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 A254573 Number of ways to write n = x*(x+1) + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z nonnegative integers 8
 1, 1, 2, 2, 1, 2, 1, 4, 2, 3, 1, 1, 3, 3, 5, 2, 2, 2, 3, 3, 4, 3, 4, 1, 4, 2, 4, 5, 4, 3, 2, 4, 5, 4, 2, 4, 2, 6, 3, 5, 3, 3, 6, 5, 5, 3, 3, 6, 2, 6, 5, 3, 4, 3, 6, 2, 4, 9, 6, 4, 4, 5, 5, 5, 7, 3, 2, 3, 8, 4, 6 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS Conjecture: a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 1, 4, 6, 10, 11, 23. This has been verified for all n = 0..10^7. We have proved that every nonnegative integer can be written as x*(x+1) + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z integers. 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. EXAMPLE a(10) = 1 since 10 = 1*2 + 2*(3*2+1)/2 + 1*(3*1-1)/2. a(11) = 1 since 11 = 2*3 + 0*(3*0+1)/2 + 2*(3*2-1)/2. a(23) = 1 since 23 = 4*5 + 1*(3*1+1)/2 + 1*(3*1-1)/2. a(34) = 2 since 34 = 3*4 + 0*(3*0+1)/2 + 4*(3*4-1)/2 = 4*5 + 1*(3*1+1)/2 + 3*(3*3-1)/2. MATHEMATICA sQ[n_]:=IntegerQ[Sqrt[4n+1]] Do[r=0; Do[If[sQ[n-y(3y+1)/2-z(3z-1)/2], r=r+1], {y, 0, (Sqrt[24n+1]-1)/6}, {z, 0, (Sqrt[24(n-y(3y+1)/2)+1]+1)/6}]; Print[n, " ", r]; Continue, {n, 0, 70}] CROSSREFS Cf. A000326, A002378, A005449, A254574. Sequence in context: A262403 A326952 A109909 * A144387 A030768 A051480 Adjacent sequences:  A254570 A254571 A254572 * A254574 A254575 A254576 KEYWORD nonn AUTHOR Zhi-Wei Sun, Feb 01 2015 STATUS approved

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Last modified November 11 22:31 EST 2019. Contains 329046 sequences. (Running on oeis4.)