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 A254661 Number of ways to write n as the sum of a triangular number, an even square and a second pentagonal number. 3
 1, 1, 1, 2, 1, 2, 2, 3, 2, 1, 3, 1, 3, 1, 2, 2, 3, 4, 2, 4, 1, 5, 3, 2, 2, 3, 4, 2, 3, 3, 3, 3, 4, 3, 3, 1, 5, 3, 3, 4, 4, 4, 3, 5, 5, 4, 5, 5, 2, 2, 2, 6, 5, 2, 4, 3, 2, 6, 3, 6, 2, 5, 5, 4, 5, 3, 7, 5, 4, 1, 4, 6, 8, 3, 5, 1, 6, 6, 5, 6, 4, 6, 6, 4, 4, 7, 3, 5, 2, 5, 2, 5, 5, 7, 6, 2, 7, 6, 4, 4, 5 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS Conjecture: (i) a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 1, 2, 4, 9, 11, 13, 20, 35, 69, 75, 188. (ii) For each a = 2,3, any nonnegative integer n can be written as x(x+1)/2 + a*y^2 + z*(3*z+1)/2 with x,y,z nonnegative integers. Compare part (i) of this conjecture with the conjecture in A160325. 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. EXAMPLE a(20) = 1 since 20 = 1*2/2 + 2^2 + 3*(3*3+1)/2. a(35) = 1 since 35 = 7*8/2 + 0^2 + 2*(3*2+1)/2. a(69) = 1 since 69 = 2*3/2 + 8^2 + 1*(3*1+1)/2. a(75) = 1 since 75 = 9*10/2 + 2^2 + 4*(3*4+1)/2. a(188) = 1 since 188 = 1*2/2 + 0^2 + 11*(3*11+1)/2. MATHEMATICA TQ[n_]:=IntegerQ[Sqrt[8n+1]] Do[r=0; Do[If[TQ[n-4y^2-z(3z+1)/2], r=r+1], {y, 0, Sqrt[n/4]}, {z, 0, (Sqrt[24(n-4y^2)+1]-1)/6}]; Print[n, " ", r]; Label[aa]; Continue, {n, 0, 100}] CROSSREFS Cf. A000217, A000290, A016742, A005449, A160325. Sequence in context: A284044 A126305 A088431 * A052304 A049874 A060501 Adjacent sequences: A254658 A254659 A254660 * A254662 A254663 A254664 KEYWORD nonn AUTHOR Zhi-Wei Sun, Feb 04 2015 STATUS approved

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Last modified December 9 13:48 EST 2023. Contains 367691 sequences. (Running on oeis4.)