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.

A016941 a(n) = (6*n + 2)^9.

Original entry on oeis.org

512, 134217728, 20661046784, 512000000000, 5429503678976, 35184372088832, 165216101262848, 618121839509504, 1953125000000000, 5416169448144896, 13537086546263552, 31087100296429568, 66540410775079424, 134217728000000000, 257327417311663616, 472161363286556672
Offset: 0

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Crossrefs

Programs

  • Magma
    [(6*n+2)^9: n in [0..25]]; // Vincenzo Librandi, May 05 2011
  • Mathematica
    (6*Range[0,20]+2)^9 (* or *) LinearRecurrence[ {10,-45,120,-210,252,-210,120,-45,10,-1},{512,134217728,20661046784,512000000000,5429503678976,35184372088832,165216101262848,618121839509504,1953125000000000,5416169448144896},20] (* Harvey P. Dale, Sep 21 2013 *)

Formula

a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10). - Harvey P. Dale, Sep 21 2013
From Amiram Eldar, Mar 29 2022: (Start)
a(n) = A016933(n)^9 = A016935(n)^3.
a(n) = 2^9*A016785(n).
Sum_{n>=0} 1/a(n) = 809*Pi^9/(14285134080*sqrt(3)) + 9841*zeta(9)/10077696. (End)

A016942 a(n) = (6*n + 2)^10.

Original entry on oeis.org

1024, 1073741824, 289254654976, 10240000000000, 141167095653376, 1125899906842624, 6278211847988224, 27197360938418176, 97656250000000000, 303305489096114176, 839299365868340224, 2113922820157210624, 4923990397355877376, 10737418240000000000, 22130157888803070976
Offset: 0

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Programs

Formula

From Amiram Eldar, Mar 29 2022: (Start)
a(n) = A016933(n)^10 = A016934(n)^5 = A016937(n)^2.
a(n) = 2^10*A016786(n).
Sum_{n>=0} 1/a(n) = PolyGamma(9, 1/3)/21941965946880. (End)

A016943 a(n) = (6*n + 2)^11.

Original entry on oeis.org

2048, 8589934592, 4049565169664, 204800000000000, 3670344486987776, 36028797018963968, 238572050223552512, 1196683881290399744, 4882812500000000000, 16985107389382393856, 52036560683837093888, 143746751770690322432, 364375289404334925824, 858993459200000000000
Offset: 0

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Programs

Formula

a(n) = A016787(n)*2^11. - Zerinvary Lajos, Jun 22 2009
From Amiram Eldar, Mar 30 2022: (Start)
a(n) = A016933(n)^9 = A016935(n)^3.
Sum_{n>=0} 1/a(n) = 1847*Pi^11/(1285662067200*sqrt(3)) + 88573*zeta(11)/362797056. (End)

A016944 a(n) = (6*n + 2)^12.

Original entry on oeis.org

4096, 68719476736, 56693912375296, 4096000000000000, 95428956661682176, 1152921504606846976, 9065737908494995456, 52654090776777588736, 244140625000000000000, 951166013805414055936, 3226266762397899821056, 9774779120406941925376, 26963771415920784510976
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Magma
    [(6*n+2)^12: n in [0..20]]; // Vincenzo Librandi, May 05 2011
  • Mathematica
    (6*Range[0,20]+2)^12 (* or *) LinearRecurrence[{13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1},{4096,68719476736,56693912375296,4096000000000000,95428956661682176,1152921504606846976,9065737908494995456,52654090776777588736,244140625000000000000,951166013805414055936,3226266762397899821056,9774779120406941925376,26963771415920784510976},20] (* Harvey P. Dale, Aug 03 2021 *)

Formula

From Amiram Eldar, Mar 30 2022: (Start)
a(n) = A016933(n)^12 = A016934(n)^6 = A016935(n)^4 = A016936(n)^3 = A016938(n)^2.
a(n) = 2^12*A016788(n).
Sum_{n>=0} 1/a(n) = PolyGamma(11, 1/3)/86890185149644800. (End)
Showing 1-4 of 4 results.