This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
DeleteDuplicates[Table[Count[Prime[n]+Prime[Range[n-1]]+1,_?PrimeQ],{n,3000}],GreaterEqual]
a:= n-> add(`if`(isprime(ithprime(n)+ithprime(i)-1), 1, 0), i=1..n-1): seq(a(n), n=1..100); # Alois P. Heinz, Apr 18 2013
Table[p = Prime[n]; c = 0; i = 1; While[i < n, If[PrimeQ[p + Prime[i] - 1], c = c + 1]; i++]; c, {n, 72}]
Table[Count[Prime[n]+Prime[Range[n-1]]-1,?PrimeQ],{n,80}] (* _Harvey P. Dale, Nov 04 2020 *)
For n=3, p=5, there are a(3)=2 solutions 2,3 since 5*2+5+2=17, 5*2-5-2=3 and 5*3+5+3=23, 5*3-5-3=7. Also for n=5, p=11, there is a(5)=1 solution in the form of 11*3+11+3=47, 11*3-11-3=19.
Table[p = Prime[n]; c = 0; i = 1; While[i < n, q1 = p*Prime[i]; q2 = p + Prime[i]; If[PrimeQ[q1 + q2] && PrimeQ[q1 - q2], c = c + 1]; i++]; c, {n, 85}]
a(3)=2 since for the third prime 5 we have 5*2-(5+2)=3 and 5*3-(5+3)=7. Also a(4)=3 since for the fourth prime 7 we have 7*2-(7+2)=5, 7*3-(7+3)=11 and 7*5-(7+5)=23.
Table[p = Prime[n]; c = 0; i = 1; While[i < n, If[PrimeQ[p*Prime[i] - (p + Prime[i])], c = c + 1] i++]; c, {n, 75}]
For n=3, p=5, there are a(3)=2 solutions from 5*2+(5+2)=17 and 5*3+(5+3)=23. For n=4, p=7, there are a(4)=3 solutions in the form of 7*2+(7+2)=23, 7*3+(7+3)=31 and 7*5+(7+5)=47.
a:= n-> (p-> add((q-> `if`(isprime((p+1)*(q+1)-1),
1, 0))(ithprime(j)), j=1..n-1))(ithprime(n)):
seq(a(n), n=1..100); # Alois P. Heinz, Jul 18 2019
Table[p = Prime[n]; c = 0; i = 1; While[i < n, If[PrimeQ[p*Prime[i] + (p + Prime[i])], c = c + 1]; i++]; c, {n, 75}]
a(n) = my(p=prime(n), q); sum(k=1, n-1, q=prime(k); isprime(p*q+(p+q))); \\ Michel Marcus, Jul 18 2019
For n=5, p=11, there are a(5)=2 solutions: 11-5+1=7 and 11-7+1=5.
Table[p = Prime[n]; c = 0; i = 1; While[i < n, If[PrimeQ[p - Prime[i] + 1], c = c + 1]; i++]; c, {n, 70}]
Table[Count[Prime[n]-Prime[Range[n-1]]+1,?PrimeQ],{n,70}] (* _Harvey P. Dale, Jan 08 2015 *)
a(n)=my(p=prime(n),s);forprime(q=2,p-1,s+=isprime(p-q+1));s \\ Charles R Greathouse IV, Apr 22 2013
For n=3, p=5, there are no primes q(<5) such that both 5+q+1 and 5+q-1 are primes and hence a(3)=0. Also for n=5, p=11, there is a(5)=1 solution 7 since 11+7+1=19, 11+7-1=17.
Table[p = Prime[n]; c = 0; i = 1; While[i < n, p1 = p + Prime[i]; If[PrimeQ[p1 + 1] && PrimeQ[p1 - 1], c = c + 1]; i++]; c, {n, 85}]
pq1[n_]:=Module[{pr1=Prime[Range[n-1]],pr2=Prime[n]},Total[ Table[ If[ AllTrue[pr2+pr1[[k]]+{1,-1},PrimeQ],1,0],{k,Length[pr1]}]]]; Array[ pq1,100] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 20 2020 *)
For n=5, p=11, there are a(5)=3 solutions: 11-3-1=7, 11-5-1=5, and 11-7-1=3.
Table[p = Prime[n]; c = 0; i = 1; While[i < n, If[PrimeQ[p - Prime[i] - 1], c = c + 1]; i++]; c, {n, 69}]
a(n)=my(p=prime(n),s);forprime(q=2,p-1,s+=isprime(p-q-1));s \\ Charles R Greathouse IV, Apr 22 2013
For n=3, p=5, there are no primes q(<5) such that both 5-q+1 and 5-q-1 are primes and hence a(3)=0. Also for n=5, p=11, there are a(5)=2 solutions 5,7 since 11-5+1=7, 11-5-1=5 and 11-7+1=5, 11-7-1=3.
Table[p = Prime[n]; c = 0; i = 1; While[i < n, p1 = p - Prime[i]; If[PrimeQ[p1 + 1] && PrimeQ[p1 - 1], c = c + 1]; i++]; c, {n, 80}]
Table[Count[Prime[p]-Prime[Range[p-1]],?(AllTrue[#+{1,-1},PrimeQ]&)],{p,80}] (* _Harvey P. Dale, Aug 12 2025 *)
a(2)=1 since second prime 3 generates 23. Also a(7)=2 since for the seventh prime 17 we have two primes 317 and 1117.
con[x_,y_] := FromDigits[Join[IntegerDigits[Prime[x]], IntegerDigits[Prime[y]]]]; t={}; Do[c=0; Do[If[PrimeQ[con[i,n]], c=c+1], {i,n}]; AppendTo[t,c], {n,78}]; t
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