A002496 -id:A002496 - cp's OEIS frontend

A054755 Odd powers of primes of the form q = x^2 + 1 (A002496).

2, 5, 8, 17, 32, 37, 101, 125, 128, 197, 257, 401, 512, 577, 677, 1297, 1601, 2048, 2917, 3125, 3137, 4357, 4913, 5477, 7057, 8101, 8192, 8837, 12101, 13457, 14401, 15377, 15877, 16901, 17957, 21317, 22501, 24337, 25601, 28901, 30977, 32401

Offset: 1

Labos Elemer, Apr 25 2000

nonn

A002496 is a subset; the odd power exponent is 1.
From Bernard Schott, Mar 16 2019: (Start)
The terms of this sequence are exactly the integers with only one prime factor and whose Euler's totient is square, so this sequence is a subsequence of A039770. The primitive terms of this sequence are the primes of the form q = x^2 + 1, which are exactly in A002496.
Additionally, the terms of this sequence also have a square cototient, so this sequence is a subsequence of A063752 and A054754.
If q prime = x^2 + 1, phi(q) = x^2, phi(q^(2k+1)) = (x*q^k)^2, and cototient(q) = 1^2, cototient(q^(2k+1)) = (q^k)^2. (End)

			a(20) = 3125 = 5^5, q = 5 = 4^2+1 and Phi(3125) = 2500 = 50^2, cototient(3125) = 3125 - Phi(3125) = 625 = 25^2.

David A. Corneth, Table of n, a(n) for n = 1..18864 (terms <= 10^11)
Bernard Schott, Subfamilies and subsequences

Cf. A000010, A051953, A039770, A063752, A054754, A334745 (with 2 distinct prime factors), A306908 (with 3 distinct prime factors).

Subsequences: A002496 (primitive primes: m^2+1), A004171 (2^(2k+1)), A013710 (5^(2k+1)), A013722 (17^(2k+1)), A262786 (37^(2k+1)).

Select[Range[10^5], And[PrimeNu@ # == 1, IntegerQ@ Sqrt@ EulerPhi@ #] &] (* Michael De Vlieger, Mar 31 2019 *)

isok(m) = (omega(m)==1) && issquare(eulerphi(m)); \\ Michel Marcus, Mar 16 2019

upto(n) = {my(res = List([2]), q); forstep(i = 2, sqrtint(n), 2, if(isprime(i^2 + 1), listput(res, i^2 + 1) ) ); q = #res; forstep(i = 3, logint(n, 2), 2, for(j = 1, q, c = res[j]^i; if(c <= n, listput(res, c) , next(2) ) ) ); listsort(res); res } \\ David A. Corneth, Mar 17 2019

A000010(a(n)) = (q^(2k))*(q-1) and A051953(a(n)) = q^(2k), where q = 1 + x^2 and is prime.

A211175 Triangle read by rows: row n gives, in increasing order, the prime divisors of all the composites of the form k^2 + 1 between the two primes A002496(n) and A002496(n+1).

Original entry on oeis.org

2, 5, 2, 13, 2, 5, 13, 41, 2, 5, 17, 29, 61, 2, 113, 2, 5, 13, 29, 181, 2, 5, 13, 17, 53, 97, 2, 313, 2, 5, 13, 17, 37, 41, 53, 73, 89, 109, 157, 421, 613, 2, 5, 17, 137, 761, 2, 5, 13, 17, 29, 37, 41, 61, 73, 149, 281, 353, 461, 541, 1013, 1201, 1301, 2, 17

Offset: 2

Michel Lagneau, Feb 01 2013

A variety of conjecturally infinite subsequences and starting with {2, 5, ...} can be shown in the graph of the sequence. If the number of primes of the form n^2 + 1 is finite, then the last subsubsequence of the graph abruptly becomes A002144(n) union {2} (odd Pythagorean primes with the number 2). In this case, the discontinued forms of the graph disappear. But this case is highly improbable.

