A033461 Number of partitions of n into distinct squares.
1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 2, 2, 0, 0, 2, 2, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 2, 1, 0, 0, 2, 2, 0, 0, 2, 3, 1, 1, 2, 2, 1, 1, 1, 1, 1, 0, 2, 3, 1, 1, 4, 3, 0, 1, 2, 2, 1, 0, 1, 4, 3, 0, 2, 4, 2, 1, 3, 2, 1, 2, 3, 3, 2, 1, 3, 6, 3, 0, 2, 5, 3, 0, 1, 3, 3, 3, 4
Offset: 0
Examples
a(50)=3 because we have [1,4,9,36], [1,49], and [9,16,25]. - _Emeric Deutsch_, Jan 26 2016 From _Gus Wiseman_, Mar 09 2019: (Start) The first 30 terms count the following integer partitions: 1: (1) 4: (4) 5: (4,1) 9: (9) 10: (9,1) 13: (9,4) 14: (9,4,1) 16: (16) 17: (16,1) 20: (16,4) 21: (16,4,1) 25: (25) 25: (16,9) 26: (25,1) 26: (16,9,1) 29: (25,4) 29: (16,9,4) 30: (25,4,1) 30: (16,9,4,1) (End)
References
- James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 288-289.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)
- M. Brack, M. V. N. Murthy, and J. Bartel, Application of semiclassical methods to number theory, University of Regensburg (Germany, 2020).
- Martin Klazar, What is an answer? — remarks, results and problems on PIO formulas in combinatorial enumeration, part I, arXiv:1808.08449 [math.CO], 2018.
- Vaclav Kotesovec, Graph - The asymptotic ratio.
- M. V. N. Murthy, Matthias Brack, Rajat K. Bhaduri, and Johann Bartel, Semi-classical analysis of distinct square partitions, arXiv:1808.05146 [cond-mat.stat-mech], 2018.
Crossrefs
Programs
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Maple
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, b(n, i-1) +`if`(i^2>n, 0, b(n-i^2, i-1)))) end: a:= n-> b(n, isqrt(n)): seq(a(n), n=0..100); # Alois P. Heinz, May 14 2014
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Mathematica
nn=10; CoefficientList[Series[Product[(1+x^(k*k)), {k,nn}], {x,0,nn*nn}], x] (* T. D. Noe, Jul 24 2006 *) b[n_, i_] := b[n, i] = If[n==0, 1, If[i<1, 0, b[n, i-1] + If[i^2 > n, 0, b[n - i^2, i-1]]]]; a[n_] := b[n, Floor[Sqrt[n]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Sep 21 2015, after Alois P. Heinz *) nmax = 20; poly = ConstantArray[0, nmax^2 + 1]; poly[[1]] = 1; poly[[2]] = 1; Do[Do[poly[[j + 1]] += poly[[j - k^2 + 1]], {j, nmax^2, k^2, -1}];, {k, 2, nmax}]; poly (* Vaclav Kotesovec, Dec 09 2016 *) Table[Length[Select[IntegerPartitions[n],Reverse[Union[#]]==Length/@Split[#]&]],{n,30}] (* Gus Wiseman, Mar 09 2019 *)
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PARI
a(n)=polcoeff(prod(k=1,sqrt(n),1+x^k^2), n)
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PARI
first(n)=Vec(prod(k=1,sqrtint(n),1+'x^k^2,O('x^(n+1))+1)) \\ Charles R Greathouse IV, Sep 03 2015
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Python
from functools import cache from sympy.core.power import isqrt @cache def b(n,i): # Code after Alois P. Heinz if n == 0: return 1 if i == 0: return 0 i2 = i*i return b(n, i-1) + (0 if i2 > n else b(n - i2, i-1)) a = lambda n: b(n, isqrt(n)) print([a(n) for n in range(1, 101)]) # Darío Clavijo, Nov 30 2023
Formula
G.f.: Product_{n>=1} ( 1+x^(n^2) ).
a(n) ~ exp(3 * 2^(-5/3) * Pi^(1/3) * ((sqrt(2)-1)*zeta(3/2))^(2/3) * n^(1/3)) * ((sqrt(2)-1)*zeta(3/2))^(1/3) / (2^(4/3) * sqrt(3) * Pi^(1/3) * n^(5/6)), where zeta(3/2) = A078434. - Vaclav Kotesovec, Dec 09 2016
See Murthy, Brack, Bhaduri, Bartel (2018) for a more complete asymptotic expansion. - N. J. A. Sloane, Aug 17 2018
Extensions
More terms from Michael Somos
Comments