20200223, 16:20  #1 
Mar 2018
17·31 Posts 
Primes of the form 30*n^21
Because I found that the prime 5879 is in some mysterious way involved in pg primes
And realizing that 5879 is a prime of the form 30*n^21, I wonder if in oesi there is a sequence of primes of this form 
20200223, 17:14  #2 
"Sam"
Nov 2016
474_{8} Posts 
You could always make one yourself if it doesn't exist:
Code:
(09:13) gp > g=[]; for(i=1,300, if(isprime(30*i^21), g=concat(g,i))) (09:13) gp > g %5 = [1, 3, 4, 10, 14, 17, 18, 21, 22, 25, 29, 34, 35, 38, 42, 43, 48, 52, 55, 56, 60, 62, 63, 67, 73, 74, 78, 80, 90, 92, 94, 95, 99, 101, 105, 108, 113, 116, 118, 119, 126, 127, 129, 130, 132, 133, 139, 140, 143, 147, 153, 154, 157, 178, 181, 183, 186, 190, 192, 195, 199, 200, 207, 213, 216, 221, 224, 225, 232, 234, 238, 242, 244, 248, 251, 256, 259, 265, 269, 273, 276, 281, 284, 288, 290, 294, 297, 300] (09:13) gp > 
20200224, 04:17  #3 
Aug 2006
13455_{8} Posts 
For a more general approach:
Code:
polyToSeqPrimes(p)=my(v=List(),len,x=variable(p),n,t); while(len<=260, if(isprime(t=subst(p,x,n)), listput(v,t); len+=#Str(t)+2); n++); v=Vec(v); v[1..#v1]; polyToSeqIndices(p)=my(v=List(),len,x=variable(p),n); while(len<=260, if(isprime(subst(p,x,n)), listput(v,n); len+=#Str(n)+2); n++); v=Vec(v); v[1..#v1]; polyToPrimeBFile(p)=my(v=vector(10^4),x=variable(p),n,i,t); while(i<#v, if(isprime(t=subst(p,x,n)), v[i++]=t); n++); v; polyToIndexBFile(p)=my(v=vector(10^4),x=variable(p),n,i); while(i<#v, if(isprime(subst(p,x,n)), v[i++]=n); n++); v; polyToSeqPrimes(30*'n^21) polyToSeqIndices(30*'n^21) 
20200225, 10:47  #4  
Mar 2018
527_{10} Posts 
...
Quote:
I note that there are pairs (21,22) (42,43) (126,127) with 21,42,126 multiple of 7 Is it unlikely that these consecutive (a,b) are infinitely many? 

20200225, 10:51  #5  
Mar 2018
17·31 Posts 
...
Quote:
I note that there are pairs (21,22) (42,43) (126,127) (62,63)...(a,b) with either a or b multiple of 7...do you believe these pairs with consecutive a,b are infinite? I call a(n) the nth term of the sequence so it seems that there are infinitely many terms a(n) such that a(n+1)=a(n)+1 a(n)=21 a(n+1)=23=a(n)+1 Last fiddled with by enzocreti on 20200225 at 11:15 

20200301, 02:10  #6  
"Sam"
Nov 2016
100111100_{2} Posts 
Quote:
We cannot prove or disprove this at the moment, but it should be implied there exist infinitely many pairs by one of Dickinson's conjecture. The pattern continues: (18:08) gp > g=[]; for(i=1,2000, if(isprime(30*i^21) & isprime(30*i^2+60*i+29) & i%7==0, g=concat(g,[[i,i+1]]))) (18:09) gp > g %12 = [[21, 22], [42, 43], [126, 127], [224, 225], [399, 400], [455, 456], [469, 470], [686, 687], [693, 694], [742, 743], [777, 778], [833, 834], [1057, 1058], [1092, 1093], [1127, 1128], [1309, 1310], [1484, 1485], [1659, 1660], [1673, 1674], [1750, 1751], [1792, 1793], [1806, 1807]] 

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