			The irregular triangle of divisors is:
[2, 5]
[2, 13]
[2, 5, 13, 41]
[2, 5, 17, 29, 61]
[2, 113]
[2, 5, 13, 17, 53, 97]
...
Row 1 is empty because there are no numbers of the form k^2 + 1 between A002496(1) = 2 and A002496(2) = 5.
row 2 = [2, 5] lists divisors of 3^2 + 1 between the primes A002496(2) and A002496(3);
row 3 = [2, 13] lists divisors of 5^2 + 1 between the primes A002496(3) and A002496(4);
row 4 = [2, 5, 13, 41] lists divisors of 7^2 + 1, 8^2 + 1, 9^2 + 1 between the primes A002496(4) and A002496(5).

Michel Lagneau, Rows n = 2..557 of irregular triangle, flattened

Cf. A002144, A002496, A002522, A134406, A181413, A206400.

with(numtheory) :lst:={}: for n from 2 to 150 do:p:=n^2+1:x:=factorset(p):lst:=lst union x:if type(p,prime)=true then print(lst minus {p}):lst:={}:else fi:od:

A238139 a(n) is the smallest prime divisor (not yet in the sequence) of all composite numbers of the form m^2+1 between the primes A002496(n) and A002496(n+1), or 0 if there is no such prime.

Original entry on oeis.org

0, 2, 13, 5, 17, 113, 29, 53, 313, 37, 137, 41, 89, 241, 61, 97, 233, 101, 73, 193, 557, 229, 601, 157, 8581, 109, 337, 293, 4993, 181, 14621, 433, 197, 149, 21013, 509, 277, 281, 521, 11329, 257, 173, 1321, 6917, 373, 389, 3037, 821, 7109, 353, 773, 397, 457

Offset: 1

Michel Lagneau, Feb 18 2014

nonn

By convention, a(1) = 0 because there are no composite number of the form m^2+1 between A002496(1)=2 and A002496(2)=5.
a(n) = 0 when all divisors of the numbers of the form m^2+1 between the primes A002496(n) and A002496(n+1) already exist in the sequence.
Note that a(n) = 0 for n = 1, 62, 149, 257, 281, 286,...(see A238138).

			a(7) = 29 because the composites of the form m^2+1 between the two primes A002496(7)= 16^2+1 = 257 and A002496(8)= 20^2+1=401 are:
17^2+1= 2*5*29;
18^2+1 = 5*5*13;
19^2+1=2*181 and the smallest prime divisor not yet in the sequence is 29 because 2, 5 and 13 are already in the sequence.

Michel Lagneau, Table of n, a(n) for n = 1..5000

Cf. A002522, A002496, A234759, A238138.

with(numtheory):lst:={}: lst2:={}:T:=array(1..2000):kk:=1:k:=0:for n from 2 by 2 to 500 do: p:=n^2+1:if type(p, prime)=true then k:=k+1:T[k]:=p:else fi:od:for i from 1 to k-1 do:lst1:={}:a:=sqrt(T[i]-1):b:=sqrt(T[i+1]-1):for j from a+1 to b-1 do:y:=factorset(j^2+1):lst1:=lst1 union y:od:lst1:=lst1 minus lst: if lst1<>{} then kk:=kk+1: printf(`%d, `,lst1[1]):lst:=lst union {lst1[1]}:else kk:=kk+1: printf(`%d, `,0):fi:od:

A211188 a(n) is the number of distinct prime divisors among all the composites of the form k^2 + 1 between the two primes A002496(n) and A002496(n+1).

Original entry on oeis.org

0, 2, 2, 4, 5, 2, 5, 6, 2, 13, 5, 17, 3, 12, 11, 15, 9, 6, 21, 11, 6, 7, 3, 7, 7, 18, 7, 10, 6, 14, 11, 7, 6, 29, 2, 6, 22, 10, 10, 6, 16, 12, 6, 5, 11, 15, 6, 24, 12, 13, 19, 21, 15, 45, 3, 17, 6, 11, 24, 15, 9, 9, 6, 28, 3, 7, 7, 26, 10, 55, 14, 21, 24, 8

Offset: 1

Michel Lagneau, Feb 03 2013

nonn

a(1)=0; for n > 1, a(n) = number of elements of each row in A211175(n).

