editing

approved

for n from 1 to 10 do for k from 1 do T := A088328(n, k) ; if T < 0 then break; else printf("%d, ", T) ; end if; end do; printf("\n") ; od: # _R. J. Mathar, _, Oct 30 2009

approved

editing

_Pierre CAMI (colettecami(AT)aol.com), _, Nov 06 2003

proposed

approved

editing

proposed

Some of the values k*primorial(n)-1 generated by k in the range 1 to prime(n+1)*prime(n+2)-1 are not lower twin primes, A001359, so the list of k that produces the n-th row of the irregular table, as shown in A088329, is not a list of necessarily consecutive integers.

are not lower twin primes, A001359, so the list of k that produces the n-th row of the

irregular table, as shown in A088329, is not a list of necessarily consecutive integers.

approved

editing

Edited by _R. J. Mathar (mathar(AT)strw.leidenuniv.nl), _, Oct 30 2009

Table read by rows where n-th row consists of primes P(n,k) such that P(n,k)=k*p(n)# -1 is the first of prime twins with 0 < k < p(n+1)*p(n+2) where p(i) denotes i-th prime and p(i)# denotes i-th primorial, starting with n=1 p(1)=2.

Table read by rows where row n contains lower twin primes of the form k*A002110(n)-1 in the range 0 < k < A006094(n+1).

Some of the values k*primorial(n)-1 generated by k in the range 1 to prime(n+1)*prime(n+2)-1

are not lower twin primes, A001359, so the list of k that produces the n-th row of the

irregular table, as shown in A088329, is not a list of necessarily consecutive integers.

If n>2 the number count of k values is near or greater than 4*log(4*p(n+1)), ; is this related to a proof of the infinity of twin prime twinspairs? k values given in one other sequence

first row n=1, k=2,3,6,9, P(1,k)=3,5,11,17

The first three rows are:

3,5,11,17; generated by k=2, 3, 6, 9

P(2,k)=5,11,17,29,41,59,71,101,107,137,149,179,191,197;

P(3,k)=29,59,149,179,239,269,419,569,599,659,809,1019,1049,1229,1289,1319,1619,1949,2129;

isA001359 := proc(n) option remember ; return isprime(n) and isprime(n+2) ; end:

A002110 := proc(n) local i ; if n = 0 then 1; else mul(ithprime(i), i=1..n) ; end if; end proc:

A006094 := proc(n) return ithprime(n)*ithprime(n+1) ; end proc:

A088328 := proc(n, k) option remember; for j from 1 to A006094(n+1)-1 do a := j*A002110(n)-1 ; if isA001359(a) and k =1 then return a ; elif isA001359(a) and a > procname(n, k-1) then return a ; end if; end do; return -1 ; end proc:

for n from 1 to 10 do for k from 1 do T := A088328(n, k) ; if T < 0 then break; else printf("%d, ", T) ; end if; end do; printf("\n") ; od: # R. J. Mathar, Oct 30 2009

nonn,tabl,uned,new

nonn,tabf

Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Oct 30 2009

Table read by rows where n-th row consists of primes P(n,k) such that P(n,k)=k*p(n)# -1 is the first of prime twins with 0 < k < p(n+1)*p(n+2) where p(i) denotes i-th prime and p(i)# denotes i-th primorial, starting with n=1 p(1)=2.

3, 5, 11, 17, 5, 11, 17, 29, 41, 59, 71, 101, 107, 137, 149, 179, 191, 197, 29, 59, 149, 179, 239, 269, 419, 569, 599, 659, 809, 1019, 1049, 1229, 1319, 1619, 1949, 2129, 419, 1049, 2309, 2729, 3359, 5879, 6089, 6299, 7349, 7559, 8819, 9239, 10499, 10709

1,1

If n>2 the number of k values is near or greater than 4*log(4*p(n+1)), proof of the infinity of prime twins? k values given in one other sequence

2*2 -1 = 3, k=2, n=1

3*2 -1 = 5, k=3, n=1

6*2 -1 = 11, k=6, n=1

9*2 -1 = 17, k=9, n=1

first row n=1, k=2,3,6,9, P(1,k)=3,5,11,17

P(2,k)=5,11,17,29,41,59,71,101,107,137,149,179,191,197

P(3,k)=29,59,149,179,239,269,419,569,599,659,809,1019,1049,1229,1289,1319,1619,1949,2129

nonn,tabl,uned

Pierre CAMI (colettecami(AT)aol.com), Nov 06 2003

approved