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A062938 a(n) = n*(n+1)*(n+2)*(n+3)+1 = (n^2 + 3*n + 1)^2.

%I A062938 #84 Jul 03 2025 10:04:03
%S A062938 1,25,121,361,841,1681,3025,5041,7921,11881,17161,24025,32761,43681,
%T A062938 57121,73441,93025,116281,143641,175561,212521,255025,303601,358801,
%U A062938 421201,491401,570025,657721,755161,863041,982081,1113025,1256641
%N A062938 a(n) = n*(n+1)*(n+2)*(n+3)+1 = (n^2 + 3*n + 1)^2.
%C A062938 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 A062938 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 A062938 Yes, in fact a(1)-a(11) are examples. - _Charles R Greathouse IV_, Jun 28 2011
%C A062938 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 A062938 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 A062938 All terms end with 1 or 5. - _Uri Geva_, Jan 06 2024
%D A062938 James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 19.
%D A062938 J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 85.
%H A062938 Harry J. Smith, <a href="/A062938/b062938.txt">Table of n, a(n) for n = 0..1000</a>
%H A062938 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A062938 a(n+1) = numerator( ((n+2)! + (n-2)!)/n! ), for n>=2. - _Artur Jasinski_, Jan 09 2007; corrected by _Michel Marcus_, Dec 25 2022
%F A062938 a(n) = A028387(n)^2. - _Jaroslav Krizek_, Oct 31 2010
%F A062938 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 A062938 G.f.: (1+20*x+6*x^2-4*x^3+x^4)/(1-x)^5. - _Colin Barker_, Jun 30 2012
%F A062938 a(n) = A052762(n+3) + 1. - _Bruce J. Nicholson_, Apr 22 2017
%F A062938 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 A062938 E.g.f.: (1 +24*x +36*x^2 +12*x^3 +x^4)*exp(x). - _G. C. Greubel_, Dec 24 2022
%t A062938 Table[(n^2+3*n+1)^2, {n,0,50}]
%t A062938 Times@@#+1&/@Partition[Range[0,50],4,1]  (* _Harvey P. Dale_, Apr 02 2011 *)
%o A062938 (PARI) j=[]; for(n=0,70,j=concat(j,(n^2+3*n+1)^2)); j
%o A062938 (PARI) { for (n=0, 1000, write("b062938.txt", n, " ", (n^2 + 3*n + 1)^2) ) } \\ _Harry J. Smith_, Aug 14 2009
%o A062938 (Magma) [(n^2+3*n+1)^2: n in [0..50]]; // _G. C. Greubel_, Dec 24 2022
%o A062938 (SageMath) [(n^2+3*n+1)^2 for n in range(51)] # _G. C. Greubel_, Dec 24 2022
%Y A062938 Cf. A028387, A052762, A082043.
%K A062938 nonn,easy
%O A062938 0,2
%A A062938 _Amarnath Murthy_, Jul 05 2001
%E A062938 More terms from _Jason Earls_, _Harvey P. Dale_ and _Dean Hickerson_, Jul 06 2001