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A243577 Integers of the form 8k+7 that can be written as a sum of four distinct 'almost consecutive' squares. 6
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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

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).

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

EXAMPLE

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.

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 *)

PROG

(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

CROSSREFS

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

Sequence in context: A249701 A039467 A250657 * A253153 A044105 A044486

Adjacent sequences:  A243574 A243575 A243576 * A243578 A243579 A243580

KEYWORD

nonn,easy

AUTHOR

Walter Kehowski, Jun 08 2014

STATUS

approved

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Last modified March 28 05:29 EDT 2020. Contains 333073 sequences. (Running on oeis4.)