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 1-3 of 3 results.

A062158 a(n) = n^3 - n^2 + n - 1 = (n-1) * (n^2 + 1).

Original entry on oeis.org

-1, 0, 5, 20, 51, 104, 185, 300, 455, 656, 909, 1220, 1595, 2040, 2561, 3164, 3855, 4640, 5525, 6516, 7619, 8840, 10185, 11660, 13271, 15024, 16925, 18980, 21195, 23576, 26129, 28860, 31775, 34880, 38181, 41684, 45395, 49320, 53465, 57836, 62439, 67280, 72365, 77700, 83291, 89144, 95265, 101660
Offset: 0

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Author

Henry Bottomley, Jun 08 2001

Keywords

Comments

Number of walks of length 4 between any two distinct vertices of the complete graph K_{n+1} (n >= 1). Example: a(2) = 5 because in the complete graph ABC we have the following walks of length 4 between A and B: ABACB, ABCAB, ACACB, ACBAB and ACBCB. - Emeric Deutsch, Apr 01 2004
1/a(n) for n >= 2, is in base n given by 0.repeat(0,0,1,1), due to (1/n^3 + 1/n^4)*(1/(1-1/n^4)) = 1/((n-1)*(n^2+1)). - Wolfdieter Lang, Jun 20 2014
For n>3, a(n) is 1220 in base n-1. - Bruno Berselli, Jan 26 2016
For odd n, a(n) * (n+1) / 2 + 1 also represents the first integer in a sum of n^4 consecutive integers that equals n^8. - Patrick J. McNab, Dec 26 2016

Examples

			a(4) = 4^3 - 4^2 + 4 - 1 = 64 - 16 + 4 - 1 = 51.
		

Crossrefs

Programs

  • Magma
    [n^3 - n^2 + n - 1 : n in [0..50]]; // Wesley Ivan Hurt, Dec 26 2016
  • Maple
    [seq(n^3-n^2+n-1,n=0..49)]; # Zerinvary Lajos, Jun 29 2006
    a:=n->sum(1+sum(n, k=1..n), k=2..n):seq(a(n), n=0...43); # Zerinvary Lajos, Aug 24 2008
  • Mathematica
    Table[n^3 - n^2 + n - 1, {n, 0, 49}] (* Alonso del Arte, Apr 30 2014 *)
  • PARI
    a(n) = { n*(n*(n - 1) + 1) - 1 } \\ Harry J. Smith, Aug 02 2009
    

Formula

a(n) = round(n^4/(n+1)) for n >= 2.
a(n) = A062160(n, 4), for n > 2.
G.f.: (4*x-1)*(1+x^2)/(1-x)^4 (for the signed sequence). - Emeric Deutsch, Apr 01 2004
a(n) = floor(n^5/(n^2+n)) for n > 0. - Gary Detlefs, May 27 2010
a(n) = -A053698(-n). - Bruno Berselli, Jan 26 2016
Sum_{n>=2} 1/a(n) = A268086. - Amiram Eldar, Nov 18 2020
E.g.f.: exp(x)*(x^3 + 2*x^2 + x - 1). - Stefano Spezia, Apr 22 2023

Extensions

More terms from Emeric Deutsch, Apr 01 2004

A339135 Decimal expansion of J = 2*log(2)/3 - Re(Psi(1/2 + i*sqrt(3)/2)), where Psi is the digamma function and i=sqrt(-1).

Original entry on oeis.org

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

Views

Author

Artur Jasinski, Nov 25 2020

Keywords

Comments

Generally in the literature there is no explicit formula for the real part of the function Psi(x + i*y) when y != 0.
Up to now there is no explicit formula expressing the constant J in terms of other mathematical constants.

Examples

			J = 0.677024679102993347...
		

Crossrefs

Programs

  • Maple
    evalf(1 + 2*log(2)/3 - Psi(0, 5/2 - sqrt(3)*I/2)/2 - Psi(0, 5/2 + sqrt(3)*I/2)/2, 100); # Vaclav Kotesovec, Nov 26 2020
  • Mathematica
    RealDigits[N[Re[2 Log[2]/3 - PolyGamma[0, 1/2 + I Sqrt[3]/2]], 110]][[1]]
    Chop[N[1 + 2*Log[2]/3 - PolyGamma[0, 5/2 - I*Sqrt[3]/2]/2 - PolyGamma[0, 5/2 + I*Sqrt[3]/2]/2, 120]] (* Vaclav Kotesovec, Nov 26 2020 *)
  • PARI
    2*log(2)/3 - real(psi(1/2 + I*sqrt(3)/2)) \\ Michel Marcus, Nov 25 2020

