A093582 -id:A093582 - cp's OEIS frontend

A000089 Number of solutions to x^2 + 1 == 0 (mod n).

1, 1, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 4, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 2, 0, 0, 0, 0

Offset: 1

N. J. A. Sloane

Number of elliptic points of order 2 for GAMMA_0(n).
The Dirichlet inverse, 1, -1, 0, 1, -2, 0, 0, -1, 0, 2, 0, 0, -2, 0,.. seems to equal A091400, apart from signs. - R. J. Mathar, Jul 15 2010
Shadow transform of A002522. - Michel Marcus, Jun 06 2013
a(n) != 0 iff n in A008784. - Joerg Arndt, Mar 26 2014
For n > 1, number of positive solutions to n = a^2 + b^2 such that gcd(a, b) = 1. - Haehun Yang, Mar 20 2022

			G.f. = x + x^2 + 2*x^5 + 2*x^10 + 2*x^13 + 2*x^17 + 2*x^25 + 2*x^26 + 2*x^29 + ...

Michael Baake, "Solution of the coincidence problem in dimensions d <= 4", in R. V. Moody, ed., Mathematics of Long-Range Aperiodic Order, Kluwer, 1997, pp. 9-44.
Goro Shimura, Introduction to the Arithmetic Theory of Automorphic Functions, Princeton, 1971, see p. 25, Eq. (2).

Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..2000 from T. D. Noe)
Michael Baake, Solution of the coincidence problem in dimensions d <= 4, arxiv:math/0605222 [math.MG] (2006).
Michael Baake and Uwe Grimm, Quasicrystalline combinatorics, 2002.
Harriet Fell, Morris Newman, and Edward Ordman, Tables of genera of groups of linear fractional transformations, J. Res. Nat. Bur. Standards Sect. B 67B 1963 61-68.
Steven R. Finch and Pascal Sebah, Squares and Cubes Modulo n, arXiv:math/0604465 [math.NT], 2006-2016.
John S. Rutherford, Sublattice enumeration. IV. Equivalence classes of plane sublattices by parent Patterson symmetry and colour lattice group type, Acta Cryst. (2009). A65, 156-163. [See Table 4].
N. J. A. Sloane, Transforms
László Tóth, Counting Solutions of Quadratic Congruences in Several Variables Revisited, J. Int. Seq. 17 (2014), Article 14.11.6.

Cf. A031358, A027748, A124010, A000095, A006278 (positions of records), A002654, A093582.

a000089 n = product $ zipWith f (a027748_row n) (a124010_row n) where
   f 2 e = if e == 1 then 1 else 0
   f p _ = if p `mod` 4 == 1 then 2 else 0
-- Reinhard Zumkeller, Mar 24 2012

with(numtheory); A000089 := proc (n) local i, s; if modp(n,4) = 0 then RETURN(0) fi; s := 1; for i in divisors(n) do if isprime(i) and i > 2 then s := s*(1+eval(legendre(-1,i))) fi od; s end: # Gene Ward Smith, May 22 2006

Array[ Function[ n, If[ EvenQ[ n ] || Mod[ n, 3 ]==2, 0, Count[ Array[ Mod[ #^2+1, n ]&, n, 0 ], 0 ] ] ], 84 ]
a[ n_] := If[ n < 1, 0, Length @ Select[ (#^2 + 1)/n & /@ Range[n], IntegerQ]]; (* Michael Somos, Aug 15 2015 *)
a[n_] := a[n] = Product[{p, e} = pe; Which[p<3 && e==1, 1, p==2 && e>1, 0, Mod[p, 4]==1, 2, Mod[p, 4]==3, 0, True, a[p^e]], {pe, FactorInteger[n]}]; Array[a, 105] (* Jean-François Alcover, Oct 18 2018, after David W. Wilson *)

{a(n) = if( n<1, 0, sum( x=0, n-1, (x^2 + 1)%n==0))}; \\ Michael Somos, Mar 24 2012

a(n)=my(o=valuation(n,2),f);if(o>1,0,n>>=o;f=factor(n)[,1]; prod(i=1,#f,kronecker(-1,f[i])+1)) \\ Charles R Greathouse IV, Jul 08 2013

from math import prod
from sympy import primefactors
def A000089(n): return prod(1 if p==2 else 2 if p&3==1 else 0 for p in primefactors(n)) if n&3 else 0 # Chai Wah Wu, Oct 13 2024

a(n) = 0 if 4|n, else a(n) = Product_{ p | N } (1 + Legendre(-1, p) ), where we use the definition that Legendre(-1, 2) = 0, Legendre(-1, p) = 1 if p == 1 mod 4, = -1 if p == 3 mod 4. This is Shimura's definition, which is different from Maple's.
Dirichlet g.f.: (1+2^(-s))*Product (1+p^(-s))/(1-p^(-s)) (p=1 mod 4).
Multiplicative with a(p^e) = 1 if p = 2 and e = 1; 0 if p = 2 and e > 1; 2 if p == 1 (mod 4); 0 if p == 3 (mod 4). - David W. Wilson, Aug 01 2001
a(3*n) = a(4*n) = a(4*n + 3) = 0. a(4*n + 1) = A031358(n). - Michael Somos, Mar 24 2012
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3/(2*Pi) = 0.477464... (A093582). - Amiram Eldar, Oct 11 2022

