washington Bomfim_, Jan 27 2011