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 A027575 a(n) = n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2. 12

%I

%S 14,30,54,86,126,174,230,294,366,446,534,630,734,846,966,1094,1230,

%T 1374,1526,1686,1854,2030,2214,2406,2606,2814,3030,3254,3486,3726,

%U 3974,4230,4494,4766,5046,5334,5630,5934,6246,6566,6894,7230,7574

%N a(n) = n^2 + (n+1)^2 + (n+2)^2 + (n+3)^2.

%C Summation of n^2 taken 4 at a time. - Al Hakanson (hawkuu(AT)gmail.com), May 20 2009

%C Terms are congruent to (2,0,0) mod 6. - _Ezhilarasu Velayutham_, Apr 04 2019

%H P. De Geest, <a href="http://www.worldofnumbers.com/sumsquare.htm">Palindromic Sums of Squares of Consecutive Integers</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>

%F a(n) = 4*n^2 + 12*n + 14. - Al Hakanson (hawkuu(AT)gmail.com), May 20 2009

%F a(n) = a(n-1)+8*(n+1) for n>0, a(0)=14. - _Vincenzo Librandi_, Nov 19 2010

%F G.f.: 2*(7-6*x+3*x^2)/(1-x)^3. - _Colin Barker_, Feb 17 2012

%F From _Jean-Christophe HervĂ©_, Nov 11 2015: (Start)

%F a(n) = (2*n+3)^2 + 5 = A016754(n+1) + 5, hence a(n) is never square.

%F The last formula defines a(n) for n < 0; then we have a(-n) = a(n-3) for all n.

%F (End)

%t f[n_]:=(n^2+(n+2)^2+(n+4)^2+(n+6)^2)/4;Table[f[n],{n,0,6!,2}] (* _Vladimir Joseph Stephan Orlovsky_, Mar 03 2010 *)

%t Table[n^2 + (n + 1)^2 + (n + 2)^2 + (n + 3)^2, {n, 0, 42}] (* _Alonso del Arte_, Feb 17 2012 *)

%t Table[Total[Range[n,n+3]^2],{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{14,30,54},50] (* _Harvey P. Dale_, Jan 23 2017 *)

%o (Sage) [i^2+(i+1)^2+(i+2)^2+(i+3)^2 for i in xrange(0,50)] # _Zerinvary Lajos_, Jul 03 2008

%o (PARI) vector(100, n, n--; n^2+(n+1)^2+(n+2)^2+(n+3)^2) \\ _Altug Alkan_, Nov 11 2015

%Y Cf. A016754, A001844, A120328, A027578, A027865, A027577.

%K nonn,easy

%O 0,1

%A _Patrick De Geest_

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Last modified October 19 12:22 EDT 2019. Contains 328221 sequences. (Running on oeis4.)