A147809 Half the number of proper divisors (> 1) of n^2 + 1, i.e., tau(n^2 + 1)/2 - 1.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 1, 1, 3, 0, 1, 0, 3, 2, 1, 0, 3, 1, 3, 0, 1, 0, 3, 1, 1, 1, 3, 2, 3, 1, 1, 0, 3, 2, 1, 0, 2, 1, 5, 1, 1, 1, 7, 1, 1, 1, 1, 1, 3, 0, 3, 0, 7, 1, 1, 1, 1, 1, 3, 1, 1, 0, 3, 3, 1, 2, 1, 3, 7, 0, 3, 1, 3, 1, 1, 1, 3, 2, 7, 0, 1, 1, 3, 1, 3, 0, 3, 1, 5, 0, 1, 1, 3, 3, 5

Offset: 1

M. F. Hasler, Dec 13 2008

For any n > 0, n^2 + 1 cannot be a square and thus has an even number of divisors which always include 1 and n^2 + 1, therefore a(n) = (half that number minus 1) is always a nonnegative integer.

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

Cf. A000005, A002522, A147810, A048691, A063647, A093582.

DivisorSigma[0,Range[100]^2+1]/2-1 (* Harvey P. Dale, Feb 11 2015 *)

PARI
```
A147809(n)=numdiv(n^2+1)/2-1
```

a(n) = A000005(A002522(n))/2 - 1 = A147810(n) - 1.

Sum_{k=1..n} a(k) ~ c * n * log(n), where c = 3/(2*Pi) = 0.477464... (A093582). - Amiram Eldar, Dec 01 2023

A147810 Half the number of divisors of n^2+1.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 2, 4, 1, 2, 1, 4, 3, 2, 1, 4, 2, 4, 1, 2, 1, 4, 2, 2, 2, 4, 3, 4, 2, 2, 1, 4, 3, 2, 1, 3, 2, 6, 2, 2, 2, 8, 2, 2, 2, 2, 2, 4, 1, 4, 1, 8, 2, 2, 2, 2, 2, 4, 2, 2, 1, 4, 4, 2, 3, 2, 4, 8, 1, 4, 2, 4, 2, 2, 2, 4, 3, 8, 1, 2, 2, 4, 2, 4, 1, 4, 2, 6, 1, 2, 2, 4, 4, 6

Offset: 1

M. F. Hasler, Dec 13 2008

For any n>0, n^2+1 cannot be a square and thus has an even number of divisors which always include 1 and n^2+1, therefore a(n) is always a positive integer.
Also number of ways to write n^2+1 as n^2+1 = x*y with 1 <= x <= y. - Michel Lagneau, Mar 10 2014
Also number of ways to write arctan(1/n) = arctan(1/x)+arctan(1/y), for integral 0 < n < x < y. - Matthijs Coster, Dec 09 2014
Number of ways that n can be expressed as (j*k-1)/(j+k) with j >= k > n. For any nonnegative integer n, the equation j*k = 1+n*(j+k) always has at least one integer solution with j >= k > n. As j >= k > n, let k=n+c (c is a positive integer), then j=n+(n^2+1)/c; we can easily conclude that c <= n, i.e., for n > 0, a(n) is the number of divisors of (n^2+1) which are <= n. - Zhining Yang, May 18 2023

			For n = 7 the a(7) = 3 solutions are (17,12), (32,9), (57,8). For  n = 13 the a(13) = 4 solutions are (30,23), (47,18), (98,15), (183,14). - _Zhining Yang_, May 18 2023

Amiram Eldar, Table of n, a(n) for n = 1..10000
Shouen Wang, The general term formula of an integer sequence.

