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, 105527255, 124002480

Offset: 0

Author

Henry Bottomley, Jan 23 2001

Keywords

• nice
• nonn
• easy

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

Examples

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

References

• Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Table 68 at p. 152.

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.
• Vladimir Gurvich and Mariya Naumova, On pairs of triangular numbers whose product is a perfect square and pairs of intervals of successive integers with equal sums of squares, arXiv:2508.06598 [math.NT], 2025. See p. 3.
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

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

Programs

Magma

[n*(n+1)*(2*n+1)*(12*n^2+12*n+1)/6 : n in [0..50]]; // Wesley Ivan Hurt, Jun 21 2014

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 *)

PARI

a(n)=n*(n+1)*(2*n+1)*(12*n^2+12*n+1)/6 \\ Charles R Greathouse IV, Jul 27 2021

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)).

From Harvey P. Dale, May 09 2011: (Start)

G.f.: (5x(1+x)(5+x(38+5x)))/(x-1)^6.

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). (End)

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

E.g.f.: exp(x)*x*(150 + 945*x + 1010*x^2 + 300*x^3 + 24*x^4)/6. - Stefano Spezia, Aug 05 2024