cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A306462 Number of ways to write n as C(2w,2) + C(x+2,3) + C(y+3,4) + C(z+4,5), where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!), w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 3, 1, 1, 4, 7, 6, 2, 2, 6, 8, 5, 1, 2, 9, 11, 5, 1, 4, 9, 12, 7, 2, 4, 10, 12, 7, 4, 6, 10, 11, 6, 5, 5, 10, 15, 8, 4, 7, 11, 14, 9, 4, 5, 11, 14, 6, 6, 10, 15, 12, 5, 7, 8, 11, 14, 7, 5, 6, 11, 14, 12, 11, 6, 11, 15, 12, 7, 9, 18, 21, 12, 5, 5, 15, 19, 11, 3, 5
Offset: 1

Views

Author

Zhi-Wei Sun, Feb 17 2019

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 5, 14, 19.
We have verified a(n) > 0 for all n = 1..5*10^6.
See also A306471 and A306477 for similar conjectures.

Examples

			a(1) = 1 with 1 = C(2,2) + C(2,3) + C(3,4) + C(4,5).
a(4) = 1 with 4 = C(2,2) + C(3,3) + C(4,4) + C(5,5).
a(5) = 1 with 5 = C(2,2) + C(4,3) + C(3,4) + C(4,5).
a(14) = 1 with 14 = C(4,2) + C(3,3) + C(4,4) + C(6,5).
a(19) = 1 with 19 = C(6,2) + C(4,3) + C(3,4) + C(4,5).
		

Crossrefs

Programs

  • Mathematica
    f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];
    HQ[n_]:=HQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==3;
    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-1-f[5,z]-f[4,y]}];Label[aa],{y,0,n-1-f[5,z]}];Label[bb],{z,0,n-1}];Label[cc];tab=Append[tab,r],{n,1,80}];Print[tab]

A306471 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)!).

Original entry on oeis.org

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, 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, 13, 20, 16, 9, 6, 8, 17, 17, 4, 2, 9, 20, 17, 6, 9, 9, 15, 23, 14, 9, 15
Offset: 0

Views

Author

Zhi-Wei Sun, Feb 17 2019

Keywords

Comments

Conjecture 1: a(n) > 1 for all n > 0.
We have verified a(n) > 0 for all n = 0..5*10^6.
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.
See also A306462 and A306477 for similar conjectures.

Examples

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

Crossrefs

Programs

  • Mathematica
    f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];
    HQ[n_]:=HQ[n]=IntegerQ[Sqrt[8n+1]]&&Mod[Sqrt[8n+1],4]==1;
    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]

A306459 Number of ways to write n as w^3 + C(x+2,3) + C(y+2,3) + C(z+2,3), where w,x,y,z are nonnegative integers with x <= y <= z, and C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 2, 1, 2, 3, 3, 3, 4, 3, 2, 2, 2, 1, 2, 2, 4, 4, 4, 2, 2, 3, 2, 1, 4, 4, 4, 4, 4, 2, 1, 3, 4, 3, 4, 4, 4, 5, 3, 2, 3, 4, 2, 4, 5, 3, 2, 4, 2, 1, 1, 3, 4, 6, 4, 2, 3, 4, 2, 3, 5, 4, 5, 7, 5, 2, 4, 4, 4, 3, 3, 4, 6, 4, 4, 2, 2, 2, 4, 3, 6, 6, 5, 4, 6, 3, 2, 3, 6, 4, 6, 4, 4, 4, 4, 3, 3
Offset: 0

Views

Author

Zhi-Wei Sun, Feb 20 2019

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0. In other words, each nonnegative integer can be written as the sum of a nonnegative cube and three tetrahedral numbers.
It seems that a(n) = 1 only for n = 0, 7, 17, 27, 34, 53, 54, 110, 118, 163, 207, 263, 270, 309, 362, 443, 1174, 1284.
We have verified a(n) > 0 for all n = 0..2*10^6.

