A059255 Both sum of n+1 consecutive squares and sum of the immediately following n consecutive squares.

0, 25, 365, 2030, 7230, 19855, 45955, 94220, 176460, 308085, 508585, 802010, 1217450, 1789515, 2558815, 3572440, 4884440, 6556305, 8657445, 11265670, 14467670, 18359495, 23047035, 28646500, 35284900, 43100525, 52243425, 62875890, 75172930, 89322755

Offset

0,2

Comments

The analog for sums of integers is A059270, and the analog for sums of triangular numbers is A222716. - Jonathan Sondow, Mar 07 2013

In 1879, Dostor gave formulas for all solutions -- see the Dickson link. - Jonathan Sondow, Jun 21 2014

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

M. Boardman, Proof Without Words: Pythagorean Runs, Math. Mag., 73 (2000), 59.

L. E. Dickson, History of the Theory of Numbers, II, p. 320.

Georges Dostor, Question sur les nombres, Archiv der Mathematik und Physik, 64 (1879), 350-352.

Greg Frederickson, Casting Light on Cube Dissections, Math. Mag., 82 (2009), 323-331.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Formula

a(n) = n*(n + 1)*(2n + 1)*(12n^2 + 12n + 1)/6.

a(n) = 4*n^5 + 10*n^4 + (25/3)*n^3 + (5/2)*n^2 + (1/6)*n. [Corrected by Ignacio Larrosa Cañestro, Nov 15 2021]

a(n) = A000330(A046092(n)) - A000330(A014107(n + 1)).

a(n) = A000330(A014106(n)) - A000330(A046092(n)).

G.f.: (5x(1+x)(5+x(38+5x)))/(x-1)^6. - Harvey P. Dale, May 09 2011

a(0)=0, a(1)=25, a(2)=365, a(3)=2030, a(4)=7230, a(5)=19855, a(n) = 6a(n-1)-15a(n-2)+20a(n-3)-15a(n-4)+6a(n-5)-a(n-6). - Harvey P. Dale, May 09 2011

a(n) = (4*T(n)-n)^2+(4*T(n)-n+1)^2+...+(4*T(n))^2 = (4*T(n)+1)^2+(4*T(n)+2)^2+...+(4*T(n)+n)^2, where T = A000217. See Boardman (2000). - Jonathan Sondow, Mar 07 2013

a(0)=0, a(n) = 25 + 340*C(n-1,1) + 1325*C(n-1,2) + 2210*C(n-1,3) + 1680*C(n-1,4) + 480*C(n-1,5) for n >= 1, where C(a,b) are binomial coefficients. - Kieren MacMillan, Sep 16 2014

Example

a(3) = 2030 = 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2.

Maple

A059255:=n->n*(n+1)*(2*n+1)*(12*n^2+12*n+1)/6; seq(A059255(n), n=0..50); # Wesley Ivan Hurt, Jun 21 2014

Mathematica

Table[1/6(-1+n)(-n+14n^2-36n^3+24n^4), {n, 40}] (* or *) LinearRecurrence[ {6, -15, 20, -15, 6, -1}, {0, 25, 365, 2030, 7230, 19855}, 40] (* Harvey P. Dale, May 09 2011 *)

Prog

(Magma) [n*(n+1)*(2*n+1)*(12*n^2+12*n+1)/6 : n in [0..50]]; // Wesley Ivan Hurt, Jun 21 2014
(PARI) a(n)=n*(n+1)*(2*n+1)*(12*n^2+12*n+1)/6 \\ Charles R Greathouse IV, Jul 27 2021

Crossrefs

The n+1 consecutive squares start with the square of A014105, while the n consecutive squares start with the square of A001844.

Cf. also A059270, A222716.

Cf. A234319 for nonexistence of analogs for sums of n-th powers, n > 2. - Jonathan Sondow, Apr 23 2014

Sequence in context: A197536 A045622 A130052 * A227024 A254376 A022749

Adjacent sequences: A059252 A059253 A059254 * A059256 A059257 A059258

Keyword

nice,nonn,easy

Author

Henry Bottomley, Jan 23 2001

Status

approved