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 A352632 Number of ways to write n as (a+b)*(a+2*b) + c^4 + 2*d^4, where a is a positive integer, and b,c,d are nonnegative integers. 5
 1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 2, 3, 2, 2, 2, 1, 3, 1, 2, 3, 2, 3, 3, 1, 3, 3, 2, 3, 2, 1, 4, 3, 4, 2, 2, 3, 4, 3, 3, 3, 3, 2, 5, 4, 2, 3, 4, 2, 3, 1, 4, 5, 4, 1, 4, 3, 2, 5, 4, 3, 4, 4, 4, 2, 2, 1, 6, 4, 3, 2, 2, 2, 4, 2, 4, 5, 6, 3, 5, 2, 3, 4, 5, 3, 5, 3, 4, 6, 1, 3, 6, 5, 4, 5, 3 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Conjecture: a(n) > 0 for all n > 0. In other words, any positive integer n can be written as u*v + x^4 + 2*y^4, where u,v,x,y are nonnegative integers with 0 < u <= v < 2*u. It seems that a(n) = 1 only for n = 1, 2, 3, 5, 8, 10, 11, 13, 14, 21, 23, 29, 35, 55, 59, 71, 94, 127, 134, 245, 329, 395, 404. See also A352627, A352628 and A352629 for similar conjectures. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 EXAMPLE a(11) = 1 with 11 = (3+0)*(3+2*0) + 0^4 + 2*1^4. a(23)= 1 with 23 = (3+1)*(3+2*1) + 1^4 + 2*1^4. a(134) = 1 with 134 = (10+1)*(10+2*1) + 0^4 + 2*1^4. a(245) = 1 with 245 = (10+1)*(10+2*1) + 3^4 + 2*2^4. a(329) = 1 with 329 = (6+6)*(6+2*6) + 3^4 + 2*2^4. a(395) = 1 with 395 = (2+11)*(2+2*11) + 3^4 + 2*1^4. a(404) = 1 with 404 = (15+2)*(15+2*2) + 3^4 + 2*0^4. MATHEMATICA SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]; tab={}; Do[r=0; Do[If[SQ[4(n-2d^4-c^4)+b^2], r=r+1], {d, 0, ((n-1)/2)^(1/4)}, {c, 0, (n-1-2d^4)^(1/4)}, {b, 0, Sqrt[(n-1-2d^4-c^4)/2]}]; tab=Append[tab, r], {n, 1, 100}]; Print[tab] CROSSREFS Cf. A000290, A000583, A352627, A352628, A352629. Sequence in context: A336532 A100429 A049710 * A025143 A174314 A237253 Adjacent sequences: A352629 A352630 A352631 * A352633 A352634 A352635 KEYWORD nonn AUTHOR Zhi-Wei Sun, Mar 25 2022 STATUS approved

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