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A002144, A002496, A002522, A134406, A181413, A206400, A211175.

with(numtheory) :lst:={}: for n from 2 to 600 do:p:=n^2+1:x:=factorset(p):lst:=lst union x:if type(p,prime)=true then m:=nops(lst minus {p}): printf(`%d, `,m):lst:={}:else fi:od:

A348598 Least prime p of the form k^2+1 such that p == A002496(n) (mod A002496(n+1)) with p>A002496(n), or 0 if no such p exists.

Original entry on oeis.org

17, 1297, 90001, 1008017, 147457, 2421137, 15952037, 1378277, 7203857, 107122501, 164968337, 34503877, 38688401, 4851958337, 1075577617, 197121601, 1044582401, 315559697, 70924211857, 730296577, 20705483237, 15103426817, 197740302401, 4587352901, 155964965777

Offset: 1

Michel Lagneau, Mar 20 2022

nonn

a(n) == 1, 5 (mod 16).
Conjecture: Consider the smallest prime p of the form k^2+1 such that p is congruent to A002496(n) modulo q, q prime of the form m^2+1 > A002496(n). Then q = A002496(n+1).
Corollary: For any pair (A002496(n), A002496(n+1)), there exist two integers m, k such that A002496(m) = A002496(n) + k*A002496(n+1), m>n+1 and n=1,2,3,...
Examples (see A352582):
  A002496(3)   = A002496(1) +    3*A002496(2),
  A002496(11)  = A002496(2) +   76*A002496(3),
  A002496(49)  = A002496(3) + 2432*A002496(4),
  A002496(113) = A002496(4) + 9980*A002496(5).

			a(2) = 1297 because 1297 == A002496(2) (mod A002496(3)) => 1297 == 5 (mod 17).

Michel Lagneau, a(n),n=1..90

Cf. A002496, A352582.

with(numtheory):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: printf(`%d, `,p)
         else
        fi:
       od:
      od:

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.

Original entry on oeis.org

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

Michel Lagneau, Mar 21 2022

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).

			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 |

Michel Lagneau, 90 first pairs (p,q)

Cf. A002496, A348598.

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:

A211189 Number of prime divisors formed by {2} and the consecutive Pythagorean primes for all the composites k^2 + 1 between the two primes A002496(n) and A002496(n+1).

Original entry on oeis.org

0, 2, 1, 3, 2, 1, 3, 4, 1, 4, 2, 7, 1, 4, 7, 6, 4, 2, 6, 4, 2, 4, 1, 2, 2, 4, 4, 3, 2, 5, 4, 3, 2, 10, 1, 2, 7, 4, 2, 3, 5, 4, 2, 2, 4, 5, 3, 4, 6, 5, 4, 7, 4, 7, 1, 5, 3, 2, 7, 5, 3, 4, 2, 8, 1, 2, 4, 7, 2, 9, 5, 4, 12, 2, 4, 6, 10, 1, 4, 1, 2, 9, 2, 5, 2, 4

Offset: 1

Michel Lagneau, Feb 03 2013

a(1)=0; for n > 1, a(n) = number of consecutive elements of the form {2, A002144(1), A002144(2), ...} of each row in A211175(n).
The immediate objective of this sequence is to show that it is difficult to obtain a large range of consecutive Pythagorean primes from the decomposition of n^2 + 1, because the growth of a(n) is very slow, for example a(351) = 29, a(22215) = 34, ...
These considerations confirm the opinion of the truthfulness of the conjecture about an infinity of primes of the form n^2 + 1. This sequence gives the length of a variety of conjecturally infinite subsequences of consecutive primes starting with {2, 5, ...}. If the number of primes of the form n^2 + 1 were finite, there should exist a last prime p such that this sequence stops abruptly from p because the length of A002144(n) is infinite. In this case, we should observe a contradictory behavior of this sequence between the stability of the slow growth of a(n) and the discontinuity from the prime p. But this case is highly improbable.

			a(8) = 4 because the set formed by the union of the prime divisors of all the numbers k^2+1 between the primes A002496(8) = 401 and A002496(9) = 577 are {2, 5, 13, 17, 53, 97} and the subset {2} union {5, 13, 17} contains 4 consecutive elements, hence 4 is in the sequence.

