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 1-6 of 6 results.

A129628 Inverse Moebius transform of A001511.

Original entry on oeis.org

1, 3, 2, 6, 2, 6, 2, 10, 3, 6, 2, 12, 2, 6, 4, 15, 2, 9, 2, 12, 4, 6, 2, 20, 3, 6, 4, 12, 2, 12, 2, 21, 4, 6, 4, 18, 2, 6, 4, 20, 2, 12, 2, 12, 6, 6, 2, 30, 3, 9, 4, 12, 2, 12, 4, 20, 4, 6, 2, 24, 2, 6, 6, 28, 4, 12, 2, 12, 4, 12, 2, 30, 2, 6, 6, 12, 4, 12, 2, 30, 5, 6, 2, 24, 4, 6, 4, 20
Offset: 1

Views

Author

Ralf Stephan, May 31 2007

Keywords

Comments

Dirichlet convolution of A000005 with A209229. - Ridouane Oudra, Jul 25 2025

Crossrefs

Programs

  • Maple
    seq(add(padic[ordp](2*d, 2), d in numtheory[divisors](n)), n=1..100); # Ridouane Oudra, Sep 30 2024
  • Mathematica
    f[p_, e_] := If[p==2, (e+1)*(e+2)/2, e+1]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Sep 30 2020 *)
  • PARI
    a(n)={sumdiv(n, d, 1 + valuation(d, 2))} \\ Andrew Howroyd, Aug 04 2018

Formula

a(2n) = a(n) + A000005(2n), a(2n+1) = A000005(2n+1).
Dirichlet g.f.: zeta(s)^2 * 2^s/(2^s-1). - Ralf Stephan, Jun 17 2007, corrected by Vaclav Kotesovec, Feb 02 2019
a(n) = Sum_{d|n} A001511(d). - Andrew Howroyd, Aug 04 2018
Sum_{k=1..n} a(k) ~ 2*n * (2*gamma - 1 + log(n/2)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Feb 02 2019
Multiplicative with a(2^e) = (e+1)*(e+2)/2, and a(p^e) = e+1 for p > 2. - Amiram Eldar, Sep 30 2020
From Ridouane Oudra, Sep 30 2024: (Start)
a(n) = Sum_{i=0..A007814(n)} tau(n/2^i).
a(n) = Sum_{d|2*n} A007814(d).
a(n) = (1/2)*A001511(n)*A099777(n).
a(n) = (1/2)*(A001511(n) + 1)*A000005(n).
a(n) = A115364(n)*A001227(n). (End)

A094290 a(n) = prime(A001511(n)), where A001511 is one more than the 2-adic valuation of n.

Original entry on oeis.org

2, 3, 2, 5, 2, 3, 2, 7, 2, 3, 2, 5, 2, 3, 2, 11, 2, 3, 2, 5, 2, 3, 2, 7, 2, 3, 2, 5, 2, 3, 2, 13, 2, 3, 2, 5, 2, 3, 2, 7, 2, 3, 2, 5, 2, 3, 2, 11, 2, 3, 2, 5, 2, 3, 2, 7, 2, 3, 2, 5, 2, 3, 2, 17, 2, 3, 2, 5, 2, 3, 2, 7, 2, 3, 2, 5, 2, 3, 2, 11, 2, 3, 2, 5, 2, 3, 2, 7, 2, 3, 2, 5, 2, 3, 2, 13, 2, 3, 2, 5, 2, 3
Offset: 1

Views

Author

Amarnath Murthy, Apr 28 2004

Keywords

Comments

Originally defined as: a(1) = 2 = prime(1). Then the first occurrence of prime(n) followed by all previous terms. i.e. If the index of first occurrence of prime(n) is k then the next k-1 terms are defined as a(k+r) = a(r), r = 1 to k-1. and a(2k) = prime(n+1) and so on.
Index of the first occurrence of prime(n)= 2^(n-1). Subsidiary sequences: If prime(n) is replaced by f(n) a large number of sequences can be obtained choosing f(n) = composite(n), f(n) = n^2,f(n) = n^r, r =3,4,5,..., f(n) = tau(n), f(n) = sigma(n), f(n) = n!, f(n) = Fibonacci(n), f(n) = T(n), triangular number, f(n) = n-th Bell, etc. each giving a distinct fascinating music.
The lexicographically earliest sequence such that no product of consecutive terms is a perfect square. - Joshua Zucker, Apr 30 2011

Crossrefs

Cf. also A115364.