Examples

			a(0) = 1 with 0 = 0^3 + C(2,3) + C(2,3) + C(2,3).
a(17) = 1 with 17 = 2^3 + C(3,3) + C(4,3) + C(4,3).
a(27) = 1 with 27 = 3^3 + C(2,3) + C(2,3) + C(2,3).
a(362) = 1 with 362 = 0^3 + C(6,3) + C(8,3) + C(13,3).
a(443) = 1 with 443 = 3^3 + C(5,3) + C(10,3) + C(13,3).
a(1174) = 1 with 1174 = 1^3 + C(9,3) + C(10,3) + C(19,3).
a(1284) = 1 with 1284 = 10^3 + C(7,3) + C(9,3) + C(11,3).
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=Binomial[n+2,3];
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    tab={};Do[r=0;Do[If[f[x]>n/3,Goto[cc]];Do[If[f[y]>(n-f[x])/2,Goto[bb]];Do[If[f[z]>n-f[x]-f[y],Goto[aa]];If[CQ[n-f[x]-f[y]-f[z]],r=r+1],{z,y,n-f[x]-f[y]}];Label[aa],{y,x,(n-f[x])/2}];Label[bb],{x,0,n/3}];Label[cc];tab=Append[tab,r],{n,0,100}];Print[tab]

A306790 Number of ways to write n as x*(3*x+1)/2 + y*(y+1)*(y+2)/2 + z*(z+1)*(z+2)/6, where x is a nonzero integer, and y and z are nonnegative integers.

Original entry on oeis.org

1, 2, 1, 1, 4, 3, 1, 3, 3, 1, 3, 3, 2, 4, 5, 3, 3, 4, 4, 2, 2, 4, 3, 3, 6, 3, 4, 4, 3, 3, 2, 4, 3, 3, 7, 5, 4, 5, 7, 3, 3, 6, 3, 4, 5, 3, 5, 4, 3, 3, 3, 6, 4, 3, 5, 3, 6, 5, 1, 5, 7, 6, 4, 5, 4, 6, 5, 2, 5, 7, 7, 5, 7, 5, 5, 3, 5, 3, 2, 6, 5, 7, 5, 1, 6, 5, 7, 3, 5, 6, 6, 6, 4, 3, 4, 7, 4, 3, 6, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 10 2019

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0. In other words, each positive integer n can be written as the sum of a positive generalized pentagonal number, a tetrahedral number and a tetrahedral number times three.
This has been verified for all n = 1..2*10^7.

Examples

			a(59) = 1 with 59 = (-3)*(3*(-3)+1)/2 + 2*3*4/2 + 5*6*7/6.
a(19694) = 1 with 19694 = 20*(3*20+1)/2 + 10*11*12/2 + 47*48*49/6.
a(19919) = 1 with 19919 = (-45)*(3*(-45)+1)/2 + 30*31*32/2 + 22*23*24/6.
a(33989) = 1 with 33989 = 55*(3*55+1)/2 + 20*21*22/2 + 52*53*54/6.
a(60769) = 1 with 60769 = 46*(3*46+1)/2 + 47*48*49/2 + 23*24*25/6.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=Binomial[n+2,3]; PQ[n_]:=PQ[n]=IntegerQ[Sqrt[24n+1]];
    tab={};Do[r=0;Do[If[f[x]>=n/3,Goto[cc]];Do[If[f[y]>=n-3*f[x],Goto[bb]];If[PQ[n-3*f[x]-f[y]],r=r+1];Label[aa],{y,0,n-1-3*f[x]}];Label[bb],{x,0,(n-1)/3}];Label[cc];tab=Append[tab,r],{n,1,100}];Print[tab]

A306571 Number of ways to write n as w*(w+1) + C(x+3,4) + C(y+5,6) + C(z+7,8) with w,x,y,z nonnegative integers, where C(m,k) denotes the binomial coefficient m!/(k!*(m-k)!).

Original entry on oeis.org

1, 3, 4, 4, 3, 2, 3, 6, 7, 5, 4, 4, 6, 7, 7, 6, 6, 6, 6, 5, 4, 8, 9, 6, 3, 4, 4, 6, 7, 7, 6, 8, 5, 5, 4, 8, 10, 9, 5, 5, 5, 5, 8, 10, 10, 9, 8, 7, 9, 9, 7, 8, 8, 7, 7, 6, 7, 12, 12, 8, 2, 3, 6, 11, 9, 8, 9, 7, 2, 4, 5, 8, 13, 14, 8, 6, 6, 8, 9, 9, 11, 8, 7, 7, 10, 9, 10, 11, 8, 7, 9, 11, 13, 11, 8, 5, 6, 7, 10, 10, 13
Offset: 0

Views

Author

Zhi-Wei Sun, Feb 24 2019

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0. In other words, each n = 0,1,2,... can be written as 2*C(w,2) + C(x,4) + C(y,6) + C(z,8) with w,x,y,z positive integers.
We have verified a(n) > 0 for all n = 0..2*10^7.
Note that 10413917 is the least positive integer not representable as w^2 + C(x,4) + C(y,6) + C(z,8) with w,x,y,z nonnegative integers.
See also A306477 for a similar conjecture.

