A243577 Integers of the form 8k+7 that can be written as a sum of four distinct 'almost consecutive' squares.

39, 71, 87, 119, 191, 255, 287, 351, 471, 567, 615, 711, 879, 1007, 1071, 1199, 1415, 1575, 1655, 1815, 2079, 2271, 2367, 2559, 2871, 3095, 3207, 3431, 3791, 4047, 4175, 4431, 4839, 5127, 5271, 5559, 6015, 6335, 6495, 6815, 7319, 7671, 7847, 8199, 8751, 9135, 9327, 9711

Offset: 1

Author

Walter Kehowski, Jun 08 2014

Keywords

• nonn
• easy

Comments

By Lagrange's Four Square Theorem, any integer n of the form 8k+7 (A004771) can be written as sum of no fewer than four squares. The initial terms 7,15,23,31 are the generating set for A004771 in the sense that if n = a^2 + b^2+ c^2 + d^2, then (a mod 4)^2 + (b mod 4)^2 + (c mod 4)^2 + (d mod 4)^2 is one of 7,15,23,31.

From now on assume that n is of the form 8k+7 and is a sum of distinct squares a,b,c,d, sorted.

We say that [a,b,c,d] is almost consecutive if the differences b-a, c-b, d-c are 1 or 2. The generating set for this sequence is

39, [1,2,3,5], with gap pattern 112,

71, [1,3,5,6], with gap pattern 221,

87, [2,3,5,7], with gap pattern 122,

119, [3,5,6,7], with gap pattern 211,

in the sense that adding [4*i,4*i,4*i,4*i], i >= 0, preserves the gap pattern. It should be noted that the four generators are all obtainable from [1,1,2,3] or [1,2,3,3] by addition of suitable vectors. Let's write it out:

[1,2,3,5] = [1,1,2,3] + [4,0,0,0] or

[1,2,3,5] = [1,1,2,3] + [0,4,0,0]

[1,3,5,6] = [1,1,2,3] + [0,4,4,0]

[2,3,5,7] = [1,2,3,3] + [4,0,4,0] or

[2,3,5,7] = [1,2,3,3] + [4,0,0,4]

[3,5,6,7] = [1,2,3,3] + [4,4,4,0] or

[3,5,6,7] = [1,2,3,3] + [4,4,0,4].

There are generators for other gap patterns, but the minimal gap patterns are of the most interest. - Walter Kehowski, Jul 07 2014

Examples

			For n=1, a(n) = 4*1^2 + 14*1 + 21 =  39 and  39 = 1^2 + 2^2 + 3^2 + 5^2.
For n=2, a(n) = 4*2^2 + 14*2 + 27 =  39 and  71 = 1^2 + 3^2 + 5^2 + 6^2.
For n=3, a(n) = 4*3^2 + 10*3 + 21 =  87 and  87 = 2^2 + 3^2 + 5^2 + 7^2.
For n=4, a(n) = 4*4^2 + 10*4 + 15 = 119 and 119 = 3^2 + 5^2 + 6^2 + 7^2.
		

Links

• Walter Kehowski, Table of n, a(n) for n = 1..1728
• J. O. Sizemore, Lagrange's Four Square Theorem
• R. C. Vaughan, Lagrange's Four Square Theorem
• Eric Weisstein's World of Mathematics, Lagrange's Four-Square Theorem
• Wikipedia, Lagrange's four-square theorem
• Index entries for linear recurrences with constant coefficients, signature (3,-5,7,-7,5,-3,1).

Crossrefs

Cf. A004771, A008586, A016813, A016825, A243578, A243579, A243580, A243581, A243582.

Programs

Maple

A243577 := proc(n::posint)
   if n mod 4 = 1 then
      return [4*n^2+14*n+21, [n,n+1,n+2,n+4]]
   elif n mod 4 = 2 then
      return [4*n^2+14*n+27, [n-1,n+1,n+3,n+4]]
   elif n mod 4 = 3 then
      return [4*n^2+10*n+21, [n-1,n,n+2,n+4]]
   else
      return [4*n^2+10*n+15, [n-1,n+1,n+2,n+3]]
   fi;
end:
# Walter A. Kehowski, Jun 08 2014

Mathematica

Rest@ CoefficientList[Series[-x (15 x^6 - 30 x^5 + 45 x^4 - 60 x^3 + 69 x^2 - 46 x + 39)/((x - 1)^3*(x^2 + 1)^2), {x, 0, 48}], x] (* Michael De Vlieger, Feb 19 2019 *)
LinearRecurrence[{3,-5,7,-7,5,-3,1},{39,71,87,119,191,255,287},50] (* Harvey P. Dale, Jul 05 2021 *)

PARI

Vec(-x*(15*x^6-30*x^5+45*x^4-60*x^3+69*x^2-46*x+39)/((x-1)^3*(x^2+1)^2) + O(x^100)) \\ Colin Barker, Jun 09 2014

Formula

If n mod 4 = 1, then a(n) = 4*n^2 + 14*n + 21.

If n mod 4 = 2, then a(n) = 4*n^2 + 14*n + 27.

If n mod 4 = 3, then a(n) = 4*n^2 + 10*n + 21.

If n mod 4 = 0, then a(n) = 4*n^2 + 10*n + 15.

a(n) = -3*(-7 + (-i)^n+i^n) - (1-i)*((-6-6*i) + (-i)^n + i*i^n)*n + 4*n^2 where i=sqrt(-1). - Colin Barker, Jun 09 2014

G.f.: -x*(15*x^6 - 30*x^5 + 45*x^4 - 60*x^3 + 69*x^2 - 46*x + 39) / ((x-1)^3*(x^2+1)^2). - Colin Barker, Jun 09 2014