A147809 -id:A147809 - cp's OEIS frontend

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

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

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

A147811 Alexandrian integers: numbers of the form n=pqr such that 1/n = 1/p - 1/q - 1/r for some integers p,q,r.

Original entry on oeis.org

6, 42, 120, 156, 420, 630, 930, 1428, 1806, 2016, 2184, 3192, 4950, 5256, 8190, 8364, 8970, 10296, 10998, 12210, 17556, 19110, 21114, 23994, 24492, 28050, 32640, 33306, 34362, 37506, 39270, 44310, 52326, 57684, 57840, 70686, 74256, 79800, 83076

Offset: 1

M. F. Hasler and Alexis Olson (AlexisOlson(AT)gmail.com), Dec 13 2008

nonn

The numbers are of the form p(p+d)(p+(p^2+1)/d), where d runs over divisors of p^2+1 and p runs over all positive integers. See also A147807..A147810. - M. F. Hasler, Jan 07 2009

			630 is an Alexandrian integer since 630 = 5(-7)(-18) and 1/630 = 1/5 - 1/7 - 1/18.

Robert Israel, Table of n, a(n) for n = 1..10000
Project Euler, Problem 221: Alexandrian integers

Cf. A147807, A147808, A147809, A147810.

N:= 10^5: # to get all terms <= N
A:= select(`<=`,{seq(seq(p*(p+d)*(p+(p^2+1)/d), d=numtheory:-divisors(p^2+1)),p=1..floor(N^(1/3)))},N):
sort(convert(A,list)); # Robert Israel, Dec 16 2018

is_A147811(n) = { my(d=divisors(n), c=#d+1); n<42 && return(n==6); for( i=2, c-3, d[i+1]^2>d[c-i] && return; d[c-i]%d[i]==1 | next; for( j=i+1, c-i,d[j]^2>d[c-i] && next(2); d[c-i]\d[j]*(d[j]-d[i]) == d[j]*d[i]+1 && return(1))) }

A152823 Largest divisor < n of n^2 + 1, a(1) = 1.

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 5, 5, 2, 1, 2, 5, 10, 1, 2, 1, 10, 13, 2, 1, 17, 5, 10, 1, 2, 1, 10, 5, 2, 17, 26, 25, 10, 13, 2, 1, 10, 17, 2, 1, 29, 5, 37, 13, 2, 29, 34, 5, 2, 41, 2, 5, 10, 1, 34, 1, 50, 5, 2, 13, 2, 5, 10, 17, 2, 1, 10, 37, 2, 29, 2, 61, 65, 1, 58, 53, 10, 5, 2, 37, 34, 25, 65, 1, 2, 13

Offset: 1

M. F. Hasler, Dec 15 2008

nonn

If a(2k) = 3, then 4k^2 + 1 = 3p with p prime. For odd n > 1, a(n) >= 2, with equality if (n^2+1)/2 is prime. Conversely, A147809(n) = 1 iff n^2 + 1 is a semiprime, which for odd n > 1 implies a(n) = 2.

a(1) = 1 by convention, which is compatible with the FORMULA (a(n) = 1 iff n^2 + 1 is prime) and also with a(n) = the floor(d/2)-th divisor of n^2+1, when d is its total number of divisors, cf. PROGRAM. - M. F. Hasler, Sep 11 2019

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

a[1] = 1; a[n_] := Max[Select[Divisors[n^2 + 1], # < n &]]; Array[a, 100] (* Amiram Eldar, Sep 12 2019 *)

A152823(n)={ n=divisors(n^2+1); n[ #n\2] }

a(n) = 1 iff n^2 + 1 is prime (iff A147809(n)=0), which can only happen for n = 1 or even n.

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A147811 Alexandrian integers: numbers of the form n=pqr such that 1/n = 1/p - 1/q - 1/r for some integers p,q,r.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

A152823 Largest divisor < n of n^2 + 1, a(1) = 1.

Views

Author

Keywords

Comments

Links

Programs

Mathematica

PARI

Formula