A339007 -id:A339007 - cp's OEIS frontend

A339008 Least k such that p = k^2 + 1 and q = (k+2n)^2 + 1 are two consecutive prime numbers of the same form with q - p square.

24, 6, 312984, 16896, 240, 734994, 10640, 10360, 1946016, 2550, 13189264, 72996, 416520, 2184336, 1584360, 202484, 232696, 1700150, 2394456, 375360, 8736504, 9237866, 53629744, 360126, 87000, 574339974, 82404216, 23237760, 1249877496, 826650, 127119344, 1527720

Offset: 1

Michel Lagneau, Nov 18 2020

nonn

4*n*(k + n) is a square. If n is a square, then k + n is also a square.
If n is prime, then n divides k.
a(n) = A339007(n) for n = 1, 2, 3, 4, 6, 7 and 9.

			a(1) = 24 because 24^2 + 1 = 577, (24 + 2)^2 + 1 = 677. The numbers 577 and 677 are two consecutive primes of the form m^2+1, and 677 - 577 = 10^2 is a square. The other values m such that p = m^2 + 1 and q = (m+2)^2 + 1 are consecutive primes with q - p square are 11024, 133224, 156024, 342224, 416024, ...
a(2) = 6 because 6^2 + 1 = 37, (6 + 4)^2 + 1 = 101. The numbers 37 and 101 are two consecutive primes of the form m^2+1, and 101 - 37 = 8^2 is a square. The other values m such that p = m^2 + 1 and q = (m+4)^2 + 1 are consecutive primes with q - p square are 16, 126, 1350, 1456, 1566, 2310, 5200, ...

Chai Wah Wu, Table of n, a(n) for n = 1..60

Cf. A002496, A096012, A193558, A206328, A216330, A339007.

for n from 1 to 25 do:
ii:=0:n1:=0:q:=2:
  for k from 2 by 2 to 10^9 while(ii=0) do:
    p:=k^2+1:
   if isprime(p)
    then
     x:=p-q:q:=p:z:=sqrt(x):
      if z=floor(z) and k-n1=2*n
       then
        ii:=1:printf(`%d %d \n`,n,n1):
         else
         n1:=k:
       fi:
    fi:
  od:
od:

consecutive(p, q) = {forprime(r = nextprime(p+1), precprime(q-1), if (isprime(r) && issquare(r-1), return(0));); return(1);}
a(n) = my(k=1); while (!(isprime(p=k^2+1) && isprime(q=(k+2*n)^2 + 1) && issquare(q-p) && consecutive(p, q)), k++); k; \\ Michel Marcus, Nov 30 2020

a(26)-a(32) from Chai Wah Wu, Dec 06 2020

A351141 Pairs of primes (p,q) = (A002496(m), A002496(m+1)) such that q-p is a power r of the product of its prime factors for some m.

Original entry on oeis.org

37, 101, 577, 677, 15877, 16901, 57601, 62501, 33988901, 34035557, 113209601, 113507717, 121528577, 121572677, 345960001, 346332101, 635040001, 635544101, 7821633601, 7823402501, 17748634177, 17749167077, 24343488577, 24344112677, 97958984257, 97962740101

Offset: 1

Michel Lagneau, Feb 02 2022

nonn

Subsequence of A002496.
The corresponding sequence of numbers q - p is a subsequence of A076292.
Conjecture: the sequence is infinite.
The corresponding powers r are given by the sequence b(n) = 6, 2, 10, 2, 6, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ... It seems that b(n) = 2 for n > 5.

			The pair (257, 401) = (16^2+1, 20^2+1) is not in the sequence because 401 - 257 = 144 = 2^4*3^2.
The pair (577, 677) = (24^2+1, 26^2+1) is in the sequence because 577 - 677 = 100 = 2^2*5^2.
The pair (33988901, 34035557) = (5830^2+1, 5834^2+1) is in the sequence because 33988901 - 34035557 = 46656 = 2^6*3^6.

Cf. A002496, A076292, A096012, A193558, A216330, A339007.

with(numtheory):
T:=array(1..26):nn:=350000:q:=5:j:=1:
for n from 4 by 2 to nn do:
  p:=n^2+1:
   if type(p, prime)=true
    then
     x:=p-q:r:=q:q:=p:
     u:=factorset(x):n0:=nops(u):ii:=0:d:=product(u[i],i=1..n0):
      for k from 2 to 20 while(ii=0) do:
       if d^k=x
        then ii=1:T[j]:=r:T[j+1]:=q:j:=j+2:
        else
       fi:
      od:
   fi:
od:
print(T):

lista(nn) = my(lastp=2); forprime(p=nextprime(lastp+1), nn, if (issquare(p-1), if (ispowerful(p-lastp), my(f=factor(p-lastp)[,2]); if (vecmin(f) == vecmax(f), print1(lastp, ", ", p, ", "));); lastp = p;);); \\ Michel Marcus, Feb 03 2022

cp's OEIS Frontend

A339008 Least k such that p = k^2 + 1 and q = (k+2n)^2 + 1 are two consecutive prime numbers of the same form with q - p square.

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