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 81-90 of 112 results. Next

A056647 a(n) = A056623(A001405(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 9, 36, 1, 4, 4, 1, 9, 9, 1, 4, 1, 4, 4, 1, 1, 4, 100, 25, 100, 25, 9, 144, 9, 9, 1, 4, 25, 100, 100, 25, 9, 36, 4, 1, 4, 1, 25, 400, 225, 900, 1764, 441, 196, 49, 49, 784, 4, 1, 1, 16, 1, 16, 16, 4, 441, 441, 49, 196, 49, 196, 36, 9, 1, 4, 4, 1, 100, 25, 1225
Offset: 1

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Author

Labos Elemer, Aug 09 2000

Keywords

Comments

Previous name "Largest unitary square divisor of central binomial coefficient" was incorrect. See A376553 for the correct sequence with this name. - Amiram Eldar, Sep 28 2024

Examples

			a(28) = A056623(binomial(28,14)) = A056623(40116600) = 25.
		

Crossrefs

Formula

a(n) = A008833(A001405(n))/A055229(A001405(n))^2 = A056057(n)/A056059(n)^2.

Extensions

Incorrect name replaced with a formula by Amiram Eldar, Sep 28 2024

A083730 Greatest prime^2 factor of n, or a(n)=1 for squarefree n.

Original entry on oeis.org

1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 25, 1, 9, 4, 1, 1, 1, 4, 1, 1, 1, 9, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 49, 25, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 4, 1, 1, 1, 4, 1, 1, 1, 9, 1, 1, 25, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 49, 9
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 14 2003

Keywords

Comments

Not multiplicative, for example a(4)*a(9) <> a(36). - R. J. Mathar, Oct 31 2011

Crossrefs

Programs

Formula

a(n) = A249740(n)^2. - Amiram Eldar, Feb 11 2021

A282344 The smallest square referenced in A086982 (Numbers n such that 10^n+1 is not squarefree).

Original entry on oeis.org

121, 49, 121, 169, 121, 49, 121, 121, 49, 169, 121, 289, 121, 49, 121, 361, 121, 49, 169, 10201, 121, 49, 237169, 121, 49, 121, 5329, 121, 49, 121, 121, 169, 49, 121, 121, 49, 841, 121, 289, 121, 49, 121, 121, 49, 121, 169, 361, 121, 49, 121, 18769, 121, 49
Offset: 1

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Author

Robert Price, Feb 12 2017

Keywords

Crossrefs

Formula

a(n) = A008833( 1+10^A086982(n) ). - R. J. Mathar, Feb 13 2017

A306300 Discriminant D of real quadratic number field Q(sqrt(D)) associated with fundamental discriminant d = A003658(n).

Original entry on oeis.org

1, 5, 2, 3, 13, 17, 21, 6, 7, 29, 33, 37, 10, 41, 11, 53, 14, 57, 15, 61, 65, 69, 73, 19, 77, 85, 22, 89, 23, 93, 97, 101, 26, 105, 109, 113, 30, 31, 129, 133, 34, 137, 35, 141, 145, 149, 38, 39, 157, 161, 165, 42, 43, 173, 177, 181, 46, 185, 47, 193, 197
Offset: 1

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Author

N. J. A. Sloane, Apr 02 2019

Keywords

Crossrefs

Formula

a(n) = A007913(A003658(n)) = A003658(n) / (A008833(A003658(n))).

A358272 Multiplicative sequence with a(p^e) = (-1)^e * p^(2*floor(e/2)) for prime p and e >= 0.

Original entry on oeis.org

1, -1, -1, 4, -1, 1, -1, -4, 9, 1, -1, -4, -1, 1, 1, 16, -1, -9, -1, -4, 1, 1, -1, 4, 25, 1, -9, -4, -1, -1, -1, -16, 1, 1, 1, 36, -1, 1, 1, 4, -1, -1, -1, -4, -9, 1, -1, -16, 49, -25, 1, -4, -1, 9, 1, 4, 1, 1, -1, 4, -1, 1, -9, 64, 1, -1, -1, -4, 1, -1, -1, -36, -1, 1, -25, -4, 1, -1, -1, -16
Offset: 1

Views

Author

Werner Schulte, Nov 07 2022

Keywords

Comments

Signed version of A008833.

