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%I #27 May 06 2019 16:31:11
%S 28,55,60,79,92,95,112,240,368,448,960,1472,1792,3840,5888,7168,15360,
%T 23552,28672,61440,94208,114688,245760,376832,458752,983040,1507328,
%U 1835008,3932160,6029312,7340032,15728640,24117248,29360128,62914560,96468992,117440512,251658240,385875968,469762048,1006632960
%N Numbers of the form 4^k*(8*j+7) that have exactly three partitions into four positive squares.
%H Colin Barker, <a href="/A274642/b274642.txt">Table of n, a(n) for n = 1..1000</a>
%H D. H. Lehmer, <a href="/A274642/a274642.jpg">Draft of Math Review of paper by Om Prakash Srivastava</a>, Sep 18 1957.
%H Om Prakash Srivastava, <a href="/A274642/a274642.pdf">On the number of representations as sum of four squares of numbers of the form 4^a(8b+7)</a>, Journal of Scientific Research, Banaras Hindu University, VI(2) (1955-1956), 278-285. [Annotated scanned copy]
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,4).
%F Consists of 55, 79, 95, and the numbers 4^k*m where k >= 1 and m is 7, 15, or 23.
%F From _Colin Barker_, Jul 10 2016: (Start)
%F a(n) = 4*a(n-3) for n>9.
%F G.f.: x*(28+55*x+60*x^2-33*x^3-128*x^4-145*x^5-204*x^6-128*x^7-12*x^8) / (1-4*x^3).
%F (End)
%e 55 is a member because we have 55 = 49+4+1+1 = 36+9+9+1 = 25+25+4+1.
%t LinearRecurrence[{0,0,4},{28,55,60,79,92,95,112,240,368},50] (* _Harvey P. Dale_, May 06 2019 *)
%o (PARI) Vec(x*(28+55*x+60*x^2-33*x^3-128*x^4-145*x^5-204*x^6-128*x^7-12*x^8) / (1-4*x^3) + O(x^50)) \\ _Colin Barker_, Jul 10 2016
%K nonn,easy
%O 1,1
%A _N. J. A. Sloane_, Jul 09 2016