Search: id:A139312 Results 1-1 of 1 results found. %I A139312 %S A139312 1,1,1,0,1,0,1,1,0,1,0,0,1,1,1,0,1,0,0,1,0,1,1,0,0,1,0,1,1,0,1,0,1,0,1, %T A139312 1,0,1,1,0,1,0,1,0,1,1,0,0,1,1,0,1,0,1,1,0,1,0,0,1,1,0,0,1,1,0,1,0,1,1, %U A139312 1,0,1,0,1,1,0,1,1,0,1,0,1,0,1,1,0,0,1,1,0,0,1,0,1,1,0,1,0 %N A139312 An binary "appearance" or frequency sequence for good and bad primes: A028388 and A130903: a(n)=If[ Prime[n]^2-Prime[n-1]*Prime[n+1]>=0, 1,0]. %C A139312 When the sequence of gaps repeats, the f[n] function %C A139312 comes up "ComplexInfinity": those are singularities of a=b in the derivation of when in the function f[n]: %C A139312 -Prime[ -1 + n] + 2 Prime[n] - Prime[1 + n] == 0 %C A139312 Those are "bad primes". %F A139312 Starting at 3: a(n)=If[ Prime[n]^2-Prime[n-1]*Prime[n+1]>=0,1,0] %t A139312 b0 = Table[If[Prime[n]^2 - Prime[n - 1]*Prime[n + 1] < 0, 1, 0], {n, 2, 100}] (*alternative formula: derived*) Solve[x^2 - (x - a)*(x + b) == 0, x] a = -Prime[ -1 + n] + Prime[n] b = -Prime[n] + Prime[1 + n] f[n_] = If[ -Prime[ -1 + n] + 2 Prime[n] - Prime[1 + n] == 0, 0, -a*b/(a - b)] a0 = Table[If[f[n] > 0, 1, 0], {n, 2, 100}] %Y A139312 Cf. A028388, A130903. %Y A139312 Sequence in context: A071023 A132194 A092079 this_sequence A071041 A140074 A090174 %Y A139312 Adjacent sequences: A139309 A139310 A139311 this_sequence A139313 A139314 A139315 %K A139312 nonn,uned %O A139312 1,1 %A A139312 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jun 07 2008 Search completed in 0.001 seconds