Examples

			a(0) = 1 with 0 = 0*1 + C(3,4) + C(5,6) + C(7,8).
a(60) = 2 with 60 = 0*1 + C(6,4) + C(5,6) + C(10,8) = 5*6 + C(4,4) + C(8,6) + C(8,8).
a(220544) = 1 with 220544 = 151*152 + C(48,4) + C(14,6) + C(9,8).
a(809165) = 1 with 809165 = 295*296 + C(63,4) + C(10,6) + C(20,8).
a(16451641) = 1 with 16451641 = 2256*2257 + C(130,4) + C(12,6) + C(10,8).
		

Crossrefs

Programs

  • Mathematica
    f[m_,n_]:=f[m,n]=Binomial[m+n-1,m];
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[4n+1]];
    tab={};Do[r=0;Do[If[f[8,z]>n,Goto[cc]];Do[If[f[6,y]>n-f[8,z],Goto[bb]];Do[If[f[4,x]>n-f[8,z]-f[6,y],Goto[aa]];If[TQ[n-f[8,z]-f[6,y]-f[4,x]],r=r+1],{x,0,n-f[8,z]-f[6,y]}];Label[aa],{y,0,n-f[8,z]}];Label[bb],{z,0,n}];Label[cc];tab=Append[tab,r],{n,0,100}];Print[tab]

A307981 Number of ways to write n as x^3 + 2*y^3 + 3*z^3 + w*(w+1)*(w+2)/6, where x,y,z,w are nonnegative integers.

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 3, 3, 2, 2, 3, 3, 3, 3, 3, 2, 2, 3, 2, 1, 5, 4, 1, 4, 4, 4, 4, 5, 6, 3, 5, 5, 2, 4, 4, 3, 5, 5, 3, 3, 4, 3, 3, 2, 5, 3, 3, 5, 2, 2, 3, 3, 5, 2, 4, 4, 3, 3, 5, 6, 3, 5, 6, 3, 4, 4, 5, 7, 5, 4, 2, 3, 2, 3, 2, 4, 3, 3, 3, 3, 4, 5, 6, 8, 7, 7, 6, 7, 8, 6, 7, 4, 5, 4, 4, 2, 2, 4, 4, 5, 4
Offset: 0

Views

Author

Zhi-Wei Sun, May 08 2019

Keywords

Comments

Conjecture: a(n) > 0 for every nonnegative integer n. In other words, we have {x^3 + 2*y^3 + 3*z^3 + w*(w+1)*(w+2)/6: x,y,z,w = 0,1,2,...} = {0,1,2,...}.
We have verified a(n) > 0 for all n = 0..2*10^6.

Examples

			a(19) = 1 with 19 = 0^3 + 2*2^3 + 3*1^3 + 0*1*2/6.
a(22) = 1 with 22 = 0^3 + 2*1^3 + 3*0^3 + 4*5*6/6.
a(112) = 1 with 112 = 3^3 + 2*0^3 + 3*3^3 + 2*3*4/6.
a(158) = 1 with 158 = 3^3 + 2*4^3 + 3*1^3 + 0*1*2/6.
a(791) = 1 with 791 = 1^3 + 2*5^3 + 3*5^3 + 9*10*11/6.
a(956) = 1 with 956 = 9^3 + 2*0^3 + 3*4^3 + 5*6*7/6.
a(6363) = 1 with 6363 = 10^3 + 2*13^3 + 3*0^3 + 17*18*19/6.
		

Crossrefs

Programs

  • Mathematica
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];f[w_]:=f[w]=Binomial[w+2,3];
    tab={};Do[r=0;w=0;Label[bb];If[f[w]>n,Goto[aa]];Do[If[CQ[n-f[w]-2y^3-3z^3],r=r+1],{y,0,((n-f[w])/2)^(1/3)},{z,0,((n-f[w]-2y^3)/3)^(1/3)}];w=w+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,0,100}];Print[tab]
Showing 1-6 of 6 results.