Formula

J = -log(2)/3 - (1/2)*Pi/cosh(Pi*sqrt(3)/2) - Re(Psi(1/4 + i*sqrt(3)/4)).
J = -log(2)/3 + (1/2)*Pi/cosh(Pi*sqrt(3)/2) - Re(Psi(3/4 + i*sqrt(3)/4)).
J = 3 + gamma + (2/3)*log(2) - (1/2)* sqrt(3)*Pi*tanh(Pi*sqrt(3)/2) - 3*(Sum_{n>=1} zeta(3*n)-1), where zeta is Riemann zeta function and gamma is Euler gamma constant see A001620.
J = -(1/2) + gamma + (2/3)*log(2) + (3/2)*(Sum_{n>=1} zeta(3*n+1)-1).
J = -1 + gamma + (2/3)*log(2) + (1/2)*sqrt(3)*Pi*tanh(Pi*sqrt(3)/2) - 3*(Sum_{n>=0} zeta(3*n+2)-1).
J = -(3/8) + gamma + (2/3)*log(2) + (3/2)*(Sum_{n>=1} zeta(6*n+1)-1).
J = 1/4 + gamma + (2/3)*log(2) - 3*(Sum_{n>=0} zeta(6*n+3)-1).
J = -(1/4) + gamma + (2/3)*log(2) - 3 (Sum_{n>=0} zeta(6*n+5)-1).
J = (11/12 - (1/4)*i*sqrt(3))*Psi(1/2 + i*sqrt(3)/2) + (-(5/4) + (1/4)*i*sqrt(3))*Psi(-(1/2) - i*sqrt(3)/2) + (-(17/24) + (1/8)*i*sqrt(3))* Psi(1/4 + i*sqrt(3)/4) + (3/8 - (1/8)*i*sqrt(3))*Psi(-(1/4) - i*sqrt(3)/4) + (-(17/24) + (1/8)*i*sqrt(3))*Psi(3/4 + i*sqrt(3)/4) + (3/8 - (1/8)*i*sqrt(3))*Psi(-(3/4) - i*sqrt(3)/4).
J = 2*log(2)/3 - Integral_{t=0..infinity} cosh(t)/t - sinh(t)/t - (cos(sqrt(3)*t)*cosh(t/2))/(1 - cosh(t) + sinh(t)) + (cos(sqrt(3)*t)*sinh(t/2))/(1 - cosh(t) + sinh(t)).
J = gamma + (1/6)*Sum_{t>=1} (6*t^3-4*t^2-4*t-1)/(t*(t+1)*(2t+1)*(t^2+t+1)).
Equals 1 + 2*log(2)/3 - Psi(0, 5/2 - i*sqrt(3)/2)/2 - Psi(0, 5/2 + i*sqrt(3)/2)/2. - Vaclav Kotesovec, Nov 26 2020

A268046 Decimal expansion of Sum_{k>0} (k+1)/(k*((k+1)^2+1)).

Original entry on oeis.org

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

Views

Author

Bruno Berselli, Jan 26 2016

Keywords

Comments

Also, decimal expansion of Integral_{x=0..1} (2 - (1+i)*x^(1-i) - (1-i)*x^(1+i))/(4 - 4*x) dx, where i is the imaginary unit.

Examples

			.8742700164962955060062206837771568219820392870536409438069097216969635...
		

Crossrefs

Cf. A268086: (1-i)*(H(1-i)+i*H(1+i))/4.

Programs

  • Maple
    ((1+I)*(harmonic(1-I)-I*harmonic(1+I)))/4:
    Re(evalf(%, 101)); # Peter Luschny, Jan 27 2016
  • Mathematica
    (1 + I)*(HarmonicNumber[1 - I] - I*HarmonicNumber[1 + I])/4 // Re // RealDigits[#, 10, 100]& // First (* Jean-François Alcover, Jan 26 2016 *)
  • Sage
    # Warning: Floating point calculation. Adjust precision as needed
    # and use some guard digits!
    from mpmath import mp, chop, psi, pi, coth
    mp.dps = 109; mp.pretty = True
    chop((psi(0,I-1)-psi(0,1)-I-1)/2-pi*(I-1)*coth(pi)/4) # Peter Luschny, Jan 27 2016

Formula

Equals (1 + i)*(H(1-i) - i*H(1+i))/4, where H(z) is a harmonic number with complex argument.
Equals (Psi(i-1)-Psi(1)-i-1)/2 - Pi*(i-1)*coth(Pi)/4, where Psi(x) is the digamma function. - Peter Luschny, Jan 27 2016
Showing 1-3 of 3 results.