A065084 Smallest prime having alternating bit sum (A065359) equal to n.

3, 7, 5, 0, 277, 1109, 0, 17749, 70997, 0, 1398037, 5526869, 0, 72701269, 357915989, 0, 5659514197, 22902297941, 0, 297784399189, 1465948394837, 0, 23456248042837, 89426945725781, 0, 1430831131612501, 6004798429418837, 0

Offset: 0

Author

Robert G. Wilson v, Nov 09 2001

Keywords

• base
• nonn

Comments

Only 3d = 11b has an alternating sum of 0 and alternated sums of 3*k are impossible for primes.

Examples

			a(4)=277 since the smallest number having alternating bit sum n is (4^n-1)/3, which for n = 4 is 85. Because 85 =(1010101)2 is composite, the next number with alternating bit sum 4 is the prime (100010101)2 = 277. - _Washington Bomfim_, Jan 21 2011
		

Links

• Washington Bomfim, Table of n, a(n) for n = 0..121

Crossrefs

Cf. A065359, A002450, A065085.

Programs

Mathematica

f[n_] := (d = Reverse[ IntegerDigits[n, 2]]; l = Length[d]; s = 0; k = 1; While[k < l + 1, s = s - (-1)^k*d[[k]]; k++ ]; s); a = Table[ f[ Prime[n]], {n, 1, 10^6} ]; b = Table[0, {13} ];
Do[ If[ a[[n]] > -1 && b[[a[[n]] + 1]] == 0, b[[a[[n]] + 1]] = Prime[n]], {n, 1, 10^6} ]; b

PARI

M(n)={return((4^n - 1)/3 + 2^(2*n) - 2^(2*n-2))};
T(n,k)={pow2=2^(2*n-2);k+=pow2; for(j=1,n-2,pow2/=4; k-=pow2;if(isprime(k),return(k),k+=pow2;)); return(k)};
T2(n,k)={pow2=2; for(j=1,n, k+=pow2;if(isprime(k),return(k),k-=pow2; pow2*=4)); return(k)};
print("0 3");print("1 7");print("2 5");print("3 0");for(n=4,127,if(n%3==0,print(n," 0"),k=M(n);if(isprime(k),print(n," ",k),k=T(n,k);if(isprime(k),print(n," ",k),k=T2(n,k);if(isprime(k),print(n," ",k),print("a(",n,") not found")))))) \\ Washington Bomfim, Jan 22 2011

Extensions

a(14)-a(27) from Washington Bomfim, Jan 21 2011