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A065084
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Smallest prime having alternating bit sum (A065359) equal to n.
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1
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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
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OFFSET
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0,1
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COMMENTS
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Only 3d = 11b has an alternating sum of 0 and alternated sums of 3*k are impossible for primes.
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LINKS
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Table of n, a(n) for n=0..27.
W. Bomfim, Table of n, a(n) for n = 1..121
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EXAMPLE
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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
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MATHEMATICA
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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
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PROG
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(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")))))) - W. Bomfim Jan 22, 2011
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CROSSREFS
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Cf. A065359, A002450.
Sequence in context: A193506 A086242 A096627 * A132742 A210641 A021910
Adjacent sequences: A065081 A065082 A065083 * A065085 A065086 A065087
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KEYWORD
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base,nonn
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AUTHOR
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Robert G. Wilson v, Nov 09 2001
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EXTENSIONS
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a(14)-a(27) from Washington Bomfim Jan 21, 2011
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STATUS
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approved
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