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.

Previous Showing 51-54 of 54 results.

A345202 Decimal expansion of gamma + zeta(2), where gamma is Euler's constant (A001620).

Original entry on oeis.org

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

Views

Author

Amiram Eldar, Jun 10 2021

Keywords

Comments

The value of the sum (see the Formula section) discovered in 1926 by the Italian mathematician and historian of science Giovanni Enrico Eugenio Vacca (1872-1953).

Examples

			2.22214973174975929707892725672842762026110923714672...
		

References

  • G. Vacca, Nuova serie per la costante di Eulero, C=0,577..., Rendiconti, Accademia Nazionale dei Lincei, Roma, Classe di Scienze Fisiche, Matematiche e Naturali, Serie 6, Vol. 3 (1926), pp. 19-20.

Crossrefs

Programs

  • Mathematica
    RealDigits[EulerGamma + Pi^2/6, 10, 100][[1]]

Formula

Equals Sum_{k>=1} (1/floor(sqrt(k))^2 - 1/k) (Vacca, 1926).
Equals Sum_{k>=1} f(k)/k^2, where f(k) = Sum_{j=1..2*k} j/(j + k^2).
Equals A001620 + A013661.

A386472 The maximum exponent in the prime factorization of the largest divisor of n whose exponents in its prime factorization are squares.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Amiram Eldar, Jul 23 2025

Keywords

Comments

All the terms are by definition squares.

Crossrefs

Programs

  • Mathematica
    a[n_] := Floor[Sqrt[Max[FactorInteger[n][[;; , 2]]]]]^2; a[1] = 0; Array[a, 100]
  • PARI
    a(n) = if(n == 1, 0, sqrtint(vecmax(factor(n)[,2]))^2);

Formula

a(n) = A051903(A386469(n)).
a(n) = A048760(A051903(n)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1 + Sum_{k>=1} (2*k+1)*(1-1/zeta((k+1)^2)) = 1.2383138701540899647042996... .

A263727 Largest square number less than or equal to the n-th Fibonacci number.

Original entry on oeis.org

0, 1, 1, 1, 1, 4, 4, 9, 16, 25, 49, 81, 144, 225, 361, 576, 961, 1521, 2500, 4096, 6724, 10816, 17689, 28561, 46225, 74529, 121104, 196249, 316969, 514089, 831744, 1345600, 2175625, 3523129, 5702544, 9223369, 14922769, 24157225, 39087504, 63234304, 102333456
Offset: 0

Views

Author

Eli Jaffe, Oct 24 2015

Keywords

Examples

			For a(8), Fibonacci(8) = 21, the largest square under 21 is 16, so a(8) = 16.
		

Crossrefs

Programs

  • Mathematica
    Floor[Sqrt[Fibonacci[Range[40]]]]^2 (* Alonso del Arte, Oct 24 2015 *)
  • PARI
    a(n) = sqrtint(fibonacci(n))^2; \\ Michel Marcus, Oct 25 2015

Formula

a(n) = floor(sqrt(Fibonacci(n)))^2.
a(n) = A061287(n)^2. - Michel Marcus, Oct 25 2015
a(n) = A048760(A000045(n)). - Michel Marcus, Nov 11 2015

A350493 a(n) = floor(sqrt(prime(n)))^2 mod n.

Original entry on oeis.org

0, 1, 1, 0, 4, 3, 2, 0, 7, 5, 3, 0, 10, 8, 6, 1, 15, 13, 7, 4, 1, 20, 12, 9, 6, 22, 19, 16, 13, 10, 28, 25, 22, 19, 4, 0, 33, 30, 27, 9, 5, 1, 40, 37, 16, 12, 8, 4, 29, 25, 21, 17, 13, 9, 36, 32, 28, 24, 20, 16, 12, 41, 37, 33, 29, 25, 56, 52, 48, 44, 40, 36
Offset: 1

Views

Author

Simon Strandgaard, Jan 01 2022

Keywords

Examples

			a(4) = A065730(4) mod 4 =  4 mod 4 = 0;
a(5) = A065730(5) mod 5 =  9 mod 5 = 4;
a(6) = A065730(6) mod 6 =  9 mod 6 = 3;
a(7) = A065730(7) mod 7 = 16 mod 7 = 2.
		

Crossrefs

Programs

  • Mathematica
    Table[PowerMod[Floor[Sqrt[Prime[n]]],2,n],{n,72}] (* Stefano Spezia, Jan 02 2022 *)
  • PARI
    a(n) = (sqrtint(prime(n))^2) % n;
    vector(20,n,a(n))
    
  • Python
    from sympy import prime, integer_nthroot
    def a(n): return (integer_nthroot(prime(n), 2)[0]**2)%n
    print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Jan 02 2022
  • Ruby
    require 'prime'
    values = []
    Prime.first(20).each_with_index do |prime, i|
        values << ((Integer.sqrt(prime) ** 2) % (i + 1))
    end
    p values
    

Formula

a(n) = A065730(n) mod n.
Previous Showing 51-54 of 54 results.