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 1-10 of 24 results. Next

A303401 Number of ways to write n as a*(3*a-1)/2 + b*(3*b-1)/2 + 3^c + 3^d with a,b,c,d nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two pentagonal numbers and two powers of 3.
a(n) > 0 for all n = 2..7*10^6. See A303434 for the numbers of the form x*(3*x-1)/2 + 3^y with x and y nonnegative integers. See also A303389 and A303432 for similar conjectures.

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    PenQ[n_]:=PenQ[n]=SQ[24n+1]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0);
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[12(n-3^j-3^k)+1],Do[If[PenQ[n-3^j-3^k-x(3x-1)/2],r=r+1],{x,0,(Sqrt[12(n-3^j-3^k)+1]+1)/6}]],{j,0,Log[3,n/2]},{k,j,Log[3,n-3^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

Formula

a(78) = 1 with 78 = 3*(3*3-1)/2 + 3*(3*3-1)/2 + 3^3 + 3^3.
a(285) = 1 with 285 = 3*(3*1-1)/2 + 11*(3*11-1)/2 + 3^3 + 3^4.
a(711) = 1 with 711 = 9*(3*9-1)/2 + 20*(3*20-1)/2 + 3^0 + 3^1.
a(775) = 1 with 775 = 7*(3*7-1)/2 + 21*(3*21-1)/2 + 3^3 + 3^3.
a(3200) = 1 with 12*(3*12-1)/2 + 44*(3*44-1)/2 + 3^3 + 3^4.
a(13372) = 1 with 13372 = 17*(3*17-1)/2 + 65*(3*65-1)/2 + 3^4 + 3^8.
a(16545) = 1 with 16545 = 0*(3*0-1)/2 + 98*(3*98-1)/2 + 3^0 + 3^7.

A303540 Number of ways to write n as a^2 + b^2 + binomial(2*c,c) + binomial(2*d,d), where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

0, 1, 2, 3, 2, 2, 3, 4, 3, 2, 3, 6, 4, 2, 2, 4, 4, 2, 2, 5, 5, 5, 4, 4, 4, 4, 5, 6, 5, 5, 4, 5, 4, 4, 3, 4, 5, 5, 6, 5, 5, 5, 4, 7, 3, 4, 5, 6, 4, 2, 4, 6, 7, 4, 4, 5, 7, 6, 2, 5, 4, 6, 3, 2, 5, 5, 5, 4, 4, 3, 7, 9, 6, 5, 6, 11, 7, 3, 4, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 25 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two squares and two central binomial coefficients.
It has been verified that a(n) > 0 for all n = 2..10^10.
See also A303539 and A303541 for related information.
Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 10^11. - Zhi-Wei Sun, Jul 30 2022

Examples

			a(2) = 1 since 2 = 0^2 + 0^2 + binomial(2*0,0) + binomial(2*0,0).
a(10) = 2 with 10 = 2^2 + 2^2 + binomial(2*0,0) + binomial(2*0,0) = 1^2 + 1^2 + binomial(2*1,1) + binomial(2*2,2).
a(2435) = 1 with 2435 = 32^2 + 33^2 + binomial(2*4,4) + binomial(2*5,5).
		

Crossrefs

Programs

  • Maple
    N:= 100: # for a(1)..a(N)
    A:= Vector(N):
    for b from 0 to floor(sqrt(N)) do
      for a from 0 to min(b, floor(sqrt(N-b^2))) do
        t:= a^2+b^2;
        for d from 0 do
          s:= t + binomial(2*d,d);
          if s > N then break fi;
          for c from 0 to d do
            u:= s + binomial(2*c,c);
            if u > N then break fi;
            A[u]:= A[u]+1;
    od od od od:
    convert(A,list); # Robert Israel, May 30 2018
  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    c[n_]:=c[n]=Binomial[2n,n];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;k=0;Label[bb];If[c[k]>n,Goto[aa]];Do[If[QQ[n-c[k]-c[j]],Do[If[SQ[n-c[k]-c[j]-x^2],r=r+1],{x,0,Sqrt[(n-c[k]-c[j])/2]}]],{j,0,k}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]

A303434 Numbers of the form x*(3*x-1)/2 + 3^y with x and y nonnegative integers.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 10, 13, 14, 15, 21, 23, 25, 27, 28, 31, 32, 36, 38, 39, 44, 49, 52, 54, 60, 62, 71, 73, 78, 79, 81, 82, 86, 93, 95, 97, 101, 103, 116, 118, 119, 120, 126, 132, 144, 146, 148, 151, 154, 172, 173, 177, 179, 185
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

The author's conjecture in A303401 has the following equivalent version: Each integer n > 1 can be written as the sum of two terms of the current sequence.
This has been verified for all n = 2..7*10^6.

Examples

			a(1) = 1 with 1 = 0*(3*0-1)/2 + 3^0.
a(2) = 2 with 2 = 1*(3*1-1)/2 + 3^0.
a(5) = 6 with 6 = 2*(3*2-1)/2 + 3^0.
a(6) = 8 with 8 = 2*(3*2-1)/2 + 3^1.
		