Crossrefs

Programs

  • Maple
    A358272 := proc(n)
        local a,pe,e,p ;
        a := 1;
        for pe in ifactors(n)[2] do
            e := op(2,pe) ;
            p := op(1,pe) ;
            a := a*(-1)^e*p^(2*floor(e/2)) ;
        end do:
        a ;
    end proc:
    seq(A358272(n),n=1..80) ; # R. J. Mathar, Jan 17 2023
  • Mathematica
    f[p_, e_] := (-1)^e * p^(2*Floor[e/2]); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Nov 07 2022 *)
  • Python
    from math import prod
    from sympy import factorint
    def A358272(n): return prod(-p**(e&-2) if e&1 else p**(e&-2) for p, e in factorint(n).items()) # Chai Wah Wu, Jan 17 2023

Formula

a(n) = lambda(n) * A008833(n) for n > 0 where lambda(n) = A008836(n).
Dirichlet g.f.: zeta(2*s-2) / zeta(s).
Dirichlet inverse b(n), n > 0, is multiplicative with b(p) = 1 and b(p^e) = 1 - p^2 for prime p and e > 1.
Dirichlet convolution with A034444 equals A008833.
Equals Dirichlet convolution of A000010 and A061019.
Conjecture: a(n) = Sum_{k=1..n} gcd(k, n) * lambda(gcd(k, n)) for n > 0.
a(n) = Sum_{d|n} lambda(d)*d*phi(n/d), where lambda(n) = A008836(n). - Ridouane Oudra, May 11 2025

A365334 The sum of exponentially odd divisors of the largest square dividing n.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 1, 3, 4, 1, 1, 3, 1, 1, 1, 11, 1, 4, 1, 3, 1, 1, 1, 3, 6, 1, 4, 3, 1, 1, 1, 11, 1, 1, 1, 12, 1, 1, 1, 3, 1, 1, 1, 3, 4, 1, 1, 11, 8, 6, 1, 3, 1, 4, 1, 3, 1, 1, 1, 3, 1, 1, 4, 43, 1, 1, 1, 3, 1, 1, 1, 12, 1, 1, 6, 3, 1, 1, 1, 11, 31, 1, 1, 3, 1
Offset: 1

Views

Author

Amiram Eldar, Sep 01 2023

Keywords

Comments

The number of these divisors is A365333(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := (p^(e + 1 - Mod[e, 2]) - p)/(p^2 - 1) + 1; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i,1]^(f[i,2] + 1 - f[i,2]%2) - f[i,1])/(f[i,1]^2 - 1) + 1);}

Formula

a(n) = A033634(A008833(n)).
a(n) = 1 if and only if n is squarefree (A005117).
Multiplicative with a(p^e) = 1 + (p^(e + 1 - (e mod 2)) - 1)/(p^2 - 1).
Dirichlet g.f.: zeta(s) * zeta(2*s-2) * Product_{p prime} (1 - 1/p^(2*s-2) + 1/p^(2*s-1)).

A365403 The sum of the unitary divisors of the largest square dividing n.

Original entry on oeis.org

1, 1, 1, 5, 1, 1, 1, 5, 10, 1, 1, 5, 1, 1, 1, 17, 1, 10, 1, 5, 1, 1, 1, 5, 26, 1, 10, 5, 1, 1, 1, 17, 1, 1, 1, 50, 1, 1, 1, 5, 1, 1, 1, 5, 10, 1, 1, 17, 50, 26, 1, 5, 1, 10, 1, 5, 1, 1, 1, 5, 1, 1, 10, 65, 1, 1, 1, 5, 1, 1, 1, 50, 1, 1, 26, 5, 1, 1, 1, 17, 82
Offset: 1

Views

Author

Amiram Eldar, Sep 03 2023

Keywords

Comments

The number of these divisors is A323308(n).
The sum of the unitary divisors of the square root of the largest square dividing n is A365404(n).

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := If[e == 1, 1, p^(2*Floor[e/2]) + 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]
  • PARI
    a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,2] == 1, 1, 1 + f[i,1]^(2*(f[i,2]\2))));}

Formula

a(n) = A034448(A008833(n)).
a(n) <= A034448(n) with equality if and only if n is a square (A000290).
a(n) >= 1 with equality if and only if n is squarefree (A005117).
Multiplicative with a(p) = 1 and a(p^e) = p^(2*floor(e/2)) + 1 for e >= 2.
Dirichlet g.f.: zeta(s) * zeta(2*s-2) / zeta(4*s-2).
Sum_{k=1..n} a(k) ~ c * n^(3/2), where c = zeta(3/2)/(3*zeta(4)) = 30*zeta(3/2)/Pi^4 = 0.804557969165... .

