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 1-2 of 2 results.

A185096 Let T(n) = n(n+1)/2 be the n-th triangular number (A000217); a(n) = T(8T(n)).

Original entry on oeis.org

0, 36, 300, 1176, 3240, 7260, 14196, 25200, 41616, 64980, 97020, 139656, 195000, 265356, 353220, 461280, 592416, 749700, 936396, 1155960, 1412040, 1708476, 2049300, 2438736, 2881200, 3381300, 3943836, 4573800, 5276376, 6056940, 6921060, 7874496, 8923200, 10073316, 11331180, 12703320, 14196456, 15817500, 17573556, 19471920, 21520080
Offset: 0

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Author

N. J. A. Sloane, Feb 18 2011

Keywords

References

  • C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See p. 4.

Crossrefs

Programs

  • Mathematica
    Table[2*n*(n + 1)*(2*n + 1)^2, {n, 0, 50}] (* G. C. Greubel, Jun 22 2017 *)
  • PARI
    for(n=0,50, print1(2*n*(n+1)*(2*n+1)^2, ", ")) \\ G. C. Greubel, Jun 22 2017

Formula

From G. C. Greubel, Jun 22 2017: (Start)
a(n) = 2*n*(n + 1)*(2*n + 1)^2.
G.f.: 12*x*(3 + 10*x + 3*x^2)/(1 - x)^5.
E.g.f.: 2*x*(18 + 57*x + 32*x^2 + 4*x^3)*exp(x). (End)

A190840 a(n+1) = 4*a(n)*(a(n)+1) for a(0) = 1.

Original entry on oeis.org

1, 8, 288, 332928, 443365544448, 786292024016459316676608, 2473020588127600939387543243786675530709484249088
Offset: 0

Views

Author

Alexander Zhukov, Aug 08 2011

Keywords

Comments

For n>0, subsequence of A132592: both a(n)/2 and a(n)+1 are squares.
All terms (n > 0) are divisible by 8, yielding all terms of A185097, which is indexed from n=1, thus having the first term A185097(1) = 1.
The next term has 98 digits. - Harvey P. Dale, Jan 01 2014
For n>0, subsequence of A060355: both a(n) and a(n)+1 are powerful numbers. - Bernard Schott, Apr 24 2023

Crossrefs

Programs

  • Mathematica
    NestList[4#(#+1)&,1,7] (* Harvey P. Dale, Jan 01 2014 *)

Formula

a(n+1) = 4*a(n)*(a(n)+1) for a(0) = 1.
a(n) = sinh(2^(n-2)*arccosh(17))^2. - Alexander R. Povolotsky, Aug 14 2011
a(n) = 8*A185097(n) for n > 0. - Alexander R. Povolotsky, Aug 14 2011
a(n) = (1 + sqrt(2))^(2^(n+1))/4 + (1 - sqrt(2))^(2^(n+1))/4 - 1/2. Therefore 2*a(n) + 1 = A001601(n+1). - Bruno Berselli, Feb 01 2017
Showing 1-2 of 2 results.