A007533 a(n) = (5*n + 1)^2 + 4*n + 1.

2, 41, 130, 269, 458, 697, 986, 1325, 1714, 2153, 2642, 3181, 3770, 4409, 5098, 5837, 6626, 7465, 8354, 9293, 10282, 11321, 12410, 13549, 14738, 15977, 17266, 18605, 19994, 21433, 22922, 24461, 26050, 27689, 29378, 31117, 32906, 34745, 36634, 38573, 40562, 42601

Offset: 0

Author

N. J. A. Sloane, Robert G. Wilson v

Keywords

• nonn
• easy

Comments

Also, numbers of the form (3*k + 1)^2 + (4*k + 1)^2. - Bruno Berselli, Dec 11 2011

The continued fraction expansion of sqrt(a(n)) is [5n+1; {2, 2, 10n+2}]. For n=0, this collapses to [1; {2}]. - Magus K. Chu, Aug 27 2022

References

• W. Sierpiński, Elementary Theory of Numbers. Państ. Wydaw. Nauk., Warsaw, 1964, p. 323.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• W. Sierpiński, Elementary Theory of Numbers, Warszawa 1964.
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A154355.

Programs

Magma

[(5*n+1)^2 + 4*n+1: n in [0..40]]; // Vincenzo Librandi, May 02 2011

Mathematica

Table[25n^2+14n+2,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{2,41,130},40] (* Harvey P. Dale, Dec 18 2013 *)

PARI

a(n)=25*n^2 + 14*n + 2 \\ Charles R Greathouse IV, May 02 2011

Formula

From Bruno Berselli, Dec 11 2011: (Start)

a(n) = 25*n^2 + 14*n + 2.

G.f.: (2 + 35*x + 13*x^2)/(1-x)^3. (End)

From Elmo R. Oliveira, Oct 31 2024: (Start)

E.g.f.: (2 + 39*x + 25*x^2)*exp(x).

a(n) = A154355(n+1).

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)