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A078359 Number of ways to write n as sum of a positive square and a positive cube.

%I A078359 #30 Jun 26 2024 09:34:21
%S A078359 0,1,0,0,1,0,0,0,1,1,0,1,0,0,0,0,2,0,0,0,0,0,0,1,0,1,0,1,0,0,1,0,1,0,
%T A078359 0,1,1,0,0,0,0,0,1,1,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0,1,0,2,0,0,1,
%U A078359 0,0,0,1,1,0,0,1,0,0,0,1,0,1,0,0,0,0,0,0,2,0,1,0,0,0,0,0,0,0,0,1,1,0,0,0,0
%N A078359 Number of ways to write n as sum of a positive square and a positive cube.
%C A078359 a(A066650(n))=0, a(A055394(n))>0, a(A078360(n))=1, a(A054402(n))>1.
%C A078359 Earliest entries with a(n)=3 are n=1737, 2089, 2628, 2817. Earliest entries with a(n)=4 are n=1025, 19225, 27289, 29025, 39329, 48025, 54225. Earliest entries with a(n)=5 are n=92025, 540900, 567225, 747225. There are no a(n)>=6 in the range n=1..700000. - _R. J. Mathar_, Aug 16 2006
%C A078359 a(3375900) = 6 and a(5472225) = 7 are the first entries with those values. - _Robert Israel_, Jun 25 2024, [but see A060835. - _Hugo Pfoertner_, Jun 26 2024]
%H A078359 Seiichi Manyama, <a href="/A078359/b078359.txt">Table of n, a(n) for n = 1..10000</a>
%F A078359 G.f.: (Sum_{k>=1} x^(k^2)) * (Sum_{k>=1} x^(k^3)). - _Seiichi Manyama_, Jun 17 2023
%e A078359 a(1025)=4, as 1025 = 5^2 + 10^3 = 30^2 + 5^3 = 31^2 + 4^3 = 32^2 + 1^3.
%p A078359 interface(prettyprint=0) : A078359 := proc(n) local resul,isq,icu ; resul := 0 ; icu := 1 ; while icu^3 < n do if issqr(n-icu^3) then resul := resul+1 ; fi ; icu := icu+1 ; od ; RETURN(resul) ; end: for n from 1 to 100000 do printf("%d %d ",n,A078359(n)) ; od ; # _R. J. Mathar_, Aug 16 2006
%t A078359 a[n_] := Which[r = Reduce[x > 0 && y > 0 && n == x^2 + y^3, {x, y}, Integers]; r === False, 0, r[[0]] === And, 1, r[[0]] === Or, Length[r], True, Print["error: ", r]];
%t A078359 Table[a[n], {n, 1, 105}] (* _Jean-François Alcover_, Feb 13 2018 *)
%o A078359 (Python)
%o A078359 from collections import Counter
%o A078359 from itertools import count, takewhile, product
%o A078359 def aupto(lim):
%o A078359   sqs = list(takewhile(lambda x: x<=lim-1, (i**2 for i in count(1))))
%o A078359   cbs = list(takewhile(lambda x: x<=lim-1, (i**3 for i in count(1))))
%o A078359   cts = Counter(sum(p) for p in product(sqs, cbs))
%o A078359   return [cts[i] for i in range(1, lim+1)]
%o A078359 print(aupto(105)) # _Michael S. Branicky_, May 29 2021
%Y A078359 Cf. A054402, A055394, A060835, A066650, A078360, A171385.
%K A078359 nonn
%O A078359 1,17
%A A078359 _Reinhard Zumkeller_, Nov 25 2002