

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 ba, cb, dc 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 FourSquare Theorem
Wikipedia, Lagrange's foursquare 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)  (1i)*((66*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) / ((x1)^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, [n1, n+1, n+3, n+4]]
elif n mod 4 = 3 then
return [4*n^2+10*n+21, [n1, n, n+2, n+4]]
else
return [4*n^2+10*n+15, [n1, 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 *)


PROG

(PARI) Vec(x*(15*x^630*x^5+45*x^460*x^3+69*x^246*x+39)/((x1)^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



