A128428 -id:A128428 - cp's OEIS frontend

A193432 Number of divisors of n^2 + 1.

1, 2, 2, 4, 2, 4, 2, 6, 4, 4, 2, 4, 4, 8, 2, 4, 2, 8, 6, 4, 2, 8, 4, 8, 2, 4, 2, 8, 4, 4, 4, 8, 6, 8, 4, 4, 2, 8, 6, 4, 2, 6, 4, 12, 4, 4, 4, 16, 4, 4, 4, 4, 4, 8, 2, 8, 2, 16, 4, 4, 4, 4, 4, 8, 4, 4, 2, 8, 8, 4, 6, 4, 8, 16, 2, 8, 4, 8, 4, 4, 4, 8, 6, 16, 2

Offset: 0

Michel Lagneau, Jul 28 2011

nonn

			a(7) = 6 because 7^2 + 1 = 50 and the 6 divisors are {1, 2, 5, 10, 25, 50}.

Seiichi Manyama, Table of n, a(n) for n = 0..10000
Anton Shakov, Polynomials in Z[x] whose divisors are enumerated by SL_2(N_0), arXiv:2405.03552 [math.NT], 2024.

Cf. A000005, A002522, A128428, A193330, A193433.

with(numtheory):for n from 0 to 110 do:n1:=nops(divisors(n^2+1)):s:=0:for m from 1 to n1 do: s:=s+1:od: printf(`%d, `, s):od:

Array[DivisorSigma[0, #^2 + 1] &, 85, 0] (* Michael De Vlieger, Mar 17 2018 *)

a(n) = numdiv(n^2+1); \\ Michel Marcus, Mar 16 2018

from sympy import divisor_count
def A193432(n): return divisor_count(n**2+1) # Chai Wah Wu, Apr 17 2025

a(n) = A000005(A002522(n)). - Michel Marcus, Mar 16 2018

A180278 Smallest nonnegative integer k such that k^2 + 1 has exactly n distinct prime factors.

Original entry on oeis.org

0, 1, 3, 13, 47, 447, 2163, 24263, 241727, 2923783, 16485763, 169053487, 4535472963, 36316463227, 879728844873, 4476534430363, 119919330795347, 1374445897718223, 106298577886531087

Offset: 0

Michel Lagneau, Jan 17 2011

			a(2) = 3 because the 2 distinct prime factors of 3^2 + 1 are {2, 5};
a(10) = 16485763 because the 10 distinct prime factors of 16485763^2 + 1 are {2, 5, 13, 17, 29, 37, 41, 73, 149, 257}.

Cf. A001221, A002144, A002522, A128428, A164511, A185952, A219017, A219108, A257366.

a[n_] := a[n] = Module[{k = 1}, If[n == 0, Return[0]]; Monitor[While[PrimeNu[k^2 + 1] != n, k++]; k, {n, k}]]; Table[a[n], {n, 0, 8}] (* Robert P. P. McKone, Sep 13 2023 *)

a(n)=for(k=0, oo, if(omega(k^2+1) == n, return(k))) \\ Andrew Howroyd, Sep 12 2023

from itertools import count
from sympy import factorint
def A180278(n):
    return next(k for k in count() if len(factorint(k**2+1)) == n) # Pontus von Brömssen, Sep 12 2023

a(n) >= sqrt(A185952(n)-1). - Charles R Greathouse IV, Feb 17 2015

a(n) <= A164511(n). - Daniel Suteu, Feb 20 2023

a(9), a(10) and example corrected; a(11) added by Donovan Johnson, Aug 27 2012
a(12) from Giovanni Resta, May 10 2017
a(13)-a(17) from Daniel Suteu, Feb 20 2023
Name clarified and incorrect programs removed by Pontus von Brömssen, Sep 12 2023
a(18) from Max Alekseyev, Feb 24 2024

A339461 Number of Fibonacci divisors of n^2 + 1.

Original entry on oeis.org

1, 2, 2, 3, 1, 3, 1, 3, 3, 2, 1, 2, 2, 4, 1, 2, 1, 3, 3, 2, 1, 4, 2, 3, 1, 2, 1, 3, 2, 2, 1, 3, 2, 3, 3, 2, 1, 3, 2, 2, 1, 2, 2, 3, 2, 2, 1, 5, 2, 2, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 2, 2, 2, 3, 1, 2, 1, 3, 2, 2, 3, 2, 2, 4, 1, 2, 1, 3, 2, 2, 1, 3, 2, 4, 1, 2, 2

Offset: 0

Michel Lagneau, Dec 06 2020

nonn

			a(13) = 4 because the divisors of 13^2 + 1 = 170 are {1, 2, 5, 10, 17, 34, 85, 170} with 4 Fibonacci divisors: 1, 2, 5 and 34.

