A024381 a(n) = sum of squares of first n positive integers congruent to 1 mod 4.

1, 26, 107, 276, 565, 1006, 1631, 2472, 3561, 4930, 6611, 8636, 11037, 13846, 17095, 20816, 25041, 29802, 35131, 41060, 47621, 54846, 62767, 71416, 80825, 91026, 102051, 113932, 126701, 140390, 155031, 170656, 187297, 204986, 223755, 243636, 264661

Offset

1,2

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

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

Formula

1^2 + 5^2 + 9^2 + ... + (4n+1)^2 = (4n+1)(n+1)+(4^2)(2n+1)(n+1)n/6, which generalizes to (0a+1)^2 + (1a+1)^2 + (2a+1)^2 + ... + (na+1)^2 = (an+1)(n+1) + (a^2)(2n+1)(n+1)n/6. Also Sum{n}_(an+1)^2 = (an+1)(n+1) + (a^2)* sum{n}_(n+1)^2. - Helmut Rasinger (helmut.rasinger(AT)wanadoo.fr), Sep 04 2003

G.f.: x*(1 + 22*x + 9*x^2) / (x-1)^4. - R. J. Mathar, Oct 08 2011

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 19 2012

Mathematica

LinearRecurrence[{4, -6, 4, -1}, {1, 26, 107, 276}, 40] (* Vincenzo Librandi, Jun 19 2012 *)
Accumulate[Range[1, 151, 4]^2] (* Harvey P. Dale, Apr 25 2020 *)

Prog

(Magma) I:=[1, 26, 107, 276]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 19 2012

Crossrefs

Sequence in context: A042324 A044277 A044658 * A262756 A262476 A156385

Adjacent sequences: A024378 A024379 A024380 * A024382 A024383 A024384

Keyword

nonn,easy

Author

Clark Kimberling

Status

approved