Cf. A048691, A093582, A290332, A290333, A359225.

with(numtheory); A147810:=n->tau(n^2+1)/2; seq(A147810(n), n=1..100); # Wesley Ivan Hurt, Mar 10 2014

Table[c=0; Do[If[i<=j && i*j==n^2+1, c++], {i, t=Divisors[n^2+1]}, {j, t}]; c, {n, 100}] (* Michel Lagneau, Mar 10 2014 *)

PARI
```
A147810(n)=numdiv(n^2+1)/2
```

from sympy import divisor_count
def A147810(n): return divisor_count(n**2+1)>>1 if n else 1 # Chai Wah Wu, Jul 09 2023

a(n) = A000005(A002522(n))/2 = A147809(n)+1.

Sum_{k=1..n} a(k) ~ c * n * log(n), where c = 3/(2*Pi) = 0.477464... (A093582). - Amiram Eldar, Dec 01 2023

A147807 Partial sums of A147810(n) = tau(n^2 + 1)/2.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 11, 13, 15, 16, 18, 20, 24, 25, 27, 28, 32, 35, 37, 38, 42, 44, 48, 49, 51, 52, 56, 58, 60, 62, 66, 69, 73, 75, 77, 78, 82, 85, 87, 88, 91, 93, 99, 101, 103, 105, 113, 115, 117, 119, 121, 123, 127, 128, 132, 133, 141, 143, 145, 147, 149, 151, 155, 157

Offset: 1

M. F. Hasler, Dec 13 2008

Also, number of inequivalent (i.e., q < r) integer solutions to 1/pqr = 1/p - 1/q - 1/r with p <= n; cf. A147811.

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

Cf. A093582, A147806, A147809, A147810, A147811.

Accumulate[DivisorSigma[0, Range[64]^2 + 1]/2] (* Amiram Eldar, Oct 25 2019 *)

s=0;A147807=vector(99,n,s+=numdiv(n^2+1))/2

a(n) = Sum_{p = 1..n} tau(1 + p^2)/2 = n + A147806(n) > n.

a(n) ~ c * n * log(n), where c = 3/(2*Pi) = 0.477464... (A093582). - Amiram Eldar, Dec 01 2023

A147806 Partial sums of A147809(n) = tau(n^2 + 1)/2 - 1.

Original entry on oeis.org

0, 0, 1, 1, 2, 2, 4, 5, 6, 6, 7, 8, 11, 11, 12, 12, 15, 17, 18, 18, 21, 22, 25, 25, 26, 26, 29, 30, 31, 32, 35, 37, 40, 41, 42, 42, 45, 47, 48, 48, 50, 51, 56, 57, 58, 59, 66, 67, 68, 69, 70, 71, 74, 74, 77, 77, 84, 85, 86, 87, 88, 89, 92, 93, 94, 94, 97, 100, 101, 103, 104, 107

Offset: 1

M. F. Hasler, Dec 13 2008

It seems that a(10^n) = (6, 168, 2754, 38561, 495569, ...) ~ 1.1*(n-0.5)*10^n; otherwise said, a(n) ~ 1.1*(log_10(n)-0.5)*n, asymptotically.

The exact value of the coefficient above is 3*log(10)/(2*Pi) = 1.09940339... . - Amiram Eldar, Dec 01 2023

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

Cf. A093582, A147807, A147809, A147810, A147811.

Accumulate[DivisorSigma[0, Range[72]^2 + 1]/2 - 1] (* Amiram Eldar, Oct 25 2019 *)

s=0;a147806=vector(99,n,s+=numdiv(n^2+1))/2
A147806(n)=sum(p=1,n,numdiv(n^2+1))/2-n

a(n) = A147807(n) - n.

a(n) ~ c * n * log(n), where c = 3/(2*Pi) = 0.477464... (A093582). - Amiram Eldar, Dec 01 2023

A289502 Mean width of the regular tetrahedron of unit length.

Original entry on oeis.org

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

Offset: 0

R. J. Mathar, Jul 07 2017

Steven R. Finch, Mean width of a regular simplex, arxiv:1111.4976 [math.MG] (2016), E(w_3).

Cf. A156546, A093582.

Maple
```
x := 3*arccos(-1/3)/2/Pi ; evalf(%) ;
```

RealDigits[3 ArcCos[-1/3]/(2*Pi), 10, 87][[1]] (* Indranil Ghosh, Jul 08 2017 *)

Equals A156546*A093582.

cp's OEIS Frontend

A000089 Number of solutions to x^2 + 1 == 0 (mod n).

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A147809 Half the number of proper divisors (> 1) of n^2 + 1, i.e., tau(n^2 + 1)/2 - 1.

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A147810 Half the number of divisors of n^2+1.

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A147807 Partial sums of A147810(n) = tau(n^2 + 1)/2.

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A147806 Partial sums of A147809(n) = tau(n^2 + 1)/2 - 1.

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A289502 Mean width of the regular tetrahedron of unit length.

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