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A139276 a(n) = n*(8*n+3). 9
0, 11, 38, 81, 140, 215, 306, 413, 536, 675, 830, 1001, 1188, 1391, 1610, 1845, 2096, 2363, 2646, 2945, 3260, 3591, 3938, 4301, 4680, 5075, 5486, 5913, 6356, 6815, 7290, 7781, 8288, 8811, 9350, 9905, 10476, 11063, 11666, 12285, 12920 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Sequence found by reading the line from 0, in the direction 0, 11,..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139272 in the same spiral.
LINKS
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
FORMULA
a(n) = 8*n^2 + 3*n.
Sequences of the form a(n)=8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n)= 3a(n-1)-3a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n+a(n-1)-5 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(5*x + 11)/(1-x)^3.
E.g.f.: (8*x^2 + 11*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/9 - (sqrt(2)-1)*Pi/6 - 4*log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3. - Amiram Eldar, Mar 17 2022
EXAMPLE
a(1)=16*1+0-5=11; a(2)=16*2+11-5=38; a(3)=16*3+38-5=81. - Vincenzo Librandi, Aug 03 2010
MATHEMATICA
a[n_]:=n*(8*n+3); a[Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3, -3, 1}, {0, 11, 38}, 50] (* Harvey P. Dale, Jul 02 2022 *)
PROG
(PARI) a(n)=n*(8*n+3) \\ Charles R Greathouse IV, Jun 16 2017
CROSSREFS
Sequence in context: A071853 A072313 A063146 * A010002 A143109 A007585
KEYWORD
easy,nonn
AUTHOR
Omar E. Pol, Apr 26 2008
STATUS
approved

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Last modified May 13 15:39 EDT 2024. Contains 372521 sequences. (Running on oeis4.)