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 A253187 Number of ordered ways to write n as the sum of a pentagonal number, a second pentagonal number and a generalized decagonal number. 2
 1, 2, 2, 2, 1, 1, 1, 3, 4, 2, 2, 1, 4, 3, 3, 4, 2, 3, 1, 3, 2, 2, 5, 3, 3, 3, 3, 6, 3, 6, 4, 2, 3, 1, 7, 2, 4, 5, 5, 4, 1, 5, 5, 2, 3, 4, 4, 5, 5, 5, 3, 5, 7, 6, 4, 3, 1, 6, 6, 8, 5, 3, 6, 4, 7, 4, 2, 6, 5, 5, 3, 4, 8, 3, 3, 3, 6, 6, 7, 9, 6, 2, 5, 6, 7, 7, 4, 6, 6, 7, 5, 3, 10, 6, 3, 4, 5, 7, 3, 10, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Conjecture: a(n) > 0 for all n. Also, for any ordered pair (k,m) among (5,7), (5,9), (5,13), (6,5), (6,7), (7,5), each nonnegative integer n can be written as the sum of a k-gonal number, a second k-gonal number and a generalized m-gonal number. See also the author's similar conjectures in A254574, A254631, A255916 and the two linked papers. 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, On universal sums a*x^2+b*y^2+f(z), a*T_x+b*T_y+f(z) and a*T_x+b*y^2+f(z), arXiv:1502.03056 [math.NT], 2015. EXAMPLE a(33) = 1 since 33 = 0*(3*0-1)/2 + 4*(3*4+1)/2 + 1*(4*1+3). a(56) = 1 since 56 = 4*(3*4-1)/2 + 2*(3*2+1)/2 + 3*(4*3+3). MATHEMATICA DQ[n_]:=IntegerQ[Sqrt[16n+9]] Do[r=0; Do[If[DQ[n-x(3x-1)/2-y(3y+1)/2], r=r+1], {x, 0, (Sqrt[24n+1]+1)/6}, {y, 0, (Sqrt[24(n-x(3x-1)/2)+1]-1)/6}]; Print[n, " ", r]; Continue, {n, 0, 100}] CROSSREFS Cf. A000326, A000384, A000566, A005449, A014105, A074377, A085787, A147875, A254574, A254631. Sequence in context: A320278 A302111 A124278 * A139755 A213366 A212648 Adjacent sequences:  A253184 A253185 A253186 * A253188 A253189 A253190 KEYWORD nonn AUTHOR Zhi-Wei Sun, Apr 07 2015 STATUS approved

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Last modified September 26 11:15 EDT 2020. Contains 337358 sequences. (Running on oeis4.)