A366523 Largest square divisor of n which is <= sqrt(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 4, 9, 1, 1, 4, 1, 1, 1, 4, 1, 9, 1, 4, 1, 1, 1, 4, 1, 1, 9, 4
Offset: 1

Views

Author

Ilya Gutkovskiy, Oct 11 2023

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Last[Select[Divisors[n], # <= Sqrt[n] && IntegerQ[Sqrt[#]] &]], {n, 100}]
  • PARI
    a(n) = {my(m=1); fordiv(sqrtint(n/core(n)), d, if(d^4 <= n, m=max(m,d))); m^2} \\ Andrew Howroyd, Oct 11 2023

A379663 a(n) is the number of integer-sided triangles whose sides are in geometric progression with smallest side n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 2, 1, 1, 1, 2, 4, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 3, 5, 4, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 4, 2, 1, 1, 1, 3, 6, 1, 1, 2, 1, 1, 1, 2
Offset: 1

Views

Author

Felix Huber, Jan 07 2025

Keywords

Comments

The integer sides of the triangles are n, n*r, n*r^2 with rational r >= 1. From the triangle inequality n + n*r >= n*r^2 follows r <= (1 + sqrt(5))/2 (golden ratio). Therefore 1 <= r = c/d < (1 + sqrt(5))/2, where c and d are coprimes and d^2 divides n.

Examples

			The a(18) = 2 integer-sided triangles whose sides form a geometric sequence are [18, 18, 18] with r = 1, [18, 24, 32] with r = 4/3.
The a(25) = 4 integer-sided triangles whose sides form a geometric sequence are [25, 25, 25] with r = 1, [25, 30, 36] with r = 6/5, [25, 35, 49] with r = 7/5, [25, 40, 64] with r = 8/5.
The a(36) = 4 integer-sided triangles whose sides form a geometric sequence are [36, 36, 36] with r = 1, [36, 54, 81] with r = 3/2, [36, 48, 64] with r = 4/3, [36, 42, 49] with r = 7/6.
See also the linked Maple program "Triangles for a given n".
		

Crossrefs

Programs

  • Maple
    A379663:=n->floor(2*expand(NumberTheory:-LargestNthPower(n,2))/(1+sqrt(5)))+1;
    seq(A379663(n),n=1..88);

Formula

a(n) = A060143(A000188(n)) + 1.

A379705 a(n) is the least integer k > n such that integers p, q exist for which n, p, k are in arithmetic and n, q, k are in geometric progression.

Original entry on oeis.org

9, 8, 27, 16, 45, 24, 63, 18, 25, 40, 99, 48, 117, 56, 135, 36, 153, 32, 171, 80, 189, 88, 207, 54, 49, 104, 75, 112, 261, 120, 279, 50, 297, 136, 315, 64, 333, 152, 351, 90, 369, 168, 387, 176, 125, 184, 423, 108, 81, 72, 459, 208, 477, 96, 495, 126, 513, 232
Offset: 1

Views

Author

Felix Huber, Jan 07 2025

Keywords

Examples

			a(9) = 25 because 9, 17, 25 are in arithmetic progression (common difference = 8) and 9, +-15, 25 are in geometric progression (common ratio = +-5/3) and there is no other integer k with 9 < k < 25 such that integers p and q exist for which 9, p, k are in arithmetic and 9, q, k are in geometric progression.
		

Crossrefs

Programs

  • Maple
    A379705:=proc(n)
       local d;
       d:=expand(NumberTheory:-LargestNthPower(n,2));
       if is(n*(1+(d+1)^2/d^2),even) then
          n*(d+1)^2/d^2
       else
          n*(d+2)^2/d^2
       fi;
    end proc;
    seq(A379705(n),n=1..58);

Formula

a(n) = n/A008833(n)*(A000188(n) + k)^2, where k = 1 if n*(1+(A000188(n)+1)^2/A008833(n)) is even or k = 2 else.
a(n) = A072905(n) if n*(1+(A000188(n)+1)^2/A008833(n)) is even.
Previous Showing 81-90 of 112 results. Next