A065084 -id:A065084 - cp's OEIS frontend

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

Robert G. Wilson v, Nov 09 2001

nonn

			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

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

Cf. A065359, A020988, A065084.

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

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

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

A184907 Let S_n be the set of the integers having alternating bit sum equal to n. There are a(n) primes among the smallest 2n+1 numbers of S_n.

Original entry on oeis.org

0, 1, 4, 0, 4, 7, 0, 4, 5, 0, 1, 6, 0, 9, 3, 0, 6, 5, 0, 8, 5, 0, 3, 2, 0, 6, 3, 0, 3, 9, 0, 5, 5, 0, 3, 6, 0, 5, 2, 0, 8, 6, 0, 8, 6, 0, 2, 12, 0, 8, 1, 0, 2, 7, 0, 2, 1, 0, 4, 5, 0, 7, 5, 0, 8, 6, 0, 7, 6, 0, 3, 9, 0, 4, 11, 0, 4, 5, 0, 5, 2

Offset: 0

Washington Bomfim, Jan 25 2011

nonn

The smallest 2n+1 numbers in the set S_n are given by (4^(n+1)-1)/3 - (-2)^k for k in 0..2*n. - Andrew Howroyd, Dec 15 2024

			The smallest 2n+1 = 5 numbers of the set S_2 of the integers having alternating bit sum 2, are 5, 17, 20, 23, and 29, so a(2)=4.

Andrew Howroyd, Table of n, a(n) for n = 0..500
Washington Bomfim, A method to find the first 2n+1 integers having alternating bit sum equal to n.

Cf. A065359, A065084, A002450, A184908.

II()={i = (4^n - 1)/3 - 2^(2*n-2) + 2^(2*n); if(isprime(i),an++)};
III()={w = 2^(2*n-2); for(j=1, n-1, i += w; w /= 4; i -= w; if(isprime(i), an++;))};
IV()={i+=3; if(isprime(i), an++); w=2; for(j=1, n-1, i -= w; w *= 4; i+=w; if(isprime(i),an++))};
print1("0, 1, 4, ");for(n=3,80, an=0; II(); III(); IV(); print1(an,", ")) \\ Washington Bomfim, Jan 25 2011

a(n)={my(m=(4^(n+1)-1)/3); sum(k=0, 2*n, isprime(bitxor(m,1<Andrew Howroyd, Dec 15 2024

a(n) = 0 for n == 0 (mod 3). - Andrew Howroyd, Dec 15 2024

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

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A184907 Let S_n be the set of the integers having alternating bit sum equal to n. There are a(n) primes among the smallest 2n+1 numbers of S_n.

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