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 A343384 Number of ways to write n as [a^3/3] + [b^3/4] + [c^3/5] + [d^6/6] with a,b,c,d positive integers, where [.] is the floor function. 5
 1, 1, 2, 2, 1, 2, 1, 3, 1, 3, 2, 3, 4, 3, 4, 3, 5, 4, 3, 4, 3, 5, 3, 4, 4, 3, 6, 5, 5, 2, 4, 5, 3, 6, 3, 3, 4, 6, 5, 2, 4, 5, 4, 7, 3, 5, 4, 4, 5, 3, 3, 4, 7, 6, 3, 6, 4, 5, 6, 5, 1, 3, 7, 3, 5, 3, 5, 3, 8, 4, 3, 2, 6, 3, 6, 4, 6, 4, 6, 5, 5, 1, 5, 5, 7, 4, 7, 6, 4, 6, 5, 2, 2, 5, 5, 5, 5, 6, 3, 7, 7 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS 3-4-5-6 Conjecture: a(n) > 0 for all n >= 0. We have verified a(n) > 0 for all n = 0..10^6. Conjecture verified up to 2*10^9. - Giovanni Resta, Apr 28 2021 LINKS Zhi-Wei Sun, Table of n, a(n) for n = 0..10000 Zhi-Wei Sun, Natural numbers represented by floor(x^2/a) + floor(y^2/b) + floor(z^2/c), arXiv:1504.01608 [math.NT], 2015. EXAMPLE a(0) = 1 with 0 = [1^3/3] + [1^3/4] + [1^3/5] + [1^6/6]. a(1) = 1 with 1 = [1^3/3] + [1^3/4] + [2^3/5] + [1^6/6]. a(4) = 1 with 4 = [2^3/3] + [2^3/4] + [1^3/5] + [1^6/6]. a(6) = 1 with 6 = [1^3/3] + [3^3/4] + [1^3/5] + [1^6/6]. a(8) = 1 with 8 = [2^3/3] + [3^3/4] + [1^3/5] + [1^6/6]. a(60) = 1 with 60 = [3^3/3] + [4^3/4] + [5^3/5] + [2^6/6]. a(81) = 1 with 81 = [2^3/3] + [6^3/4] + [5^3/5] + [1^6/6]. a(300) = 1 with 300 = [7^3/3] + [5^3/4] + [9^3/5] + [2^6/6]. a(4434) = 1 with 4434 = [11^3/3] + [4^3/4] + [19^3/5] + [5^6/6]. MATHEMATICA CQ[n_]:=CQ[n]=n>0&&IntegerQ[n^(1/3)]; tab={}; Do[r=0; Do[If[CQ[3(n-Floor[x^6/6]-Floor[y^3/5]-Floor[z^3/4])+s], r=r+1], {s, 0, 2}, {x, 1, (6n+5)^(1/6)}, {y, 1, (5(n-Floor[x^6/6])+4)^(1/3)}, {z, 1, (4(n-Floor[x^6/6]-Floor[y^3/5])+3)^(1/3)}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab] CROSSREFS Cf. A000578, A001014, A343326, A343368, A343387. Sequence in context: A305432 A305298 A298824 * A237615 A343190 A256132 Adjacent sequences:  A343381 A343382 A343383 * A343385 A343386 A343387 KEYWORD nonn AUTHOR Zhi-Wei Sun, Apr 13 2021 STATUS approved

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Last modified September 23 08:02 EDT 2021. Contains 347610 sequences. (Running on oeis4.)