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

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

0,2

Comments

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

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

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

Mathematica

LinearRecurrence[{4, -6, 4, -1}, {1, 5, 25, 77}, 50] (* Harvey P. Dale, Nov 22 2011 *)

Prog

(PARI) a(n)=4*n*(1+2*n^2)/3+1 \\ Charles R Greathouse IV, Jul 07 2011
(Magma) [1+4*n*(1+2*n^2)/3: n in [0..40]]; // Vincenzo Librandi, Aug 05 2011

Crossrefs

Sequence in context: A078234 A056374 A301912 * A366158 A243303 A238449

Adjacent sequences: A171269 A171270 A171271 * A171273 A171274 A171275

Keyword

nonn,easy

Author

Paul Curtz, Dec 06 2009

Status

approved