

A085001


a(n) = (3*n+1)*(3*n+4).


3



4, 28, 70, 130, 208, 304, 418, 550, 700, 868, 1054, 1258, 1480, 1720, 1978, 2254, 2548, 2860, 3190, 3538, 3904, 4288, 4690, 5110, 5548, 6004, 6478, 6970, 7480, 8008, 8554, 9118, 9700, 10300, 10918, 11554, 12208, 12880, 13570, 14278, 15004
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OFFSET

0,1


COMMENTS

1/4 + 1/28 + 1/70 + ... = 1/3; 1/4 + 1/28 + 1/70 + ... n terms = (n+1)/(3n+4). [Jolley].  Gary W. Adamson, Jan 03 2007 [Corrected by Gary Detlefs, Mar 14 2018]


REFERENCES

L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 38


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,3,1).


FORMULA

Sum_{k=0..n} 3/a(k) = 3*(n+1)/(3n+4). [Corrected by Gary Detlefs, Mar 14 2018]
Sum_{k>=0} 3/a(k) = 1.
G.f.: 2*(2+8*xx^2)/(1x)^3.  R. J. Mathar, Sep 17 2008
a(n) = 3*a(n1)  3*a(n2) + a(n3).  Vincenzo Librandi, Jul 08 2012


MATHEMATICA

CoefficientList[Series[2*(2+8xx^2)/(1x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 08 2012 *)
Table[(3n+1)(3n+4), {n, 0, 40}] (* or *) LinearRecurrence[{3, 3, 1}, {4, 28, 70}, 50] (* Harvey P. Dale, Apr 07 2019 *)


PROG

(MAGMA) [(3*n+1)*(3*n+4): n in [0..50]]; // Vincenzo Librandi, Jul 08 2012
(PARI) a(n)=(3*n+1)*(3*n+4) \\ Charles R Greathouse IV, Jun 17 2017


CROSSREFS

Sequence in context: A197542 A203280 A085024 * A153784 A030117 A005634
Adjacent sequences: A084998 A084999 A085000 * A085002 A085003 A085004


KEYWORD

nonn,easy


AUTHOR

Gary W. Adamson, Jun 17 2003


EXTENSIONS

Edited by Don Reble, Nov 13 2005


STATUS

approved



