A024940 Number of partitions of n into distinct triangular numbers.
1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 0, 2, 3, 1, 1, 3, 2, 1, 4, 3, 0, 3, 3, 2, 4, 3, 3, 3, 2, 3, 3, 2, 4, 6, 4, 2, 5, 4, 2, 6, 5, 3, 7, 6, 3, 5, 5, 5, 6, 5, 4, 7, 7, 6, 8, 6, 5, 9, 7, 4, 9, 9, 6, 10, 9, 4, 9, 10, 8, 11, 11, 9, 10, 10, 9, 10, 10, 9, 14, 14, 7, 14, 14, 7, 15, 15, 8, 15, 17, 13
Offset: 0
Keywords
Examples
a(31) counts these partitions: [28,3], [21,10], [21,6,3,1], [15,10,6] _Clark Kimberling_, Mar 09 2014
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)
Programs
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Haskell
a024940 = p $ tail a000217_list where p _ 0 = 1 p (k:ks) m = if m < k then 0 else p ks (m - k) + p ks m -- Reinhard Zumkeller, Jun 28 2013
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Mathematica
Drop[ CoefficientList[ Series[ Product[(1 + x^(k*(k + 1)/2)), {k, 1, 15}], {x, 0, 102}], x], 1] (* also *) t = Table[n (n + 1)/2, {n, 1, 200}] ; p[n_] := IntegerPartitions[n, All, t]; Table[p[n], {n, 0, 12}] (*shows unrestricted partitions*) d[n_] := Select[p[n], Max[Length /@ Split@#] == 1 &]; Table[d[n], {n, 1, 31}] (*shows strict partitions*) Table[Length[d[n]], {n, 1, 70}] (* Clark Kimberling, Mar 09 2014 *) nmax = 100; nn = Floor[Sqrt[8*nmax + 1]/2] + 1; poly = ConstantArray[0, nn*(nn+1)/2 + 1]; poly[[1]] = 1; poly[[2]] = 1; Do[Do[poly[[j + 1]] += poly[[j - k*(k+1)/2 + 1]], {j, nn*(nn+1)/2, k*(k+1)/2, -1}];, {k, 2, nn}]; Take[poly, nmax + 1] (* Vaclav Kotesovec, Dec 10 2016 *)
Formula
For n>0: a(n) = b(n, 1) where b(n, k) = if n>k*(k+1)/2 then b(n-k*(k+1)/2, k+1) + b(n, k+1) else (if n=k*(k+1)/2 then 1 else 0). - Reinhard Zumkeller, Aug 26 2003
a(n) ~ exp(3*Pi^(1/3) * ((sqrt(2)-1)*Zeta(3/2))^(2/3) * n^(1/3) / 2^(4/3)) * ((sqrt(2)-1)*Zeta(3/2))^(1/3) / (2^(5/3) * sqrt(3) * Pi^(1/3) * n^(5/6)). - Vaclav Kotesovec, Jan 02 2017
G.f.: prod_{i>=1} (1+x^A000217(i)). - R. J. Mathar, Sep 20 2020