cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

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

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Author

M. F. Hasler, Dec 13 2008

Keywords

Comments

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

Crossrefs

Programs

  • Mathematica
    Accumulate[DivisorSigma[0, Range[64]^2 + 1]/2] (* Amiram Eldar, Oct 25 2019 *)
  • PARI
    s=0;A147807=vector(99,n,s+=numdiv(n^2+1))/2

Formula

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

Views

Author

M. F. Hasler, Dec 13 2008

Keywords

Comments

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

Crossrefs

Programs

  • Mathematica
    Accumulate[DivisorSigma[0, Range[72]^2 + 1]/2 - 1] (* Amiram Eldar, Oct 25 2019 *)
  • PARI
    s=0;a147806=vector(99,n,s+=numdiv(n^2+1))/2
    A147806(n)=sum(p=1,n,numdiv(n^2+1))/2-n

Formula

a(n) = A147807(n) - n.
a(n) ~ c * n * log(n), where c = 3/(2*Pi) = 0.477464... (A093582). - Amiram Eldar, Dec 01 2023
Showing 1-2 of 2 results.