Programs

Formula

a(n) = A000040(A001511(n)). - Omar E. Pol, Sep 13 2013

Extensions

Replaced the name with a formula given by Omar E. Pol, which is equivalent to the original definition. - Antti Karttunen, Nov 02 2018

A316631 Expansion of A(x) = x*(1+3*x^2+x^3+3*x^4+x^6)/(1-x^4)^2.

Original entry on oeis.org

0, 1, 0, 3, 1, 5, 0, 7, 2, 9, 0, 11, 3, 13, 0, 15, 4, 17, 0, 19, 5, 21, 0, 23, 6, 25, 0, 27, 7, 29, 0, 31, 8, 33, 0, 35, 9, 37, 0, 39, 10, 41, 0, 43, 11, 45, 0, 47, 12, 49, 0, 51, 13, 53, 0, 55, 14, 57, 0, 59, 15, 61, 0, 63, 16, 65, 0, 67, 17, 69, 0, 71, 18, 73, 0, 75, 19, 77, 0, 79, 20
Offset: 0

Views

Author

Werner Schulte, Jul 09 2018

Keywords

Examples

			a(22) = 0 since 22 mod 4 = 2; a(23) = 23 for 23 mod 2 = 1; a(24) = 6 because 24 mod 4 = 0 and 24/4 = 6.
		

Crossrefs

Programs

  • GAP
    a:=[0,1,0,3,1,5,0,7];; for n in [9..85] do a[n]:=2*a[n-4]-a[n-8]; od; a; # Muniru A Asiru, Jul 20 2018
  • Maple
    seq(coeff(series(x*(1+3*x^2+x^3+3*x^4+x^6)/(1-x^4)^2, x,n+1),x,n),n=0..80); # Muniru A Asiru, Jul 20 2018
  • Mathematica
    CoefficientList[Series[x (1 + 3 x^2 + x^3 + 3 x^4 + x^6)/(1 - x^4)^2, {x, 0, 80}], x] (* Michael De Vlieger, Jul 20 2018 *)
    LinearRecurrence[{0, 0, 0, 2, 0, 0, 0, -1}, {0, 1, 0, 3, 1, 5, 0, 7}, 81] (* Robert G. Wilson v, Jul 21 2018 *)
  • PARI
    concat(0, Vec((x*(1+3*x^2+x^3+3*x^4+x^6)/(1-x^4)^2) + O(x^80))) \\ Felix Fröhlich, Jul 09 2018
    
  • PARI
    {my(N=79); concat([0], dirdiv(vector(N,n,n), vector(N, n, my(k=valuation(n, 2)); if(n==2^k, k+1, 0))))} \\ Andrew Howroyd, Jul 09 2018
    

Formula

a(n) = n/4 if n mod 4 = 0, and a(n) = 0 if n mod 4 = 2, and a(n) = n if n mod 2 = 1.
Linear recurrence: a(n) = 2*a(n-4) - a(n-8) for n > 7.
a(n) for n > 0 is multiplicative with a(2^e) = 1 - e if e < 2 and a(2^e) = 2^(e-2) if e > 1 otherwise a(p^e) = p^e for prime p > 2 and e >= 0.
Dirichlet g.f.: Sum_{n>0} a(n)/n^s = (1-1/2^s)^2 * zeta(s-1).
Dirichlet inverse b(n) for n > 0 is multiplicative with b(2^e) = 1 - e and for prime p > 2: b(p) = -p and b(p^e) = 0 if e > 1.
Dirichlet convolution with A104117(n) yields A000027(n).
Dirichlet convolution with A115364(n) yields A000203(n).
Sum_{k=1..n} a(k) ~ (9/32) * n^2. - Amiram Eldar, Nov 20 2022

A373438 Expansion of Sum_{k>=1} k * x^(3^(k-1)) / (1 - x^(3^(k-1))).

Original entry on oeis.org

1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 10, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 10, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 15, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3, 1, 1, 6, 1, 1, 3, 1, 1, 3
Offset: 1

Views

Author

Ilya Gutkovskiy, Jun 05 2024

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 105; CoefficientList[Series[Sum[k x^(3^(k - 1))/(1 - x^(3^(k - 1))), {k, 1, Floor[Log[3, nmax]] + 1}], {x, 0, nmax}], x] // Rest
    Table[Binomial[IntegerExponent[3 n, 3] + 1, 2], {n, 1, 105}]
  • PARI
    a(n) = {my(e = valuation(n, 3)); (e+1)*(e+2)/2;} \\ Amiram Eldar, Jun 27 2024

Formula

a(n) = A000217(A051064(n)).
From Vaclav Kotesovec, Jun 25 2024: (Start)
Dirichlet g.f.: zeta(s) * (3^s/(3^s-1))^2.
Sum_{k=1..n} a(k) ~ 9*n/4 - log(n)*(log(n) + 2*log(6*Pi))/(4*log(3)^2). (End)
Multiplicative with a(p^e) = (e+1)*(e+2)/2 if p = 3 and 1 if p != 3. - Amiram Eldar, Jun 27 2024

A115363 ((1,x)-(x,x^2))^(-2) (using Riordan array notation).

