A089120 -id:A089120 - cp's OEIS frontend

A252096 Largest prime divisor of n^2+1 - smallest prime divisor of n^2+1.

0, 0, 3, 0, 11, 0, 3, 8, 39, 0, 59, 24, 15, 0, 111, 0, 27, 8, 179, 0, 15, 92, 51, 0, 311, 0, 71, 152, 419, 36, 35, 36, 107, 76, 611, 0, 135, 12, 759, 0, 27, 348, 35, 136, 1011, 44, 15, 456, 1199, 20, 1299, 536, 279, 0, 87, 0, 11, 668, 1739, 264, 1859, 764, 395

Offset: 1

Michel Lagneau, Dec 14 2014

			a(5)= 11 because 5^2+1 = 2*13 and 13-2 = 11.

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

Cf. A002522 (n^2+1), A005574 (n^2+1 is prime).

Cf. A014442 (largest prime factor of n2+1), A089120 (smallest prime factor).

with(numtheory):
a:= n-> (f-> max(f[])-min(f[]))(factorset(n^2+1)):
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 07 2015

f[n_]:=Transpose[FactorInteger[n^2+1]][[1]]; Table[Last[f[n]-First[f[n]]], {n, 200}]

a(n) = {my(f = factor(n^2+1)); f[#f~,1] - f[1,1];} \\ Michel Marcus, Dec 15 2014

a(n) = A014442(n) - A089120(n).

a(A005574(n)) = 0. - Michel Marcus, Dec 15 2014

A252890 Number of times the greatest prime factor of n^2 + 1 is a factor in all numbers <= n.

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 4, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 6, 1, 1, 4, 1, 3, 1, 1, 2, 7, 1, 1, 2, 1, 1, 1, 1, 2, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 6, 1, 3, 4, 1, 2, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1

Offset: 1

Michel Lagneau, Dec 24 2014

nonn

The greatest prime factor is counted with multiplicity (see the example).

a(n)=1 iff n^2 + 1 is prime.

			a(7)=4 because 7^2 + 1 = 50 and 5 is 4 times a factor:
2^2+1 = 5;
3^2+1 = 10 = 2*5;
7^2+1 = 50 = 2*5*5 (two times).

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

Cf. A089120, A014442, A078897.

with(numtheory): with(padic,ordp):
f:= proc(n) local p ,q, n0;
  q:=factorset(n^2+1);n0:=nops(q);p:= q[n0];
  add(ordp(k^2+1, p), k=1..n);
end proc:
seq(f(n), n=1.. 100);
# Using code from Robert Israel adapted for this sequence. See A078897.

A348594 Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.

Original entry on oeis.org

8, 50, 1250, 1800, 2450, 9800, 14450, 20000, 24200, 101250, 105800, 135200, 162450, 168200, 204800, 304200, 336200, 451250, 480200, 490050, 530450, 696200, 924800, 966050, 1008200, 1125000, 1155200, 1428050, 1805000, 2332800, 2420000, 2576450, 2761250, 2832200

Offset: 1

Michel Lagneau, Jan 26 2022

nonn

Subsequence of A085722.
The corresponding pairs (p, q) of the sequence are (5, 13), (41, 61), (1201, 1301), (1741, 1861), (2381, 2521), (9661, 9941), (14281, 14621), (19801, 20201), (23981, 24421), (100801, 101701), ...
Property:
a(n) = 2* A109306(n)^2 and a(n) == 0 (mod 50) for n > 1. Proof:
From the relations:
(1)        m^2 + 1 = p*q
(2)        (p + q)/2 = m + 1
We obtain:
(3)        p = m + 1 - sqrt(8*m)/2
(4)        q = m + 1 + sqrt(8*m)/2
with m = 2*k^2 we obtain:
(5)        p = k^2 + (k-1)^2
(6)        q = k^2 + (k+1)^2
For n > 1, A109306(n) == 0 (mod 5) => 2*A109306(n)^2 == 0 (mod 50).

			50 = 2*5^2 is in the sequence because 50^2 + 1 = 41*61 with 50 = (41 + 61)/2 - 1.

Cf. A014442, A085722, A089120, A109306, A144255.

with(numtheory):nn:=250:printf(`%d, `,8):
for k from 0 to nn do:
n:=50*k^2:d:=factorset(n^2+1):
  if bigomega(n^2+1)=2 and (d[1]+d[2])/2 - 1 = n
   then
    printf(`%d, `,n):
    else
  fi:
od:

q[n_] := Module[{f = FactorInteger[n^2 + 1]}, f[[;; , 2]] == {1, 1} && f[[1, 1]] + f[[2, 1]] == 2*n + 2]; Select[Range[3*10^5], q] (* Amiram Eldar, Jan 26 2022 *)

isok(m) = my(x); if (bigomega(x=m^2+1)==2, my(f=factor(x)); (f[1,1]+f[2,1] == 2*(m+1))); \\ Michel Marcus, Jan 26 2022

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A252096 Largest prime divisor of n^2+1 - smallest prime divisor of n^2+1.

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A252890 Number of times the greatest prime factor of n^2 + 1 is a factor in all numbers <= n.

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Maple

A348594 Numbers m such that m^2 + 1 = p*q with p, q primes and m = (p + q)/2 - 1.

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Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI