A352582 Two-column array read by rows, where the n-th row is the least pair of integers (p, q) such that f(p) = f(n) + q*f(n+1) where f(n) = A002496(n) is the n-th prime of the form k^2+1.

3, 3, 11, 76, 49, 2432, 113, 9980, 55, 748, 166, 9420, 384, 39780, 130, 2388, 271, 10640, 867, 82592, 1054, 103040, 548, 11828, 578, 12332, 4874, 1113600, 2461, 196380, 1137, 27932, 2426, 128944, 1393, 35708, 16086, 5861020, 2052, 54268, 9154, 1437780, 7981, 982208

Offset: 1

Author

Michel Lagneau, Mar 21 2022

Keywords

• nonn
• tabf

Comments

For given n, it seems there is an infinity of pairs (p,q) = (p0,q0), (p1, q1), (p2, q2), ... where p is the smallest p and q the smallest q: p=p0=min(p0, p1, p2, ...) and q = q0=min(q0, q1, ...).

Conjecture: Given an integer n, there always exists a pair (p, q) such that f(p) = f(n) + q*f(n+1).

Consequence: if the conjecture is true, then the set of prime numbers of the form k^2+1 is infinite because, by induction, there exists a pair (p', q') such that f(p') = f(p-1) + q'*f(p), f(p') > f(p).

Examples

			The pair (11, 76) is in the sequence because A002496(11) = A002496(2) + 76*A002496(3) and 1297=5+76*17.
+----+------+-----+------+---------------------------------------------+
|  n | f(n) |   p |    q |            f(p)=f(n)+q*f(n+1)               |
+----+------+-----+------+----------------------+----------------------+
|  1 |   2  |   3 |    3 | f(3)=f(1)+3*f(2)     |      17=2+3*5        |
|  2 |   5  |  11 |   76 | f(11)=f(2)+76*f(3)   |    1297=5+76*17      |
|  3 |  17  |  49 | 2432 | f(49)=f(3)+2432*f(4) |   90001=17+2432*37   |
|  4 |  37  | 113 | 9980 | f(113)=f(4)+9980*f(5)| 1008017=37+9980*101  |
|  5 | 101  |  55 |  748 | f(55)=f(5)+748*f(6)  |  147457=101+748*197  |
|  6 | 197  | 166 | 9420 | f(166)=f(6)+9420*f(7)| 2421137=197+9420*257 |
		

Links

• Michel Lagneau, 90 first pairs (p,q)

Crossrefs

Cf. A002496, A348598.

Programs

Maple

T:=array(1..30000):k:=0:
nn:=500000:
  for m from 1 to nn do:
   if isprime(m^2+1)
    then
     k:=k+1:T[k]:=m^2+1:
    else
   fi:
  od:
  for n from 1 to 32 do:
  ii:=0:r:=T[n]:q:=T[n+1]:
   for i from 1 to k while(ii=0) do:
     p:=T[i]:r1:=irem(p,q):
        if r1=r and p>q
         then
         ii:=1:x:=(T[i]-T[n])/T[n+1]:printf(`%d, `,i):
         printf(`%d, `,x):
         else
        fi:
       od:
      od: