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.

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A231535 Decimal expansion of Pi^4/15.

Original entry on oeis.org

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

Views

Author

Stanislav Sykora, Nov 12 2013

Keywords

Comments

Under proper scaling, the radiation distribution density function in terms of frequency is given by prl(x) = x^3/(exp(x)-1), the Planck's radiation law. This constant is the integral of prl(x) from 0 to infinity and leads to the total amount of electromagnetic radiation emitted by a body.
Also, in an 8-dimensional unit-radius hypersphere, equals one-fifth of its surface (A164109), and twice the integral of r^2 over its volume.

Examples

			6.4939394022668291490960221792470074166485057115123614460978572926647...
		

Crossrefs

Programs

Formula

Equals 6*zeta(4), see A013662. - Bruno Berselli, Nov 12 2013
Equals Gamma(4)*zeta(4) = 2*3*Product_{prime p} (p^4/(p^4-1)). - Stanislav Sykora, Oct 20 2014
Equals Sum_{n, m >= 1} H(n+m)/(n*m*(n+m)) where H(n) is the n-th harmonic number. See Aliev and Dil. - Michel Marcus, Aug 07 2020
From Amiram Eldar, Aug 14 2020: (Start)
Equals Integral_{x=0..oo} x^3/(exp(x)-1) dx.
Equals Integral_{x=0..1} log(x)^3/(x-1) dx. (End)
Equals psi'''(1), the third derivative of the digamma function at 1. - R. J. Mathar, Aug 29 2023

A240976 Decimal expansion of 3*zeta(3)/(2*Pi^2), a constant appearing in the asymptotic evaluation of the average LCM of two integers chosen independently from the uniform distribution [1..n].

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Aug 07 2014

Keywords

Comments

15*zeta(3)/Pi^2 = 10 * (this constant) equals the asymptotic mean of the abundancy index of the squares (Jakimczuk and Lalín, 2022). - Amiram Eldar, May 12 2023

Examples

			0.18269074235035962468150918282692865988200290126984361751783...
		

Crossrefs

Programs

  • Mathematica
    RealDigits[3*Zeta[3]/(2*Pi^2), 10, 102] // First

Formula

Equals zeta(3)/(4*zeta(2)) = 3*zeta(3)/(2*Pi^2).
From Amiram Eldar, Jan 25 2024: (Start)
Equals (1/10) * Sum_{k>=1} A000188(k)/k^2.
Equals (1/10) * Sum_{k>=1} A048250(k)/k^3. (End)

A248227 Least k such that zeta(4) - sum{1/h^4, h = 1..k} < 1/n^3.

Original entry on oeis.org

1, 1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 17, 18, 19, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 37, 37, 38, 39, 40, 40, 41, 42, 42, 43, 44, 44, 45, 46, 46, 47
Offset: 1

Views

Author

Clark Kimberling, Oct 05 2014

Keywords

Comments

This sequence and A248230 provide insight into the manner of convergence of sum{1/h^4, h = 0..k}. Since a(n+1) - a(n) is in {0,1} for n >= 1, A248228 and A248229 are complementary.

Examples

			Let s(n) = sum{1/h^4, h = 1..n}.  Approximations are shown here:
n ... zeta(4) - s(n) ... 1/n^3
1 ... 0.0823232 .... 1
2 ... 0.0198232 .... 0.125
3 ... 0.0074775 .... 0.037
4 ... 0.0035713 .... 0.015
5 ... 0.0019713 .... 0.008
6 ... 0.0011997 .... 0.005
a(6) = 4 because zeta(4) - s(4) < 1/216 < zeta(4) - s(3).
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = Infinity; z = 400; p[k_] := p[k] = Sum[1/h^4, {h, 1, k}];
    N[Table[Zeta[4] - p[n], {n, 1, z/10}]]
    f[n_] := f[n] = Select[Range[z], Zeta[4] - p[#] < 1/n^3 &, 1];
    u = Flatten[Table[f[n], {n, 1, z}]]   (* A248227 *)
    Flatten[Position[Differences[u], 0]]  (* A248228 *)
    Flatten[Position[Differences[u], 1]]  (* A248229 *)
    f = Table[Floor[1/(Zeta[4] - p[n])], {n, 1, z}]  (* A248230 *)

Formula

a(n) ~ 3^(-1/3) * n. - Vaclav Kotesovec, Oct 09 2014

A282213 Coefficients in q-expansion of (E_2^3*E_4 - 3*E_2^2*E_6 + 3*E_2*E_4^2 - E_4*E_6)/3456, where E_2, E_4, E_6 are the Eisenstein series shown in A006352, A004009, A013973, respectively.

Original entry on oeis.org

0, 1, 72, 756, 4672, 15750, 54432, 117992, 299520, 551853, 1134000, 1772892, 3532032, 4829006, 8495424, 11907000, 19173376, 24142482, 39733416, 47052740, 73584000, 89201952, 127648224, 148048056, 226437120, 246109375, 347688432, 402320520, 551258624, 594847710
Offset: 0

Views

Author

Seiichi Manyama, Feb 09 2017

Keywords

Comments

Multiplicative because A001158 is. - Andrew Howroyd, Jul 25 2018

Examples

			a(6) = 1^6*6^3 + 2^6*3^3 + 3^6*2^3 + 6^6*1^3 = 54432.
		

Crossrefs

Cf. A282211 (phi_{4, 3}), this sequence (phi_{6, 3}).
Cf. A006352 (E_2), A004009 (E_4), A013973 (E_6), A282586 (E_2^3*E_4), A282595 (E_2^2*E_6), A282101 (E_2*E_4^2), A013974 (E_4*E_6 = E_10).
Cf. A001158 (sigma_3(n)), A281372 (n*sigma_3(n)), A282099 (n^2*sigma_3(n)), this sequence (n^3*sigma_3(n))
Cf. A013662.

Programs

  • Mathematica
    terms = 30;
    E2[x_] = 1 - 24*Sum[k*x^k/(1 - x^k), {k, 1, terms}];
    E4[x_] = 1 + 240*Sum[k^3*x^k/(1 - x^k), {k, 1, terms}];
    E6[x_] = 1 - 504*Sum[k^5*x^k/(1 - x^k), {k, 1, terms}];
    (E2[x]^3*E4[x] - 3 E2[x]^2*E6[x] + 3 E2[x] E4[x]^2 - E4[x] E6[x])/3456 + O[x]^terms // CoefficientList[#, x]&
    (* or: *)
    Table[n^3*DivisorSigma[3, n], {n, 0, terms-1}] (* Jean-François Alcover, Feb 27 2018 *)
    nmax = 30; CoefficientList[Series[Sum[k^6*x^k*(x^(2*k) + 4*x^k + 1)/(1 - x^k)^4, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 02 2025 *)
  • PARI
    a(n) = if (n, n^3*sigma(n, 3), 0); \\ Michel Marcus, Feb 27 2018

Formula

G.f.: phi_{6, 3}(x) where phi_{r, s}(x) = Sum_{n, m>0} m^r * n^s * x^{m*n}.
a(n) = (A282586(n) - 3*A282595(n) + 3*A282101(n) - A013974(n))/3456. - Seiichi Manyama, Feb 19 2017
a(n) = n^3*A001158(n) for n > 0. - Seiichi Manyama, Feb 19 2017
Sum_{k=1..n} a(k) ~ zeta(4) * n^7 / 7. - Amiram Eldar, Sep 06 2023
From Amiram Eldar, Oct 31 2023: (Start)
Multiplicative with a(p^e) = p^(3*e) * (p^(3*e+3)-1)/(p^3-1).
Dirichlet g.f.: zeta(s-3)*zeta(s-6). (End)
G.f.: Sum_{k>=1} k^6*x^k*(x^(2*k) + 4*x^k + 1)/(1 - x^k)^4. - Vaclav Kotesovec, Aug 02 2025

A363604 Expansion of Sum_{k>0} x^(2*k)/(1-x^k)^4.

Original entry on oeis.org

0, 1, 4, 11, 20, 40, 56, 95, 124, 186, 220, 336, 364, 512, 584, 775, 816, 1129, 1140, 1526, 1600, 1992, 2024, 2720, 2620, 3290, 3400, 4176, 4060, 5280, 4960, 6231, 6208, 7362, 7216, 9195, 8436, 10280, 10248, 12270, 11480, 14432, 13244, 16192, 15884, 18240
Offset: 1

Views

Author

Seiichi Manyama, Jun 11 2023

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := (DivisorSigma[3, n] - DivisorSigma[1, n])/6; Array[a, 50] (* Amiram Eldar, Jul 25 2023 *)
  • PARI
    my(N=50, x='x+O('x^N)); concat(0, Vec(sum(k=1, N, x^(2*k)/(1-x^k)^4)))
    
  • PARI
    a(n) = my(f = factor(n)); (sigma(f, 3) - sigma(f))/6; \\ Amiram Eldar, Dec 30 2024

Formula

a(n) = (sigma_3(n) - sigma(n))/6 = A092348(n)/6.
G.f.: Sum_{k>0} binomial(k+1,3) * x^k/(1 - x^k).
From Amiram Eldar, Dec 30 2024: (Start)
Dirichlet g.f.: zeta(s) * (zeta(s-3) - zeta(s-1)) / 6.
Sum_{k=1..n} a(k) ~ (zeta(4)/24) * n^4. (End)

A372952 a(n) = Sum_{1 <= x_1, x_2, x_3 <= n} n/gcd(x_1, x_2, x_3, n).

Original entry on oeis.org

1, 15, 79, 239, 621, 1185, 2395, 3823, 6397, 9315, 14631, 18881, 28549, 35925, 49059, 61167, 83505, 95955, 130303, 148419, 189205, 219465, 279819, 302017, 388121, 428235, 518155, 572405, 707253, 735885, 923491, 978671, 1155849, 1252575, 1487295, 1528883
Offset: 1

Views

Author

Seiichi Manyama, May 18 2024

Keywords

Crossrefs

Column k=3 of A372968.

Programs

  • Mathematica
    f[p_, e_] := (p^(4*e+4) - p^(4*e+1) + p - 1)/(p^4-1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, May 21 2024 *)
  • PARI
    a(n) = sumdiv(n, d, moebius(n/d)*n/d*sigma(d,4));

Formula

a(n) = Sum_{d|n} mu(n/d) * (n/d) * sigma_4(d).
From Amiram Eldar, May 21 2024: (Start)
Multiplicative with a(p^e) = (p^(4*e+4) - p^(4*e+1) + p - 1)/(p^4-1).
Dirichlet g.f.: zeta(s)*zeta(s-4)/zeta(s-1).
Sum_{k=1..n} a(k) ~ c * n^5 / 5, where c = zeta(5)/zeta(4) = 0.958057374... . (End)
a(n) = Sum_{d|n} phi(n/d) * (n/d) * sigma_4(d^2)/sigma_2(d^2). - Seiichi Manyama, May 24 2024
a(n) = Sum_{1 <= x_1, x_2, x_3 <= n} ( gcd(x_1, x_2, n)/gcd(x_1, x_2, x_3, n) )^3. - Seiichi Manyama, May 25 2024

A015634 Number of ordered quadruples of integers from [ 1..n ] with no global factor.

Original entry on oeis.org

1, 4, 13, 29, 63, 106, 189, 289, 444, 626, 911, 1203, 1657, 2130, 2766, 3462, 4430, 5359, 6688, 7992, 9670, 11405, 13704, 15840, 18730, 21548, 25037, 28521, 33015, 37067, 42522, 47690, 53940, 60108, 67760, 74748, 83886, 92433, 102629, 112469, 124809, 135763, 149952
Offset: 1

Views

Author

Keywords

Crossrefs

Column k=4 of A177976.
Partial sums of A117108.

Programs

  • Mathematica
    a[n_] := Sum[DivisorSum[k, MoebiusMu[k/#]*Binomial[# + 2, 3] &], {k, 1, n}]; Array[a, 45] (* Amiram Eldar, Jun 07 2025 *)
  • PARI
    a(n) = sum(k=1, n, sumdiv(k, d, moebius(k/d)*binomial(d+2, 3))); \\ Seiichi Manyama, Jun 12 2021
    
  • PARI
    a(n) = binomial(n+3, 4)-sum(k=2, n, a(n\k)); \\ Seiichi Manyama, Jun 12 2021
    
  • PARI
    my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, moebius(k)*x^k/(1-x^k)^4)/(1-x)) \\ Seiichi Manyama, Jun 12 2021
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A015634(n):
        if n == 0:
            return 0
        c, j = 1, 2
        k1 = n//j
        while k1 > 1:
            j2 = n//k1 + 1
            c += (j2-j)*A015634(k1)
            j, k1 = j2, n//j2
        return n*(n+1)*(n+2)*(n+3)//24-c+j-n # Chai Wah Wu, Apr 18 2021
    

Formula

G.f.: (1/(1 - x)) * Sum_{k>=1} mu(k) * x^k / (1 - x^k)^4. - Ilya Gutkovskiy, Feb 14 2020
a(n) = n*(n+1)*(n+2)*(n+3)/24 - Sum_{j=2..n} a(floor(n/j)) = A000332(n+3) - Sum_{j=2..n} a(floor(n/j)). - Chai Wah Wu, Apr 18 2021
a(n) ~ n^4 / (24*zeta(4)). - Amiram Eldar, Jun 08 2025

A030140 The nonsquares squared.

Original entry on oeis.org

4, 9, 25, 36, 49, 64, 100, 121, 144, 169, 196, 225, 289, 324, 361, 400, 441, 484, 529, 576, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2500, 2601, 2704, 2809, 2916, 3025
Offset: 1

Views

Author

Keywords

Comments

The complement of the fourth powers A000583 within the squares A000290. - Peter Munn, Aug 20 2019

Examples

			a(1)=2^2, a(2)=3^2, a(3)=5^2, a(4)=6^2, a(5)=7^2, ..., a(n)=(integer which is not a perfect square)^2.
		

Crossrefs

Positions of 2's in A352080.
Related to A016945 via A225546.

Programs

  • Magma
    [(n + Floor(1/2 + Sqrt(n)))^2: n in [1..60]]; // Vincenzo Librandi, Apr 06 2020
    
  • Maple
    a:=proc(n) if type(sqrt(n),integer)=false then n^2 else fi end: seq(a(n),n=1..70); # Emeric Deutsch, Apr 11 2007
  • Mathematica
    a[n_] := (n + Floor[1/2 + Sqrt[n]])^2;
    Array[a, 50] (* Jean-François Alcover, Apr 05 2020 *)
  • Python
    from math import isqrt
    def A030140(n): return (n+(k:=isqrt(n))+int(n>=k*(k+1)+1))**2 # Chai Wah Wu, Jun 17 2024

Formula

a(n) = A000037(n)^2.
Sum_{n>=1} 1/a(n) = zeta(2) - zeta(4) = A013661 - A013662 = 0.5626108331... - Amiram Eldar, Nov 14 2020
{a(n) : n >= 1} = {A225546(6m+3) : m >= 0}. - Peter Munn, Nov 17 2022

Extensions

Edited by N. J. A. Sloane, Jul 02 2008 at the suggestion of R. J. Mathar

A052277 a(n) = (4n+2)!/2^(2n+1).

Original entry on oeis.org

1, 90, 113400, 681080400, 12504636144000, 548828480360160000, 49229914688306352000000, 8094874872198213459360000000, 2252447502438386084347676160000000, 997586474354936812896742294502400000000, 669959124447288464805194190141921792000000000
Offset: 0

Views

Author

N. J. A. Sloane, Feb 05 2000

Keywords

Crossrefs

Cf. A002432 (denominators of zeta(2*n)/Pi^(2*n)).
Cf. A068447, A067912, A013662 (zeta(4)).

Programs

  • Mathematica
    Table[(4n+2)!/2^(2n+1), {n, 0, 10}] (* Amiram Eldar, Feb 25 2022 *)
  • PARI
    a(n) = (4*n+2)!/2^(2*n+1); \\ Michel Marcus, Feb 20 2022

Formula

sin(x)*sinh(x) = Sum_{n>=0} (-1)^n*x^(4n+2)/a(n). - Benoit Cloitre, Feb 02 2002
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=0} 1/a(n) = (cosh(sqrt(2)) - cos(sqrt(2)))/2.
Sum_{n>=0} (-1)^n/a(n) = sin(1)*sinh(1). (End)

A183700 Decimal expansion of zeta(3)*zeta(4), the product of two Riemann zeta values.

Original entry on oeis.org

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

Views

Author

R. J. Mathar, Jan 06 2011

Keywords

Comments

Equals the Dirichlet zeta-function sum_{n>=1} A000203(n)/n^s at s=4.

Examples

			Equals 1.301014114532488574154...
		

Crossrefs

Cf. A183699 (s=3).

Programs

  • Maple
    evalf(Zeta(3)*Zeta(4)) ;
  • Mathematica
    RealDigits[N[Zeta[3] * Zeta[4], 75]][[1]] (* Alonso del Arte, Jan 06 2011 *)

Formula

Equals A002117 * A013662.
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