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 1-4 of 4 results.

A129194 a(n) = (n/2)^2*(3 - (-1)^n).

Original entry on oeis.org

0, 1, 2, 9, 8, 25, 18, 49, 32, 81, 50, 121, 72, 169, 98, 225, 128, 289, 162, 361, 200, 441, 242, 529, 288, 625, 338, 729, 392, 841, 450, 961, 512, 1089, 578, 1225, 648, 1369, 722, 1521, 800, 1681, 882, 1849, 968, 2025, 1058, 2209, 1152, 2401, 1250, 2601, 1352
Offset: 0

Views

Author

Paul Barry, Apr 02 2007

Keywords

Comments

The numerator of the integral is 2,1,2,1,2,1,...; the moments of the integral are 2/(n+1)^2. See 2nd formula.
The sequence alternates between twice a square and an odd square, A001105(n) and A016754(n).
Partial sums of the positive elements give the absolute values of A122576. - Omar E. Pol, Aug 22 2011
Partial sums of the positive elements give A212760. - Omar E. Pol, Dec 28 2013
Conjecture: denominator of 4/n - 2/n^2. - Wesley Ivan Hurt, Jul 11 2016
Multiplicative because both A000290 and A040001 are. - Andrew Howroyd, Jul 25 2018

References

  • G. Pólya and G. Szegő, Problems and Theorems in Analysis II (Springer 1924, reprinted 1976), Part Eight, Chap. 1, Sect. 7, Problem 73.

Links

Crossrefs

Cf. A000290, A001105, A007434, A010713, A016742, A016754, A040001.
Cf. A061038, A061040, A061050, A105398, A129204, A152020, A222171, A309337.

Programs

Formula

G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4)/(1-x^2)^3.
a(n+1) = denominator((1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*n*t)(-((Pi-t)/i)^2)), i=sqrt(-1).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n > 5. - Paul Curtz, Mar 07 2011
a(n) is the numerator of the coefficient of x^4 in the Maclaurin expansion of exp(-n*x^2). - Francesco Daddi, Aug 04 2011
O.g.f. as a Lambert series: x*Sum_{n >= 1} J_2(n)*x^n/(1 + x^n), where J_2(n) denotes the Jordan totient function A007434(n). See Pólya and Szegő. - Peter Bala, Dec 28 2013
From Ilya Gutkovskiy, Jul 11 2016: (Start)
E.g.f.: x*((2*x + 1)*sinh(x) + (x + 2)*cosh(x))/2.
Sum_{n>=1} 1/a(n) = 5*Pi^2/24. [corrected by Amiram Eldar, Sep 11 2022] (End)
a(n) = A000290(n) / A040001(n). - Andrew Howroyd, Jul 25 2018
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 (A222171). - Amiram Eldar, Sep 11 2022
From Peter Bala, Jan 16 2024: (Start)
a(n) = Sum_{1 <= i, j <= n} (-1)^(1 + gcd(i,j,n)) = Sum_{d | n} (-1)^(d+1) * J_2(n/d), that is, the Dirichlet convolution of the pair of multiplicative functions f(n) = (-1)^(n+1) and the Jordan totient function J_2(n) = A007434(n). Hence this sequence is multiplicative. Cf. A193356 and A309337.
Dirichlet g.f.: (1 - 2/2^s)*zeta(s-2). (End)
a(n) = Sum_{1 <= i, j <= n} (-1)^(n + gcd(i, n)*gcd(j, n)) = Sum_{d|n, e|n} (-1)^(n+e*d) * phi(n/d)*phi(n/e). - Peter Bala, Jan 22 2024

Extensions

More terms from Michel Marcus, Dec 28 2013

A308422 a(n) = n^2 if n odd, 3*n^2/4 if n even.

Original entry on oeis.org

0, 1, 3, 9, 12, 25, 27, 49, 48, 81, 75, 121, 108, 169, 147, 225, 192, 289, 243, 361, 300, 441, 363, 529, 432, 625, 507, 729, 588, 841, 675, 961, 768, 1089, 867, 1225, 972, 1369, 1083, 1521, 1200, 1681, 1323, 1849, 1452, 2025, 1587, 2209, 1728, 2401, 1875, 2601, 2028, 2809, 2187, 3025
Offset: 0

Views

Author

Ilya Gutkovskiy, May 26 2019

Keywords

Comments

Moebius transform of A076577.

Links

Crossrefs

Cf. A000290, A007434, A016754, A026741, A033428, A076577, A308418, A309337.

Programs

  • Mathematica
    a[n_] := If[OddQ[n], n^2, 3 n^2/4]; Table[a[n], {n, 0, 55}]
nmax = 55; CoefficientList[Series[x (1 + 3 x + 6 x^2 + 3 x^3 + x^4)/(1 - x^2)^3, {x, 0, nmax}], x]
LinearRecurrence[{0, 3, 0, -3, 0, 1}, {0, 1, 3, 9, 12, 25}, 56]
Table[(7 - (-1)^n) n^2/8, {n, 0, 55}]

Formula

G.f.: x*(1 + 3*x + 6*x^2 + 3*x^3 + x^4)/(1 - x^2)^3.
G.f.: Sum_{k>=1} J_2(k)*x^k/(1 - x^(2*k)), where J_2() is the Jordan function (A007434).
E.g.f.: x*((4 + 3*x)*cosh(x) + (3 + 4*x)*sinh(x))/4.
Dirichlet g.f.: zeta(s-2)*(1 - 1/2^s).
a(n) = (7 - (-1)^n)*n^2/8.
a(n) = Sum_{d|n, n/d odd} J_2(d).
a(2*k+1) = A016754(k), a(2*k) = A033428(k).
Sum_{n>=1} 1/a(n) = 13*Pi^2/72 = 1.7820119057522453061...
Sum_{n>=1} (-1)^(n+1)/a(n) = 5*Pi^2/72 = 0.68538919452009434853...
Multiplicative with a(2^e) = 3*2^(2*e-2), and a(p^e) = p^(2*e) for odd primes p. - Amiram Eldar, Oct 26 2020
For n >= 1, n*a(n) = A309337(n) = Sum_{d divides n} (-1)^(d+1) * J(3, n/d), where the Jordan totient function J_3(n) = A059376. - Peter Bala, Jan 21 2024

A129204 The denominator of 2/n^3.

