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%I #55 Feb 17 2022 00:12:18
%S 0,25,365,2030,7230,19855,45955,94220,176460,308085,508585,802010,
%T 1217450,1789515,2558815,3572440,4884440,6556305,8657445,11265670,
%U 14467670,18359495,23047035,28646500,35284900,43100525,52243425,62875890,75172930,89322755
%N Both sum of n+1 consecutive squares and sum of the immediately following n consecutive squares.
%C The analog for sums of integers is A059270, and the analog for sums of triangular numbers is A222716. - _Jonathan Sondow_, Mar 07 2013
%C In 1879, Dostor gave formulas for all solutions -- see the Dickson link. - _Jonathan Sondow_, Jun 21 2014
%H Vincenzo Librandi, <a href="/A059255/b059255.txt">Table of n, a(n) for n = 0..1000</a>
%H M. Boardman, <a href="http://www.jstor.org/stable/2691496">Proof Without Words: Pythagorean Runs</a>, Math. Mag., 73 (2000), 59.
%H L. E. Dickson, <a href="https://archive.org/details/HistoryOfTheTheoryOfNumbersVolII/page/n345/mode/2up">History of the Theory of Numbers, II</a>, p. 320.
%H Georges Dostor, <a href="https://www.digitale-sammlungen.de/en/view/bsb11390652?page=372">Question sur les nombres</a>, Archiv der Mathematik und Physik, 64 (1879), 350-352.
%H Greg Frederickson, <a href="http://www.jstor.org/stable/27765930">Casting Light on Cube Dissections</a>, Math. Mag., 82 (2009), 323-331.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F a(n) = n*(n + 1)*(2n + 1)*(12n^2 + 12n + 1)/6.
%F 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]
%F a(n) = A000330(A046092(n)) - A000330(A014107(n + 1)).
%F a(n) = A000330(A014106(n)) - A000330(A046092(n)).
%F G.f.: (5x(1+x)(5+x(38+5x)))/(x-1)^6. - _Harvey P. Dale_, May 09 2011
%F 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
%F 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
%F 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 a(3) = 2030 = 21^2 + 22^2 + 23^2 + 24^2 = 25^2 + 26^2 + 27^2.
%p 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
%t 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 *)
%o (Magma) [n*(n+1)*(2*n+1)*(12*n^2+12*n+1)/6 : n in [0..50]]; // _Wesley Ivan Hurt_, Jun 21 2014
%o (PARI) a(n)=n*(n+1)*(2*n+1)*(12*n^2+12*n+1)/6 \\ _Charles R Greathouse IV_, Jul 27 2021
%Y The n+1 consecutive squares start with the square of A014105, while the n consecutive squares start with the square of A001844.
%Y Cf. also A059270, A222716.
%Y Cf. A234319 for nonexistence of analogs for sums of n-th powers, n > 2. - _Jonathan Sondow_, Apr 23 2014
%K nice,nonn,easy
%O 0,2
%A _Henry Bottomley_, Jan 23 2001
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