Michel Marcus, Table of n, a(n) for n = 0..10000

Cf. A000045, A002522, A005086, A005478, A005574, A128428, A338762, A338794.

with(numtheory):with(combinat,fibonacci):nn:=100:F:={}:
for k from 1 to nn do:
  F:=F union {fibonacci(k)}:
od:
   for n from 0 to 90 do:
    f:=n^2+1:d:=divisors(f):
    lst:= F intersect d: n1:=nops(lst):printf(`%d, `,n1):
   od:

Array[DivisorSum[#^2 + 1, 1 &, Or @@ Map[IntegerQ@ Sqrt[#] &, 5 #^2 + 4 {-1, 1}] &] &, 105, 0] (* Michael De Vlieger, Dec 07 2020 *)

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || issquare(k-8);
a(n) = sumdiv(n^2+1, d, isfib(d)); \\ Michel Marcus, Dec 06 2020

a(A005574(n)) = 1 for n > 2.

a(n) = A005086(A002522(n)). - Michel Marcus, Dec 06 2020

A344869 Number of distinct prime factors of n^n+1.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 3, 3, 3, 4, 3, 6, 4, 5, 5, 6, 2, 3, 7, 5, 3, 6, 4, 8, 6, 7, 5, 11, 6, 7, 10, 7, 4, 11, 6, 13, 5, 7, 7, 8, 9, 6, 10, 8, 8, 14, 8, 10, 6, 7, 10, 11, 5, 8, 14, 11, 7, 13, 13, 9, 12, 8, 7, 18, 4, 12, 8, 7, 7, 16, 9, 8, 12, 4, 8, 24, 7, 9, 14, 7, 5, 12, 6, 12, 8, 13, 10, 12, 10, 6, 23, 15, 6, 9, 11, 16, 3, 8, 17, 23, 7

Offset: 0

Seiichi Manyama, May 31 2021

nonn

Amiram Eldar, Table of n, a(n) for n = 0..148
Eric Weisstein's World of Mathematics, Sierpinski Number of the First Kind.

Cf. A001221, A014566, A085723, A128428, A344859, A344870.

Magma
```
[#PrimeDivisors(n^n+1): n in [0..100]];
```

a[0] = 1; a[n_] := PrimeNu[n^n + 1]; Array[a, 45, 0] (* Amiram Eldar, May 31 2021 *)

PARI
```
a(n) = omega(n^n+1);
```

a(n) = A001221(A014566(n)).

a(67)-a(79) from Jon E. Schoenfield, May 31 2021

a(80)-a(100) from Seiichi Manyama, May 31 2021

A067268 Numbers k such that k and k^2+1 have the same number of distinct prime factors.

Original entry on oeis.org

2, 4, 12, 15, 16, 18, 22, 28, 34, 35, 38, 39, 44, 45, 46, 48, 50, 51, 52, 58, 62, 65, 68, 69, 76, 80, 82, 85, 86, 88, 92, 95, 96, 100, 104, 105, 106, 108, 118, 132, 136, 138, 141, 144, 145, 152, 158, 159, 164, 166, 171, 174, 175, 178, 188, 194, 196, 201, 202, 205

Offset: 1

Benoit Cloitre, Feb 21 2002

			2 is a term since omega(2) = omega(2^2+1) = 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001221, A002522, A128428, A272044.

[k:k in [1.. 210 ]| #PrimeDivisors(k) eq #PrimeDivisors(k^2+1)]; // Marius A. Burtea, Feb 18 2020

Select[Range[250],PrimeNu[#]==PrimeNu[#^2+1]&] (* Harvey P. Dale, Feb 07 2019 *)

Numbers k such that omega(k) = omega(k^2+1).

A260258 T(n,k) is the array read by rows, n>0 and k=1..q (with q = number of prime distinct divisors of n^2+1) giving the number of occurrences of the k-th prime divisor of n^2+1 counted from the prime divisors of m^2+1 for m=1..n.

