A065085 -id:A065085 - cp's OEIS frontend

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

Robert G. Wilson v, Nov 09 2001

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

			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

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

Cf. A065359, A002450, A065085.

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

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

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

A184908 Let S_n be the set of the integers having alternating bit sum equal to -n. There are a(n) primes among the smallest 3n+5 odd numbers of S_n.

Original entry on oeis.org

1, 7, 5, 0, 7, 5, 0, 7, 10, 0, 6, 3, 0, 5, 9, 0, 7, 7, 0, 7, 9, 0, 5, 6, 0, 6, 7, 0, 10, 7, 0, 4, 5, 0, 11, 7, 0, 10, 9, 0, 9, 4, 0, 4, 8, 0, 2, 6, 0, 9, 5, 0, 10, 9, 0, 8, 6, 0, 4, 3, 0, 4, 11, 0, 9, 3, 0, 5, 8, 0, 6, 3, 0, 11, 7, 0, 6, 8, 0, 5, 6

Offset: 0

Washington Bomfim, Jan 27 2011

			The smallest 3n+5 = 8 odd numbers of the set S_1 of the integers having alternating bit sum -1 are 11, 35, 41, 47, 59, 107, 131, and 137, so a(1)=7.

W. Bomfim, A method to find the first 3n+5 odd integers with alternating bit sum -n

Cf. A065359, A065085, A184907, A020988.

A065359 := proc(n) local dgs ; dgs := convert(n,base,2) ; add( -op(i,dgs)*(-1)^i,i=1..nops(dgs)) ; end proc:
S := proc(n) local ads,k; ads := {} ; for k from 1 by 2 do if A065359(k) = -n then ads := ads union {k} ; end if; if nops(ads) = 3*n+5 then return ads; end if; end do: end proc:
A184908 := proc(n) local ads,a,p; a := 0 ; for p in S(n) do if isprime(p) then a := a+1 ; end if; end do: a ; end proc:
for n from 0 do print(A184908(n)); end do: # slow! R. J. Mathar, Feb 11 2011

II()={i = (2/3)*(4^n-1) + 1 + 2^(2*n+1); if(isprime(i),an++)};
III()={w = 2^(2*n+3); for(j=1, n+1, i += w; w /= 4; i -= w; if(isprime(i), an++ ))};
IV()={i+=6; if(isprime(i), an++ ); w=4; for(j=1, n, i -= w; w *= 4; i += w; if(isprime(i), an++))};
V()={i += 2^(2*n+4) - 2^(2*n+2); if(isprime(i),an++ );w = i + 2^(2*n+5) - 2^(2*n+4); i = w - 2^(2*n+3) - 2^(2*n+1); if(isprime(i),an++ );w = 2^(2*n+1);for(j=1, n,i += w; w /= 4; i -= w;if(isprime(i),an++ ))};
print1("1, 7, ");for(n=2,80, an=0; II(); III(); IV(); V(); print1(an,", ")) \\ Washington Bomfim, Feb 06 2011

cp's OEIS Frontend

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

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