A147810 -id:A147810 - cp's OEIS frontend

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

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

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

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))) }

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

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

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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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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.

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

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