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A065084
Smallest prime having alternating bit sum (A065359) equal to n.
3
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
0,1
COMMENTS
Only 3d = 11b has an alternating sum of 0 and alternated sums of 3*k are impossible for primes.
LINKS
EXAMPLE
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
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
PROG
(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
CROSSREFS
KEYWORD
base,nonn
AUTHOR
Robert G. Wilson v, Nov 09 2001
EXTENSIONS
a(14)-a(27) from Washington Bomfim, Jan 21 2011
STATUS
approved