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

1, 25, 121, 361, 841, 1681, 3025, 5041, 7921, 11881, 17161, 24025, 32761, 43681, 57121, 73441, 93025, 116281, 143641, 175561, 212521, 255025, 303601, 358801, 421201, 491401, 570025, 657721, 755161, 863041, 982081, 1113025, 1256641

Offset: 0

Author

Amarnath Murthy, Jul 05 2001

Keywords

• nonn
• easy

Comments

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

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

Yes, in fact a(1)-a(11) are examples. - Charles R Greathouse IV, Jun 28 2011

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

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

All terms end with 1 or 5. - Uri Geva, Jan 06 2024

References

• James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 19.
• J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 85.

Links

• Harry J. Smith, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A028387, A052762, A082043.

Programs

Magma

[(n^2+3*n+1)^2: n in [0..50]]; // G. C. Greubel, Dec 24 2022

Mathematica

Table[(n^2+3*n+1)^2, {n,0,50}]
Times@@#+1&/@Partition[Range[0,50],4,1]  (* Harvey P. Dale, Apr 02 2011 *)

PARI

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

PARI

{ for (n=0, 1000, write("b062938.txt", n, " ", (n^2 + 3*n + 1)^2) ) } \\ Harry J. Smith, Aug 14 2009

SageMath

[(n^2+3*n+1)^2 for n in range(51)] # G. C. Greubel, Dec 24 2022

Formula

a(n+1) = numerator( ((n+2)! + (n-2)!)/n! ), for n>=2. - Artur Jasinski, Jan 09 2007; corrected by Michel Marcus, Dec 25 2022

a(n) = A028387(n)^2. - Jaroslav Krizek, Oct 31 2010

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

G.f.: (1+20*x+6*x^2-4*x^3+x^4)/(1-x)^5. - Colin Barker, Jun 30 2012

a(n) = A052762(n+3) + 1. - Bruce J. Nicholson, Apr 22 2017

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

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

Extensions

More terms from Jason Earls, Harvey P. Dale and Dean Hickerson, Jul 06 2001