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 A262955 Number of ordered pairs (x,y) with x >= 0 and y > 0 such that n - x^4 - y*(y+1)/2 is a pentagonal number (A000326) or twice a pentagonal number. 13
 1, 2, 3, 3, 2, 3, 3, 3, 2, 1, 4, 4, 3, 2, 3, 5, 4, 3, 3, 3, 4, 5, 5, 4, 3, 5, 6, 5, 5, 3, 6, 4, 4, 4, 1, 4, 5, 7, 6, 2, 6, 3, 3, 3, 5, 8, 5, 4, 3, 5, 4, 4, 4, 5, 5, 5, 7, 4, 3, 3, 7, 3, 3, 2, 2, 8, 5, 6, 2, 3, 5, 7, 6, 2, 1, 4, 4, 3, 6, 7, 6, 3, 5, 4, 3, 2, 6, 6, 6, 4, 6, 8 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 10, 35, 75, 134, 415, 515, 1465, 2365, 3515, 4140. LINKS Zhi-Wei Sun, Table of n, a(n) for n = 1..10000 Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396. EXAMPLE a(1) = 1 since 1 = 0^4 + 1*2/2 + p_5(0), where p_5(n) denotes the pentagonal number n*(3*n-1)/2. a(10) = 1 since 10 = 0^4 + 4*5/2 + p_5(0). a(35) = 1 since 35 = 1^4 + 4*5/2 + 2*p_5(3). a(75) = 1 since 75 = 2^4 + 5*6/2 + 2*p_5(4). a(134) = 1 since 134 = 2^4 + 1*2/2 + p_5(9). a(415) = 1 since 415 = 0^4 + 21*22/2 + 2*p_5(8). a(515) = 1 since 515 = 0^4 + 6*7/2 + 2*p_5(13). a(1465) = 1 since 1465 = 5^4 + 35*36/2 + p_5(12). a(2365) = 1 since 2365 = 5^4 + 8*9/2 + 2*p_5(24). a(3515) = 1 since 3515 = 5^4 + 51*52/2 + 2*p_5(23). a(4140) = 1 since 4140 = 1^4 + 90*91/2 + 2*p_5(4). MATHEMATICA PenQ[n_]:=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1, 6]==0) PQ[n_]:=PenQ[n]||PenQ[n/2] Do[r=0; Do[If[PQ[n-x^4-y(y+1)/2], r=r+1], {x, 0, n^(1/4)}, {y, 1, (Sqrt[8(n-x^4)+1]-1)/2}]; Print[n, " ", r]; Continue, {n, 1, 100}] CROSSREFS Cf. A000217, A000326, A000583, A262941, A262944, A262945, A262954. Sequence in context: A131294 A276869 A102313 * A276859 A007538 A242285 Adjacent sequences:  A262952 A262953 A262954 * A262956 A262957 A262958 KEYWORD nonn AUTHOR Zhi-Wei Sun, Oct 05 2015 STATUS approved

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Last modified April 5 20:19 EDT 2020. Contains 333260 sequences. (Running on oeis4.)