A101096 Third differences of fifth powers (A000584).
1, 29, 150, 390, 750, 1230, 1830, 2550, 3390, 4350, 5430, 6630, 7950, 9390, 10950, 12630, 14430, 16350, 18390, 20550, 22830, 25230, 27750, 30390, 33150, 36030, 39030, 42150, 45390, 48750, 52230, 55830, 59550, 63390, 67350, 71430, 75630, 79950, 84390, 88950
Offset: 1
Links
- Danny Rorabaugh, Table of n, a(n) for n = 1..10000
- Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
- David J. Pengelley, The bridge between the continuous and the discrete via original sources in Study the Masters: The Abel-Fauvel Conference [pdf], Kristiansand, 2002, (ed. Otto Bekken et al.), National Center for Mathematics Education, University of Gothenburg, Sweden, 2003.
- Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Archive machine link]
- Cecilia Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube [Cached copy, May 15 2013]
- Eric Weisstein, Link to section of MathWorld: Worpitzky's Identity of 1883.
- Eric Weisstein, Link to section of MathWorld: Eulerian Number.
- Eric Weisstein, Link to section of MathWorld: Nexus number.
- Eric Weisstein, Link to section of MathWorld: Finite Differences.
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Programs
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Magma
m:=40; R
:=PowerSeriesRing(Integers(), m); Coefficients(R!( x*(x^4+26*x^3+66*x^2+26*x+1)/(1-x)^3)); // G. C. Greubel, Dec 01 2018 -
Mathematica
MagicNKZ=Sum[(-1)^j*Binomial[n+1-z, j]*(k-j+1)^n, {j, 0, k+1}];Table[MagicNKZ, {n, 5, 5}, {z, 3, 3}, {k, 0, 34}] CoefficientList[Series[(-z^4-26z^3-66z^2-26z-1)/(z-1)^3, {z, 0, 200}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 19 2011 *) Join[{1,29},Differences[Range[0,40]^5,3]] (* or *) LinearRecurrence[{3,-3,1},{1,29,150,390,750},40] (* Harvey P. Dale, Feb 02 2017 *) -
PARI
a(n)=if(n>2,60*n^2-180*n+150,28*n-27) \\ Charles R Greathouse IV, Oct 11 2015
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Sage
[sum([(-1)^j*binomial(3, j)*(k-j+1)^5 for j in range(min(k+2,4))]) for k in range(40)] # Danny Rorabaugh, Apr 27 2015
Formula
a(k+1) = MagicNKZ(5,k,3) where MagicNKZ(n,k,z) = Sum_{j=0..k+1} (-1)^j*binomial(n+1-z,j)*(k-j+1)^n. (Cf. A101095.)
a(n+1) = 30*(1 - 2*n + 2*n^2) for n>2.
a(n+3) = A069477(n). - Vladimir Joseph Stephan Orlovsky, Jun 19 2011
G.f.: x*(x^4+26*x^3+66*x^2+26*x+1)/(1-x)^3. - Colin Barker, Oct 17 2012
Sum_{n>=1} 1/a(n) = (Pi/60)*tanh(Pi/2) + 871/870. - Amiram Eldar, Jan 27 2022
Extensions
MagicNKZ material edited and SeriesAtLevelR material removed by Danny Rorabaugh, Apr 27 2015
Comments