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a(n) = n*(n+1)*(n+2)*(n+3)+1 = (n^2 +3*n + 1)^2.
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%I #81 Jan 07 2024 09:16:33

%S 1,25,121,361,841,1681,3025,5041,7921,11881,17161,24025,32761,43681,

%T 57121,73441,93025,116281,143641,175561,212521,255025,303601,358801,

%U 421201,491401,570025,657721,755161,863041,982081,1113025,1256641

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

%C a(n) = product of first four terms of an arithmetic progression + n^4, where the first term is 1 and the common difference is n. E.g. a(1) = 1*2*3*4 +1^4 =25, a(4) = 1*5*9*13 + 4^4= 841 etc. - _Amarnath Murthy_, Sep 19 2003

%C Is it possible for one of the squares to be the sum of two or more lesser squares each used only once? - _J. M. Bergot_, Feb 17 2011

%C Yes, in fact a(1)-a(11) are examples. [_Charles R Greathouse IV_, Jun 28 2011]

%C This sequence demonstrates that the product of any 4 consecutive integers plus 1 is a square. The square roots are in A028387. [_Harvey P. Dale_, Oct 19 2011]

%C The sum of three consecutive terms of the sequence is divisible by 3. The quotient is a square number: [a(n)+a(n+1)+a(n+2)]/3=(n^2+5*n+7)^2. - _Carmine Suriano_, Jan 23 2012

%C All terms end with 1 or 5. - _Uri Geva_, Jan 06 2024

%D J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 85.

%H Harry J. Smith, <a href="/A062938/b062938.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n+1) = Numerator of ((n+2)! + (n-2)!)/n!, for n>=2. - _Artur Jasinski_, Jan 09 2007; corrected by _Michel Marcus_, Dec 25 2022

%F a(n) = A028387(n)^2. - _Jaroslav Krizek_, Oct 31 2010

%F a(n) = n*(n+1)*(n+2)*(n+3)+1^4 = 1*(1+n)*(1+2*n)*(1+3*n)+n^4 =(n^2+3*n+1)^2; in general, n*(n+k)*(n+2*k)*(n+3*k)+k^4 = k*(k+n)*(k+2*n)*(k+3*n)+n^4 = (n^2+3*k*n+k^2)^2. - _Charlie Marion_, Jan 13 2011

%F G.f.: (1+20*x+6*x^2-4*x^3+x^4)/(1-x)^5. - _Colin Barker_, Jun 30 2012

%F a(n) = A052762(n+3) + 1. - _Bruce J. Nicholson_, Apr 22 2017

%F Sum_{n>=0} 1/a(n) = (Pi^2/5)*(1+t^2) - 2*sqrt(5)*Pi*t/25 - 1, where t = tan(Pi*sqrt(5)/2). - _Amiram Eldar_, Apr 03 2022

%F E.g.f.: (1 +24*x +36*x^2 +12*x^3 +x^4)*exp(x). - _G. C. Greubel_, Dec 24 2022

%t Table[(n^2+3*n+1)^2, {n,0,50}]

%t Times@@#+1&/@Partition[Range[0,50],4,1] (* _Harvey P. Dale_, Apr 02 2011 *)

%o (PARI) j=[]; for(n=0,70,j=concat(j,(n^2+3*n+1)^2)); j

%o (PARI) { for (n=0, 1000, write("b062938.txt", n, " ", (n^2 + 3*n + 1)^2) ) } \\ _Harry J. Smith_, Aug 14 2009

%o (Magma) [(n^2+3*n+1)^2: n in [0..50]]; // _G. C. Greubel_, Dec 24 2022

%o (SageMath) [(n^2+3*n+1)^2 for n in range(51)] # _G. C. Greubel_, Dec 24 2022

%Y Cf. A028387, A052762, A082043.

%K nonn,easy

%O 0,2

%A _Amarnath Murthy_, Jul 05 2001

%E More terms from _Jason Earls_, _Harvey P. Dale_ and _Dean Hickerson_, Jul 06 2001