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 1-3 of 3 results.

A153876 a(n) = Sum_{i=2^(n-1)..2^n-1} sigma_0(i), sigma_0(i) number of divisors of n, n positive integer.

Original entry on oeis.org

1, 4, 11, 29, 68, 160, 364, 820, 1813, 3981, 8674, 18782, 40387, 86443, 184232, 391188, 827787, 1746443, 3674573, 7712561, 16151933, 33757505, 70422235, 146659055, 304947023, 633152157, 1312820598, 2718674046, 5623413203, 11618957217, 23982175093, 49452872529
Offset: 1

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Author

Ctibor O. Zizka, Jan 03 2009

Keywords

Comments

This sequence tells how many binary numbers with n digits are there in the multiplication matrix [1,...,2^n -1]x[1,...,2^n -1]. In general, counting how many base-B numbers of length n are there in the multiplication matrix [1,...,B^n -1]x[1,...,B^n -1] gives a(n)= sum_{i=B^(n-1),(B^n)-1} sigma_0(i). Besides this motivation it is interesting to see the behavior of partial sums of sigma_0(i) on growing intervals : a(n)= sum_{i=f(n-1),f(n)} sigma_0(i).

Crossrefs

Programs

  • PARI
    a(n) = sum(i=2^(n-1), 2^n-1, numdiv(i)); \\ Michel Marcus, Oct 10 2021
    
  • Python
    from math import isqrt
    def A153876(n): return ((t:=isqrt(b:=(1<Chai Wah Wu, Oct 23 2023

Formula

a(n) = A085831(n) - A085831(n-1)-1. - R. J. Mathar, Jan 05 2009
a(n) = Sum_{k>=1} k * A346730(n,k). - Alois P. Heinz, Aug 01 2021

Extensions

a(14)-a(32) from Alois P. Heinz, Aug 01 2021

A153816 a(n) = Sum_{i=1..(10^n-1)/9} floor(((10^n-1)/9)/i).

Original entry on oeis.org

1, 29, 542, 7967, 105225, 1308095, 15639310, 181976675, 2075608136, 23314508721, 258729364359, 2843136431305, 30989792180446, 335482200606705, 3610664794156597, 38665075822637767, 412235037037411453, 4378193158484415385, 46340359465948601163
Offset: 1

Views

Author

Ctibor O. Zizka, Jan 02 2009

Keywords

Comments

Generalized subsequences of A006218(n) are a(n) = A006218(T*A002275(n)), where T >= 1, a(n) = Sum_{i=1...n} floor(T*(10^n - 1)/9*i). For T=9 we have A095256, for T=1 this sequence. The motivation for such sequences is to count the elements of length n in a multiplication matrix m*m in base (T+1). In base 10 this gives T=9 and the number of elements of the multiplication matrix m*m of the length n=1,2,3,... digits is given by the sequence b(n) = a(n) - a(n-1), n >= 2, a(1)=23.

Crossrefs

Programs

  • Mathematica
    A153816[n_] := Sum[Floor[((10^n - 1)/9)/i], {i, (10^n - 1)/9}]; Array[A153816, 7] (* JungHwan Min, Feb 05 2017 *)
  • PARI
    a(n) = sum(i=1, (10^n-1)/9, ((10^n-1)/9)\i); \\ Michel Marcus, Jun 08 2018
    
  • Python
    def a(n): t = (10**n-1)//9; return sum(t//i for i in range(1, t+1))
    print([a(n) for n in range(1, 10)]) # Michael S. Branicky, Jan 29 2021
    
  • Python
    from math import isqrt
    def A153816(n): return -(s:=isqrt(m:=(10**n-1)//9))**2+(sum(m//k for k in range(1,s+1))<<1) # Chai Wah Wu, Oct 23 2023

Formula

a(n) = A006218(A002275(n)).

Extensions

Formula corrected by Giovanni Resta, Feb 05 2009
a(9)-a(17) from Donovan Johnson, Sep 06 2010
a(18)-a(19) from Hiroaki Yamanouchi, Jul 06 2014

A174425 Total number of divisors of all n-digit numbers.

Original entry on oeis.org

23, 450, 6580, 86590, 1073071, 12803271, 148755315, 1694786187, 19020186047, 210925125565, 2316483913054, 25237165712764, 273094922940644, 2938181887791268, 31454145461543161, 335264720452385137, 3559879862893130917, 37671125212625723995, 397434517963203503069
Offset: 1

Views

Author

Jaroslav Krizek, Nov 28 2010

Keywords

Comments

Partial sums are A095256.

Examples

			For n = 1; a(1) = 23 because tau (r) of 1-digit numbers r = 1 to 9: {1, 2, 2, 3, 2, 4, 2, 4, 3}. Sum is 23.
		

Crossrefs

Programs

  • PARI
    \\ too slow for n > 20; here b(n) is A006218(n).
    b(n)={sum(k=1, sqrtint(n), n\k)*2 - sqrtint(n)^2}
    a(n)={b(10^n-1)-b(10^(n-1)-1)} \\ Andrew Howroyd, Jan 13 2020

Formula

From Andrew Howroyd, Jan 13 2020: (Start)
a(n) = A006218(10^n-1) - A006218(10^(n-1)-1).
a(n) = A057494(n) - A057494(n-1) - 2*n - 1. (End)

Extensions

Terms a(11) and beyond from Andrew Howroyd, Jan 13 2020
Showing 1-3 of 3 results.