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.

A168574 a(n) = (4*n + 3)*(1 + 2*n^2)/3.

Original entry on oeis.org

1, 7, 33, 95, 209, 391, 657, 1023, 1505, 2119, 2881, 3807, 4913, 6215, 7729, 9471, 11457, 13703, 16225, 19039, 22161, 25607, 29393, 33535, 38049, 42951, 48257, 53983, 60145, 66759, 73841, 81407, 89473, 98055, 107169, 116831, 127057, 137863, 149265, 161279
Offset: 0

Views

Author

Paul Curtz, Nov 30 2009

Keywords

Comments

Binomial transform of quasi-finite sequence 1, 6, 20, 16, 0, 0, ... (0 continued).
a(n+1) is the sum of the first and last number at the bottom (2nd row) of each block in A172002, 3+4, 13+20, 39+56, ...

Links

Crossrefs

Cf. A168547, A168234.

Programs

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 16.
a(n) = A168582(2*n+1) .
a(n+1) = A166911(n) + A002492(n+1).
G.f.: (1 + 3*x + 11*x^2 + x^3)/(1 - x)^4.
E.g.f.: (1/3)*(3 + 18*x + 30*x^2 + 8*x^3)*exp(x). - G. C. Greubel, Jul 26 2016

Extensions

Edited and extended by R. J. Mathar, Mar 25 2010

A168582 a(n) = (4*n^3 - 6*n^2 + 8*n + 9 + 3*(-1)^n)/12.

Original entry on oeis.org

1, 1, 3, 7, 17, 33, 59, 95, 145, 209, 291, 391, 513, 657, 827, 1023, 1249, 1505, 1795, 2119, 2481, 2881, 3323, 3807, 4337, 4913, 5539, 6215, 6945, 7729, 8571, 9471, 10433, 11457, 12547, 13703, 14929, 16225, 17595, 19039, 20561
Offset: 0

Views

Author

Paul Curtz, Nov 30 2009

Keywords

Comments

Starting with a(2), the sum of the first and last term in row n-1 of the Janet table A172002.

Links

Crossrefs

Cf. A137928 (first differences).

Programs

  • Magma
    [2*n/3 +3/4 -n^2/2 +n^3/3 +(-1)^n/4: n in [0..40]]; // Vincenzo Librandi, Aug 06 2011
  • Mathematica
    Table[(4*n^3 - 6*n^2 + 8*n + 9 + 3*(-1)^n)/12, {n,0,50}] (* G. C. Greubel, Jul 26 2016 *)
  • PARI
    a(n)=(4*n^3-6*n^2+8*n+9+3*(-1)^n)/12 \\ Charles R Greathouse IV, Jul 26 2016

Formula

a(n+2) = A168388(n) + A168380(n), n >= 0.
a(2n) = A168547(n);
a(2n+1) = A168574(n).
G.f.: (1 - 2*x + x^4 + 2*x^2 + 2*x^3)/((1+x)*(x-1)^4). - R. J. Mathar, Jun 27 2011
E.g.f.: (1/12)*((4*x^3 + 6*x^2 + 6*x + 9)*exp(x) + 3*exp(-x)). - G. C. Greubel, Jul 26 2016

A171272 a(n) = 1 + 4*n*(1 + 2*n^2)/3.

Original entry on oeis.org

1, 5, 25, 77, 177, 341, 585, 925, 1377, 1957, 2681, 3565, 4625, 5877, 7337, 9021, 10945, 13125, 15577, 18317, 21361, 24725, 28425, 32477, 36897, 41701, 46905, 52525, 58577, 65077, 72041, 79485, 87425, 95877, 104857, 114381, 124465, 135125, 146377, 158237, 170721, 183845
Offset: 0

Views

Author

Paul Curtz, Dec 06 2009

Keywords

Comments

Binomial transform of quasi-finite sequence 1,4,16,16,0,0,... (0 continued).

Links

Programs

Formula

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
First differences: a(n+1) - a(n) = A108099(n).
Second differences: a(n+2) - 2*a(n+1) + a(n) = A008598(n+1).
Third differences: a(n+3) - 3*a(n+2) + 3*a(n+1) - a(n) = 16.
a(n) = (A168574(n) + A168547(n))/2. - This formula is the link to the Janet table of the PSE.
G.f.: ( 1 + x + 11*x^2 + 3*x^3 ) / (x-1)^4. - R. J. Mathar, Jul 07 2011
E.g.f.: (3 +12*x +24*x^2 +8*x^3)*exp(x)/3. - G. C. Greubel, Nov 02 2018
Showing 1-3 of 3 results.

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