A092641
Absolute value of difference between counts of uninterrupted runs of 2 primes in A092639 and A092640.
Original entry on oeis.org
0, 3, 6, 15, 29, 117, 500, 1136, 4475, 3065, 8440
Offset: 1
a(2)=3 because in A092639 the count is 2 and in A092640 the count is 5. 5 - 3 = 2.
A093186
Number of consecutive runs of 2 odd nonprimes congruent to 3 mod 4 below 10^n.
Original entry on oeis.org
0, 4, 36, 307, 2848, 25651, 231031, 2106565, 19307362, 177948719, 1649246163, 15360077721, 143683073300
Offset: 1
a(3)=36 because 36 nonprime runs of 2 occur below 10^3, each run interrupted by a nonprime congruent to 1 mod 4.
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A091113 = Select[4 Range[0, 10^4] + 1, ! PrimeQ[#] &];
A091236 = Select[4 Range[0, 10^4] + 3, ! PrimeQ[#] &];
lst = {}; Do[If[Length[s = Select[A091236, Between[{A091113[[i]], A091113[[i + 1]]}]]] == 2, AppendTo[lst, Last[s]]], {i, Length[A091113] - 1}]; Table[Count[lst, x_ /; x < 10^n], {n, 4}] (* Robert Price, May 30 2019 *)
A093398
Absolute value of difference between counts of uninterrupted runs of 2 nonprimes in A093185 and A093186.
Original entry on oeis.org
1, 0, 3, 2, 43, 85, 42, 36, 3823, 192, 5114, 2797, 148220
Offset: 1
a(3)=3 because in A093185 the count is 33 and in A093186 the count is 36. 36-33=3.
A092639
Number of consecutive prime runs of 2 primes congruent to 1 mod 4 below 10^n.
Original entry on oeis.org
0, 2, 14, 116, 780, 6066, 49510, 417230, 3631524, 32070665, 287366058
Offset: 1
a(4)=116 because 116 pairs of primes occur below 10^4, each run interrupted by a prime congruent to 3 mod 4.
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A002145 = Join[{0}, Select[4 Range[0, 10^4] + 3, PrimeQ[#] &]];
A002144 = Select[4 Range[0, 10^4] + 1, PrimeQ[#] &];
lst = {}; Do[If[Length[s = Select[A002144, Between[{A002145[[i]], A002145[[i + 1]]}]]] == 2, AppendTo[lst, Last[s]]], {i, Length[A002145] - 1}]; Table[Count[lst, x_ /; x < 10^n], {n, 4}] (* Robert Price, Jun 09 2019 *)
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ispRun(p1)={ local(p2,p3) ; if(!isprime(p1) || (p1 %4 ==3) || (precprime(p1-1) % 4 ==1), return(0), p2=nextprime(p1+1) ; if( p2 %4 == 3, return(0), p3=nextprime(p2+1) ; if( p3 %4 == 3, return(1), return(0) ) ; ) ; ) ; } { an=0 ; n=1 ; p=prime(1) ; while(1, if( (p<10^n) && (nextprime(p+1) >= 10^n), print(an); n++ ; ) ; an += ispRun(p) ; p=nextprime(p+1) ; ) } \\ R. J. Mathar, Sep 25 2006
Showing 1-4 of 4 results.