Original entry on oeis.org

1, 1, 4, 27, 32, 125, 108, 343, 256, 729, 500, 1331, 864, 2197, 1372, 3375, 2048, 4913, 2916, 6859, 4000, 9261, 5324, 12167, 6912, 15625, 8788, 19683, 10976, 24389, 13500, 29791, 16384, 35937, 19652, 42875, 23328, 50653, 27436, 59319, 32000
Offset: 0

Views

Author

Paul Barry, Apr 03 2007

Keywords

Comments

Take two consecutive triangular numbers t1 and t2 and create a triangle using (0,0), (t1,t2) and (t2,t1). The area of this triangle will be ((n+1)^3)/2 for t1 = n*(n+1)/2. - J. M. Bergot, May 08 2012
Multiplicative because both A000578 and A040001 are. - Andrew Howroyd, Jul 26 2018

Links

Crossrefs

Cf. A000578, A040001, A129194, A309337.

Programs

  • Magma
    [1] cat [Denominator(2/n^3): n in [1..40]]; // Vincenzo Librandi, Jul 26 2018
  • Mathematica
    Join[{1}, Table[Denominator[2 / n^3], {n, 100}]] (* Vincenzo Librandi, Jul 26 2018 *)
  • PARI
    a(n) = if(n < 1, n==0, lcm(2, n^3)/2) \\ Andrew Howroyd, Jul 25 2018

Formula

G.f.: (1+x+23x^2+22x^4+23x^5+x^7+x^8)/(1-x^2)^4.
a(n) = 0^n + n^3*(3/4 - (-1)^n/4).
a(n+1) = A129196(n)*(5/3 + (4/3)*cos(2*Pi*(n+1)/3)).
a(2n) = 4n^3, a(2n+1) = (2n+1)^3.
a(n) = A000578(n) / A040001(n). - Andrew Howroyd, Jul 25 2018
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + 9*zeta(3)/8.
Sum_{n>=0} (-1)^n/a(n) = 1 - 5*zeta(3)/8. (End)
From Peter Bala, Jan 21 2024: (Start)
For n >= 1, a(n) = n*A129194(n) = n*Sum_{d divides n} (-1)^(d+1)*J(2,n/d), where the Jordan totient function J_2(n) = A007434(n). Cf. A309337.
Dirichlet g.f. for sequence without the a(0) term: (1 - 4/2^s)*zeta(s-3). (End)

Extensions

More terms from Vincenzo Librandi, Jul 26 2018

A309338 a(n) = n^4 if n odd, 7*n^4/8 if n even.

Original entry on oeis.org

0, 1, 14, 81, 224, 625, 1134, 2401, 3584, 6561, 8750, 14641, 18144, 28561, 33614, 50625, 57344, 83521, 91854, 130321, 140000, 194481, 204974, 279841, 290304, 390625, 399854, 531441, 537824, 707281, 708750, 923521, 917504, 1185921, 1169294, 1500625, 1469664, 1874161, 1824494
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 24 2019

Keywords

Comments

Moebius transform of A284900.

Links

Crossrefs

Cf. A000583, A016756, A059377, A129194, A193356, A284900, A309337.

Programs

  • Mathematica
    a[n_] := If[OddQ[n], n^4, 7 n^4/8]; Table[a[n], {n, 0, 38}]
nmax = 38; CoefficientList[Series[x (1 + 14 x + 76 x^2 + 154 x^3 + 230 x^4 + 154 x^5 + 76 x^6 + 14 x^7 + x^8)/(1 - x^2)^5, {x, 0, nmax}], x]
LinearRecurrence[{0, 5, 0, -10, 0, 10, 0, -5, 0, 1}, {0, 1, 14, 81, 224, 625, 1134, 2401, 3584, 6561}, 39]
Table[n^4 (15 - (-1)^n)/16, {n, 0, 38}]

Formula

G.f.: x * (1 + 14*x + 76*x^2 + 154*x^3 + 230*x^4 + 154*x^5 + 76*x^6 + 14*x^7 + x^8)/(1 - x^2)^5.
G.f.: Sum_{k>=1} J_4(k) * x^k/(1 + x^k), where J_4() is the Jordan function (A059377).
Dirichlet g.f.: zeta(s-4) * (1 - 2^(1-s)).
a(n) = n^4 * (15 - (-1)^n)/16.
a(n) = Sum_{d|n} (-1)^(n/d + 1) * J_4(d).
Sum_{n>=1} 1/a(n) = 113*Pi^4/10080 = 1.091986834012130496797...
Multiplicative with a(2^e) = 7*2^(4*e-3), and a(p^e) = p^(4*e) for odd primes p. - Amiram Eldar, Oct 26 2020
Showing 1-4 of 4 results.

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