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

%I A065084 #17 Dec 08 2023 11:15:14
%S A065084 3,7,5,0,277,1109,0,17749,70997,0,1398037,5526869,0,72701269,
%T A065084 357915989,0,5659514197,22902297941,0,297784399189,1465948394837,0,
%U A065084 23456248042837,89426945725781,0,1430831131612501,6004798429418837,0
%N A065084 Smallest prime having alternating bit sum (A065359) equal to n.
%C A065084 Only 3d = 11b has an alternating sum of 0 and alternated sums of 3*k are impossible for primes.
%H A065084 Washington Bomfim, <a href="/A065084/b065084.txt">Table of n, a(n) for n = 0..121</a>
%e A065084 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
%t A065084 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} ];
%t A065084 Do[ If[ a[[n]] > -1 && b[[a[[n]] + 1]] == 0, b[[a[[n]] + 1]] = Prime[n]], {n, 1, 10^6} ]; b
%o A065084 (PARI)M(n)={return((4^n - 1)/3 + 2^(2*n) - 2^(2*n-2))};
%o A065084 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)};
%o A065084 T2(n,k)={pow2=2; for(j=1,n, k+=pow2;if(isprime(k),return(k),k-=pow2; pow2*=4)); return(k)};
%o A065084 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
%Y A065084 Cf. A065359, A002450, A065085.
%K A065084 base,nonn
%O A065084 0,1
%A A065084 _Robert G. Wilson v_, Nov 09 2001
%E A065084 a(14)-a(27) from _Washington Bomfim_, Jan 21 2011