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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 *) 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^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 November 28 05:56 EST 2022. Contains 358407 sequences. (Running on oeis4.)