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 1, 1, 4, 3, 4, 2, 5, 1, 1, 6, 1, 5, 1, 7, 6, 2, 1, 8, 1, 1, 9, 7, 2, 8, 3, 10, 1, 1, 11, 4, 3, 9, 1, 12, 10, 1, 1, 13, 1, 1, 14, 11, 1, 12, 1, 15, 1, 4, 2, 16, 5, 2, 13, 2, 17, 14, 1, 6, 1, 18, 1, 1, 19, 15, 1, 16, 5, 20, 1, 1, 21, 3, 17, 1

Offset: 1

Michel Lagneau, Jul 21 2015

A002313(n) are the numbers such that T(n,k)>1 for all k=1..q.

T(2n-1,1)=n and T(m,1)=1 if m =1, 2, 4, 6, 10, 14, ... = A005574(n)(numbers n such that n^2 + 1 is prime). The length of row n is A128428(n).

			T(13,k) = [7,6,2] for k = 1,2,3 because 13^2+1 = 2*5*17 =>
The number of occurrences of the prime divisor 2 is 7: 1^2+1=2, 3^2+1=2*5, 5^2+1=2*13, 7^2+1=2*5^2, 9^2+1=2*41, 11^2+1=2*61 and 13^2+1=2*5*17;
The number of occurrences of the prime divisor 5 is 6: 2^2+1=5, 3^2+1=2*5, 7^2+1=2*5^2, 8^2+1=5*13, 12^2+1=5*29;
The number of occurrences of the prime divisor 17 is 2: 4^2+1=17 and 13^2+1=2*5*17.
The array begins:
  [1]
  [1]
  [2,2]
  [1]
  [3,1]
  [1]
  [4,3]
  [4,2]
  [5,2]
  [1]
  ...

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

Cf. A005574, A002312, A002313, A128428.

with(numtheory):lst:={2}:nn:=1000:T:=array(1..270,[0$270]):
for j from 1 to nn do:
   p:=4*j+1:
   if isprime(p)
   then
   lst:=lst union {p}:
   fi:
od:
   nn0:=nops(lst):
   for n from 1 to 60 do:
     q:=factorset(n^2+1):n0:=nops(q):
     for k from 1 to n0 do:
      for m from 1 to 270 do:
      if q[k]=lst[m] then T[m]:=T[m]+1:printf(`%d, `, T[m]):
      fi:
     od:
    od:
od:

A261609 Number of composite divisors of n^2+1.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 3, 1, 1, 0, 1, 1, 4, 0, 1, 0, 4, 3, 1, 0, 4, 1, 4, 0, 1, 0, 4, 1, 1, 1, 4, 3, 4, 1, 1, 0, 4, 3, 1, 0, 3, 1, 8, 1, 1, 1, 11, 1, 1, 1, 1, 1, 4, 0, 4, 0, 12, 1, 1, 1, 1, 1, 4, 1, 1, 0, 4, 5, 1, 3, 1, 4, 11, 0, 4, 1, 4, 1, 1, 1, 4, 3, 11, 0, 1, 1

Offset: 1

Michel Lagneau, Aug 26 2015

			a(7) = 3 because the composite divisors of 7^2+1 are 10, 25, 50.

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

Cf. A002522, A055212, A128428, A193330, A193432.

Table[ Count[ PrimeQ[ Divisors[n^2+1] ], False] - 1, {n, 1, 105} ]
Table[Count[Divisors[n^2+1],?CompositeQ],{n,100}] (* _Harvey P. Dale, Dec 16 2024 *)

a(n) = A055212(A002522(n)).

A284477 Pairs of integers (x, y), such that x^2 + 1 and y^2 + 1, 1 < y < x, have the same distinct prime factors.

Original entry on oeis.org

7, 3, 18, 8, 117, 43, 239, 5, 378, 132, 843, 377, 2207, 987, 2943, 73, 4443, 53, 4662, 1568, 6072, 5118, 8307, 743, 8708, 2112, 9872, 2738, 31561, 4929, 103682, 46368, 271443, 121393, 853932, 76378, 1021693, 91383, 3539232, 41218, 3699356, 473654