Original entry on oeis.org

1, 2, 1, 0, 0, 1, 3, 2, 0, 1, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 4, 3, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 3, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1
Offset: 0

Views

Author

Paul Barry, Jan 21 2006

Keywords

Comments

Square of number triangle A115361. Row sums are A115364.

Examples

			Triangle begins
1,
2, 1,
0, 0, 1,
3, 2, 0, 1,
0, 0, 0, 0, 1,
0, 0, 2, 0, 0, 1,
0, 0, 0, 0, 0, 0, 1,
4, 3, 0, 2, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 0, 0, 2, 0, 0, 0, 0, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
0, 0, 3, 0, 0, 2, 0, 0, 0, 0, 0, 1,
		

A349693 Dirichlet convolution of the ruler function (A001511) with itself.

Original entry on oeis.org

1, 4, 2, 10, 2, 8, 2, 20, 3, 8, 2, 20, 2, 8, 4, 35, 2, 12, 2, 20, 4, 8, 2, 40, 3, 8, 4, 20, 2, 16, 2, 56, 4, 8, 4, 30, 2, 8, 4, 40, 2, 16, 2, 20, 6, 8, 2, 70, 3, 12, 4, 20, 2, 16, 4, 40, 4, 8, 2, 40, 2, 8, 6, 84, 4, 16, 2, 20, 4, 16, 2, 60, 2, 8, 6, 20, 4, 16, 2, 70
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 25 2021

Keywords

Comments

Dirichlet convolution of A000005 with A104117. - Ridouane Oudra, Jul 23 2025

Crossrefs

Programs

  • Maple
    a:= n-> (f-> add(f(d)*f(n/d), d=numtheory[divisors](n)))(k-> padic[ordp](2*k, 2)):
    seq(a(n), n=1..80);  # Alois P. Heinz, Nov 25 2021
  • Mathematica
    Table[Sum[IntegerExponent[2 d, 2] IntegerExponent[2 n/d, 2], {d, Divisors[n]}], {n, 1, 80}]
    f[p_, e_] := If[p == 2, Binomial[e + 3, 3], e + 1]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 80] (* Amiram Eldar, Nov 25 2021 *)
  • PARI
    A001511(n) = (1+valuation(n,2));
    A349693(n) = sumdiv(n,d,A001511(n/d)*A001511(d)); \\ Antti Karttunen, Nov 25 2021
    
  • Python
    from sympy import divisor_count
    def A349693(n): return divisor_count(n)*(m:=(n&-n).bit_length()+1)*(m+1)//6 # Chai Wah Wu, Jul 13 2022

Formula

Dirichlet g.f.: zeta(s)^2 * 4^s / (2^s-1)^2.
a(n) = Sum_{d|n} A001511(d) * A001511(n/d).
a(n) = Sum_{d|n} A000217(A001511(d)).
Multiplicative with a(p^e) = binomial(e+3,3) if p = 2 and e+1 otherwise. - Amiram Eldar, Nov 25 2021
Sum_{k=1..n} a(k) ~ 4*n*(log(n) - 1 + 2*gamma - 2*log(2)), where gamma is the Euler-Mascheroni constant A001620. - Vaclav Kotesovec, Nov 26 2021
From Ridouane Oudra, Jul 23 2025: (Start)
a(n) = Sum_{i=0..A007814(n)} (i+1)*tau(n/2^i).
a(n) = Sum_{d|n} A115364(d).
a(n) = (1/6)*A090739(n)*A085058(n-1)*A000005(n).
a(n) = (1/6)*A001511(n)*A090739(n)*A099777(n).
a(n) = (1/3)*A115364(n)*A372784(n).
a(n) = A001227(n)*A000292(A001511(n)).
a(2*n+1) = tau(2*n+1).
a(2^k*(2*n+1)) = binomial(k+3, 3)*tau(2*n+1), for k, n >= 0. (End)
Showing 1-6 of 6 results.