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 A255934 Number of ways to write n as the sum of four unordered generalized octagonal numbers. 3
 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 2, 3, 2, 3, 2, 1, 3, 3, 3, 3, 2, 4, 2, 1, 3, 2, 3, 4, 3, 4, 2, 2, 4, 4, 3, 4, 3, 6, 5, 2, 4, 3, 4, 5, 4, 6, 4, 1, 4, 5, 4, 5, 5, 7, 4, 1, 5, 5, 5, 6, 5, 8, 5, 3, 4, 6, 6, 6, 6, 7, 6, 3, 6, 6, 5, 6, 6, 10, 7, 1, 5, 8, 7, 7, 7, 8, 5, 3, 6, 7, 6, 8, 7, 10, 8, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,9 COMMENTS I have proved that a(n) > 0 for all n, i.e., any nonnegative integer can be expressed as the sum of four generalized octagonal numbers. I can show that a(n) = 1 if 3*n+4 is among 7, 13, 19, 31, 43, 2^(2k), 5*2^(2k+1), 11*2^(2k+1), 23*2^(2k+1) (k = 0,1,2,...), and conjecture the converse. I also conjecture that each nonnegative integer can be written as the sum of two heptagonal numbers, a second heptagonal number and a generalized heptagonal number. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 0..10000 Zhi-Wei Sun, On universal sums of polygonal numbers, arXiv:0905.0635 [math.NT], 2009-2015. Zhi-Wei Sun, A result similar to Lagrange's theorem, arXiv:1503.03743 [math.NT], 2015. EXAMPLE a(60) = 1 since 60 = 1*(3*1-2) + (-1)*(3*(-1)-2) + 3*(3*3-2) + (-3)*(3*(-3)-2). a(1876) = 1 since 1876 = (-5)*(3*(-5)-2) + (-5)*(3*(-5)-2) + 11*(3*11-2) + (-21)*(3*(-21)-2). a(15700) = 1 since 15700 = 11*(3*11-2) + (-21)*(3*(-21)-2) + 43*(3*43-2) + (-53)*(3*(-53)-2). a(21844) = 1 since 21844 = 43*(3*43-2) + 43*(3*43-2) + 43*(3*43-2) + 43*(3*43-2). a(30036) = 1 since 30036 = (-21)*(3*(-21)-2) + (-21)*(3*(-21)-2) + 43*(3*43-2) + (-85)*(3*(-85)-2). MATHEMATICA T[n_]:=Union[Table[x(3x-2), {x, -Floor[(Sqrt[3n+1]-1)/3], Floor[(Sqrt[3n+1]+1)/3]}]] Do[r=0; Do[If[n-Part[T[n], x]-Part[T[n], y]-Part[T[n], z]

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Last modified January 16 11:32 EST 2019. Contains 319188 sequences. (Running on oeis4.)