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A002144, A002496, A002522, A134406, A181413, A206400, A211175, A211188.

with(numtheory) :lst:={2}:lst1:={}:
     for k from 1 to 1000 do: q:=4*k+1:
          if type(q,prime)=true then
          lst:=lst union {q}:else fi:
     od:
  L:=subsop(lst):
        for n from 2 to 1000 do:p:=n^2+1:x:=factorset(p):lst1:=lst1 union x:
          if type(p,prime)=true then
          z:=lst1 minus {p}: n1:=nops(z): jj:=0: d0:=0:
            for j from 1 to n1 while(jj=0) do:
               d:=nops(z intersect L[1..j]): if d>d0 then
              d0:=d:
              else
              jj:=1:fi:
            od:
            lst1:={}: printf(`%d, `,d0):
           fi:
          od:

A247592 Numbers n such that A002496(n) mod A002496(n-1) is a perfect square.

Original entry on oeis.org

2, 8, 10, 25, 42, 147, 160, 169, 238, 260, 491, 544, 869, 890, 923, 1140, 1337, 1386, 1465, 1643, 1927, 3371, 4614, 5038, 5086, 5225, 5832, 5909, 5995, 7118, 7157, 8540, 9859, 12543, 13505, 13795, 13841, 14211, 15347, 17079, 17263, 18643, 20211, 21184, 21245

Offset: 1

Michel Lagneau, Sep 20 2014

nonn

A002496 : primes of form n^2+1.
The prime numbers of the sequence are 2, 491, 3371, 9859, 13841,...
The corresponding squares A002496(n) mod A002496 (n-1) are : {1, 144, 100, 1024, 4900, 10816, 11664, 12544,...} = {1} union {A216330} minus {64}.

			a(3)=10 because A002496(10) mod A002496(9)= 677 mod 577 = 10^2.

Chai Wah Wu, Table of n, a(n) for n = 1..200

Cf. A002496, A193558, A216330.

with(numtheory):nn:=360000:T:=array(1..nn):kk:=0:
for n from 1 to nn do:
  if type(n^2+1,prime)=true then
   kk:=kk+1:T[kk]:=n^2+1:
   else
   fi:
od:
    for m from 1 to kk-1 do:
     r:=irem(T[m+1],T[m]):z:=sqrt(r):
      if z=floor(z)
       then printf(`%d, `, m+1):
       else
      fi:
    od:

lst={};lst1={};nn=400000;Do[If[PrimeQ[n^2+1],AppendTo[lst,n^2+1]],{n,1,nn}];nn1:=Length[lst];
Do[If[IntegerQ[Sqrt[Mod[lst[[m]],lst[[m-1]]]]],AppendTo[lst1,m]],{m,2,nn1}];lst1

from gmpy2 import t_mod, is_square, is_prime
A247592_list, A002496_list, m, c = [], [2], 2, 2
for n in range(1, 10**7):
    m += 2*n+1
    if is_prime(m):
        if is_square(t_mod(m, A002496_list[-1])):
            A247592_list.append(c)
        A002496_list.append(m)
        c += 1 # Chai Wah Wu, Sep 20 2014

A327741 Terms of A002496 that are the average of two distinct terms of A002496.

Original entry on oeis.org

101, 21317, 24337, 462401, 1073297, 1123601, 1263377, 1887877, 1943237, 2446097, 2604997, 2890001, 3422501, 4202501, 4343057, 5354597, 6330257, 7862417, 8386817, 8410001, 9156677, 10536517, 10719077, 11383877, 12068677, 12110401, 12503297, 16273157, 18062501, 19219457, 21771557, 22429697

Offset: 1

J. M. Bergot and Robert Israel, Sep 23 2019

nonn

Primes of the form x^2+1 such that 2*x^2=y^2+z^2 where y^2+1 and z^2+1 are primes.
Some terms of the sequence are the average of more than one pair of terms of A002496. E.g., 2890001 = (115601 + 5664401)/2 = (2016401 + 3763601)/2, while 5354597 = (42437 + 10666757)/2 = (1136357 + 9572837)/2 = (1552517 + 9156677)/2.
Primes of the form u^2*(s^2 + t^2)^2 + 1 where u^2*(s^2 + 2*s*t - t^2)^2 + 1 and u^2*(-s^2 + 2*s*t + t^2)^2 + 1 are prime, (sqrt(2) - 1)*s < t < s. The generalized Bunyakovsky conjecture implies there are infinitely many terms for each such pair (s,t).

			a(3)=24337 is in the sequence because 24337=(7057+41617)/2 with 7057, 24337 and 41617 all terms of A002496, i.e., they are primes and 7057=84^2+1, 24337=156^2+1 and 41617=204^2+1.

