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Number of ways to write n as C(2w+1,2) + C(x+2,3) + C(y+3,4) + C(z+4,5) with w,x,y,z nonnegative integers, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).
6

%I #20 Feb 18 2019 14:29:06

%S 1,3,3,2,4,6,5,4,4,5,7,8,6,4,5,8,8,5,4,6,7,10,10,6,6,12,13,8,7,7,6,11,

%T 9,4,3,8,16,12,8,9,9,13,14,10,7,9,18,12,6,5,4,11,10,4,2,5,19,21,11,9,

%U 13,20,16,9,6,8,17,17,4,2,9,20,17,6,9,9,15,23,14,9,15

%N Number of ways to write n as C(2w+1,2) + C(x+2,3) + C(y+3,4) + C(z+4,5) with w,x,y,z nonnegative integers, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!).

%C Conjecture 1: a(n) > 1 for all n > 0.

%C We have verified a(n) > 0 for all n = 0..5*10^6.

%C Conjecture 2: For each r = 0, 1, any positive integer can be written as w^2 + C(x,3) + C(y,4) + C(z,5), where w,x,y,z are nonnegative integers with w - r even.

%C See also A306462 and A306477 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A306471/b306471.txt">Table of n, a(n) for n = 0..10000</a>

%e a(0) = 1 with 0 = C(1,2) + C(2,3) + C(3,4) + C(4,5).

%e a(3) = 2 with 3 = C(3,2) + C(2,3) + C(3,4) + C(4,5) = C(1,2) + C(3,3) + C(4,4) + C(5,5).

%e a(54) = 2 with 54 = C(3,2) + C(7,3) + C(6,4) + C(5,5) = C(3,2) + C(5,3) + C(7,4) + C(6,5).

%e a(69) = 1 with 69 = C(3,2) + C(5,3) + C(7,4) + C(7,5) = C(3,2) + C(5,3) + C(3,4) + C(8,5).

%t f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];

%t HQ[n_]:=HQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==1;

%t tab={};Do[r=0;Do[If[f[5,z]>n,Goto[cc]];Do[If[f[4,y]>n-f[5,z],Goto[bb]];Do[If[f[3,x]>n-f[5,z]-f[4,y],Goto[aa]];If[HQ[n-f[5,z]-f[4,y]-f[3,x]],r=r+1],{x,0,n-f[5,z]-f[4,y]}];Label[aa],{y,0,n-f[5,z]}];Label[bb],{z,0,n}];Label[cc];tab=Append[tab,r],{n,0,80}];Print[tab]

%Y Cf. A000217, A000290, A000292, A000332, A000389, A000797, A014105, A262813, A306460, A306462, A306477.

%K nonn

%O 0,2

%A _Zhi-Wei Sun_, Feb 17 2019