Logo

Greetings from The On-Line Encyclopedia of Integer Sequences!

Hints

Search: id:A139312
Displaying 1-1 of 1 results found. page 1
     Format: long | short | internal | text      Sort: relevance | references | number      Highlight: on | off
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]. +0
1
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, 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, 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 (list; graph; listen)
OFFSET

1,1

COMMENT

When the sequence of gaps repeats, the f[n] function

comes up "ComplexInfinity": those are singularities of a=b in the derivation of when in the function f[n]:

-Prime[ -1 + n] + 2 Prime[n] - Prime[1 + n] == 0

Those are "bad primes".

FORMULA

Starting at 3: a(n)=If[ Prime[n]^2-Prime[n-1]*Prime[n+1]>=0,1,0]

MATHEMATICA

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}]

CROSSREFS

Cf. A028388, A130903.

Sequence in context: A071023 A132194 A092079 this_sequence A071041 A140074 A090174

Adjacent sequences: A139309 A139310 A139311 this_sequence A139313 A139314 A139315

KEYWORD

nonn,uned

AUTHOR

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jun 07 2008

page 1

Search completed in 0.002 seconds

Lookup | Welcome | Find friends | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Transforms | Puzzles | Hot | Classics
More pages | Superseeker | Maintained by N. J. A. Sloane (njas@research.att.com)

Last modified December 16 17:18 EST 2009. Contains 170825 sequences.


AT&T Labs Research