Robert Israel, Table of n, a(n) for n = 1..1055

Cf. A002496.

N:= 10^8: # to get terms <= N
P:= select(isprime, [seq(x^2+1, x=2..floor(sqrt(N-1)),2)]):
nP:= nops(P):
R:= NULL:
for i from 1 to nP do
  x:= P[i];
  for j from 1 to i-1 do
    z:= 2*x-P[j];
    if issqr(z-1) and isprime(z) then R:= R, x; break fi
  od
od:
R;

A260576 Least k such that the product of the first n primes of the form m^2+1 (A002496) divides k^2+1.

Original entry on oeis.org

1, 3, 13, 327, 36673, 950117, 801495893, 5896798453, 760999599793, 3828797295053127, 520910599208391893, 2418812764637100821917, 793123421312468129647727, 6936392582189824489589830053, 31170731920863007986026123435697, 5284787778858696936313058199017107

Offset: 1

Michel Lagneau, Jul 29 2015

nonn

Conjecture: the sequence is infinite.
From Robert Israel, Jun 23 2025: (Start)
Consider any finite set of primes p(i) = m(i)^2 + 1, i = 1 .. n.
Then k^2 + 1 == 0 (mod p(i)) if k == m(i) (mod p(i)).
By the Chinese Remainder Theorem, there exists k such that k == m(i) (mod p(i))
for i = 1 .. n.  Thus the conjecture is true, and all terms a(n) exist.
(End)
Let b(n) = Product_{k=1..n} A002496(k): 2, 10, 170, 6290, 635290, ...
b(1) divides k^2+1 for k = 1, 3, 5, ...
b(2) divides k^2+1 for k = 3, 7, 13, 17, 23, 27, 33, 37, 43, 47, 53, 57, 63, 67, 73, 77, 83, ...
b(3) divides k^2+1 for k = 13, 47, 123, 157, 183, 217, 293, 327, 353, 387, 463, 497, 523, ...
b(4) divides k^2+1 for k = 327, 1067, 2707, 2843, 3447, 3583, 5223, 5963, 6617, 7357, 8997, 9133, 9737, 9873, ...
b(5) divides k^2+1 for k = 36673, 38067, 66347, 141087, 217443, 240087, 292183, 314827, 320463, ...

Cf. A002496, A005574.

with(numtheory):lst:={2}:nn:=100:
for i from 1 to nn do:
   p:=i^2+1:
   if isprime(p)
   then
   lst:=lst union {p}:
   else fi:
od:
  pr:=1:
  for n from 1 to 7 do:
  pr:=pr*lst[n]:ii:=0:
   for j from 1 to 10^9 while(ii=0) do:
   if irem(j^2+1,pr)=0
   then
   ii:=1:
   printf("%d %d \n",n,j):
   fi:
   od:
  od:

a(8)-a(17) from Hiroaki Yamanouchi, Aug 15 2015

cp's OEIS Frontend

A054755 Odd powers of primes of the form q = x^2 + 1 (A002496).

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A211175 Triangle read by rows: row n gives, in increasing order, the prime divisors of all the composites of the form k^2 + 1 between the two primes A002496(n) and A002496(n+1).

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A238139 a(n) is the smallest prime divisor (not yet in the sequence) of all composite numbers of the form m^2+1 between the primes A002496(n) and A002496(n+1), or 0 if there is no such prime.

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A211188 a(n) is the number of distinct prime divisors among all the composites of the form k^2 + 1 between the two primes A002496(n) and A002496(n+1).

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A348598 Least prime p of the form k^2+1 such that p == A002496(n) (mod A002496(n+1)) with p>A002496(n), or 0 if no such p exists.

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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.

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A211189 Number of prime divisors formed by {2} and the consecutive Pythagorean primes for all the composites k^2 + 1 between the two primes A002496(n) and A002496(n+1).

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A247592 Numbers n such that A002496(n) mod A002496(n-1) is a perfect square.

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