Offset: 1

Michel Lagneau, Mar 27 2017

nonn

The sequence appears to thin out quite abruptly; however, by solving the Diophantine equation x^2 + 1 = p (y^2 + 1) for a suitable prime p and selecting the solutions (x, y) for which p divides y^2 + 1, it is easy to generate larger pairs, such as (423222288438379883442890018716361, 66096216900526495715353522199871). - Giovanni Resta, Mar 27 2017
A very interesting property: the sequence contains a subsequence of pairs (Lucas numbers L(i), Fibonacci numbers F(i)) for i = 4, 6, 14, 16, 24, 36, ... These pairs are (L(4), F(4)), (L(6), F(6)), (L(14), F(14)), (L(16), F(16)), (L(24), F(24)), (L(26), F(26)), (L(34), F(34)), ... = (7, 3), (18, 8), (843, 377), (2207, 987), (103682, 46368), (271443, 121393), (12752043, 5702887), ... It seems that {i} = A090773(n) (numbers that are congruent to {4, 6} mod 10). - Michel Lagneau, Mar 28 2017
This is because L(i)^2+1 = 5*(F(i)^2+1) for even i, and 5 | F(i)^2 + 1 for i== 3,4,6,7 (mod 10).  In fact (L(i), F(i)) for i in A090773 are the solutions of the generalized Pell equation x^2 + 1 = 5 (y^2 + 1) for which 5 | y^2 + 1. - Robert Israel, Apr 10 2017

			The pair (843, 377) is in the sequence because the prime factors of 843^2 + 1 and 377^2 + 1 are 2, 5, 61 and 233.

Giovanni Resta, Table of n, a(n) for n = 1..68 (terms with x < 1.5*10^8)

Cf. A002496, A089122, A128428.

A:= NULL:
for x from 2 to 10^5 do
  P:= numtheory:-factorset(x^2+1);
  if not assigned(R[P]) then R[P]:= x
  else A:= A, op(map(t -> (x,t), [R[P]]));
       R[P]:= R[P],x
  fi
od:
A; # Robert Israel, Apr 10 2017

d[n_] := First /@ FactorInteger[n]; Flatten@ Reap[ Do[ dx = d[x^2+1]; Do[ If[ dx == d[y^2+1], Sow[{x, y}]], {y, x-1}], {x, 1, 10^4}]][[2, 1]]

upto(n) = {my(l = List(), res=List()); for(i=1, n, f = factor(i^2+1)[, 1]; listput(l, [f, i])); listsort(l); for(i=1, n-1, if(l[i][1]==l[i+1][1], listput(res, [l[i+1][2], l[i][2]]))); listsort(res); res} \\ David A. Corneth, Mar 28 2017

a(29)-a(34) from Giovanni Resta, Mar 27 2017

a(35)-a(42) from David A. Corneth, Mar 28 2017

A381605 Number of distinct prime divisors of n^3+1.

Original entry on oeis.org

1, 1, 2, 2, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 2, 2, 4, 2, 3, 3, 3, 2, 3, 3, 3, 3, 4, 2, 4, 3, 3, 3, 4, 3, 3, 3, 4, 4, 3, 3, 4, 2, 4, 3, 4, 3, 4, 3, 4, 4, 3, 3, 3, 4, 3, 4, 4, 2, 5, 2, 4, 4, 2, 4, 5, 3, 3, 4, 5, 2, 3, 2, 4, 3, 5, 3, 4, 2, 3, 3, 3, 4, 4, 4, 4, 3, 4, 4, 5, 3, 3, 3, 4, 4, 4, 3, 4, 3, 4, 2

Offset: 1

Joost de Winter, Mar 01 2025

nonn

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A001093, A001221, A128428, A366580.

function result = a(n)
result = numel(unique(factor(sym(n^3+1))));
end

f:= proc(n) nops(numtheory:-factorset(n^3+1)) end proc:
map(f, [$1..50]); # Robert Israel, Apr 03 2025

Mathematica
```
Table[PrimeNu[n^3 + 1], {n, 1, 30}]
```

a(n) = omega(n^3+1); \\ Michel Marcus, Mar 01 2025

a(n)=if(n%3==2, omega((n+1)/3^valuation(n+1,3)) + omega((n^2-n+1)/3^valuation(n^2-n+1,3)) + 1, omega(n+1) + omega(n^2-n+1)) \\ Charles R Greathouse IV, Mar 03 2025

a(n) = A001221(A001093(n)).

a(n) > 1 for n > 2. - Charles R Greathouse IV, Mar 03 2025

cp's OEIS Frontend

A193432 Number of divisors of n^2 + 1.

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A180278 Smallest nonnegative integer k such that k^2 + 1 has exactly n distinct prime factors.

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A339461 Number of Fibonacci divisors of n^2 + 1.

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A344869 Number of distinct prime factors of n^n+1.

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Magma

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A067268 Numbers k such that k and k^2+1 have the same number of distinct prime factors.

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Magma

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A260258 T(n,k) is the array read by rows, n>0 and k=1..q (with q = number of prime distinct divisors of n^2+1) giving the number of occurrences of the k-th prime divisor of n^2+1 counted from the prime divisors of m^2+1 for m=1..n.

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Maple

A261609 Number of composite divisors of n^2+1.

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