Crossrefs

Programs

  • Mathematica
    PenQ[n_]:=PenQ[n]=IntegerQ[Sqrt[24n+1]]&&(n==0||Mod[Sqrt[24n+1]+1,6]==0);
    tab={};Do[Do[If[PenQ[m-3^k],n=n+1;tab=Append[tab,m];Goto[aa]],{k,0,Log[3,m]}];Label[aa],{m,1,185}];Print[tab]

A303656 Number of ways to write n as a^2 + b^2 + 3^c + 5^d, where a,b,c,d are nonnegative integers with a <= b.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 3, 2, 4, 3, 4, 2, 4, 4, 3, 2, 4, 4, 3, 2, 4, 3, 4, 1, 4, 5, 6, 4, 6, 5, 5, 6, 6, 5, 8, 4, 6, 6, 5, 4, 7, 5, 7, 5, 6, 4, 5, 3, 4, 7, 6, 7, 8, 5, 4, 7, 5, 5, 9, 3, 6, 5, 6, 4, 6, 5, 7, 7, 4, 5, 5, 5, 4, 6, 5, 6, 10, 5, 4, 5, 7, 4, 9, 2, 9, 8, 5, 6, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 27 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two squares, a power of 3 and a power of 5.
It has been verified that a(n) > 0 for all n = 2..2*10^10.
It seems that any integer n > 1 also can be written as the sum of two squares, a power of 2 and a power of 3.
The author would like to offer 3500 US dollars as the prize for the first proof of his conjecture that a(n) > 0 for all n > 1. - Zhi-Wei Sun, Jun 05 2018
Jiao-Min Lin (a student at Nanjing University) has verified a(n) > 0 for all 1 < n <= 2.4*10^11. - Zhi-Wei Sun, Jul 30 2022

Examples

			a(2) = 1 with 2 = 0^2 + 0^2 + 3^0 + 5^0.
a(5) = 1 with 5 = 0^2 + 1^2 + 3^1 + 5^0.
a(25) = 1 with 25 = 1^2 + 4^2 + 3^1 + 5^1.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[n-3^k-5^m],Do[If[SQ[n-3^k-5^m-x^2],r=r+1],{x,0,Sqrt[(n-3^k-5^m)/2]}]],{k,0,Log[3,n]},{m,0,If[n==3^k,-1,Log[5,n-3^k]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A303539 Number of ordered pairs (k, m) with 0 <= k <= m such that n - binomial(2*k,k) - binomial(2*m,m) can be written as the sum of two squares.

Original entry on oeis.org

0, 1, 2, 3, 2, 2, 3, 4, 3, 2, 3, 6, 4, 2, 2, 4, 4, 2, 2, 5, 5, 5, 4, 4, 4, 4, 4, 5, 4, 5, 4, 4, 3, 4, 3, 4, 4, 5, 6, 5, 5, 5, 4, 7, 3, 3, 4, 6, 4, 2, 3, 5, 6, 3, 4, 5, 6, 5, 2, 5, 4, 5, 3, 2, 4, 5, 4, 3, 3, 3, 6, 7, 5, 5, 6, 10, 6, 3, 4, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 25 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
a(n) > 0 for all n = 2..10^10.
See also A303540 and A303541 for related sequences.

Examples

			a(2) = 1 with 2 - binomial(2*0,0) - binomial(2*0,0) = 0^2 + 0^2.
a(3) = 2 with 3 - binomial(2*0,0) - binomial(2*0,0) = 0^2 + 1^2 and 3 - binomial(2*0,0) - binomial(2*1,1) = 0^2 + 0^2.
a(5) = 2 with 5 - binomial(2*0,0) - binomial(2*1,1) = 1^2 + 1^2 and 5 - binomial(2*1,1) - binomial(2*1,1) = 0^2 + 1^2.
		

Crossrefs

Programs

  • Mathematica
    c[n_]:=c[n]=Binomial[2n,n];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;k=0;Label[bb];If[c[k]>n,Goto[aa]];Do[If[QQ[n-c[k]-c[j]],r=r+1],{j,0,k}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]

A303541 Numbers of the form k^2 + binomial(2*m,m) with k and m nonnegative integers.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 10, 11, 15, 17, 18, 20, 21, 22, 24, 26, 27, 29, 31, 36, 37, 38, 42, 45, 50, 51, 55, 56, 65, 66, 69, 70, 71, 74, 79, 82, 83, 84, 86, 87, 95, 101, 102, 106, 119, 120, 122, 123, 127, 134
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 25 2018

Keywords

Comments

The conjecture in A303540 has the following equivalent version: Each integer n > 1 can be written as the sum of two terms of the current sequence.
This has been verified for all n = 2..10^10.

Examples

			a(1) = 1 with 0^2 + binomial(2*0,0) = 1.
a(7) = 10 with 2^2 + binomial(2*2,2) = 10.
a(8) = 11 with 3^2 + binomial(2*1,1) = 11.
		

Crossrefs

Programs

  • Mathematica
    c[n_]:=c[n]=Binomial[2n,n];
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};n=0;Do[k=0;Label[bb];If[c[k]>m,Goto[aa]];If[SQ[m-c[k]],n=n+1;tab=Append[tab,m];Goto[aa],k=k+1;Goto[bb]];Label[aa],{m,1,134}];Print[tab]

A303543 Number of ways to write n as a^2 + b^2 + C(k) + C(m) with 0 <= a <= b and 0 < k <= m, where C(k) denotes the Catalan number binomial(2k,k)/(k+1).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 25 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be written as the sum of two squares and two Catalan numbers.
This is similar to the author's conjecture in A303540. It has been verified that a(n) > 0 for all n = 2..10^9.

Examples

			a(2) = 1 with 2 = 0^2 + 0^2 + C(1) + C(1).
a(3) = 2 with 3 = 0^2 + 1^2 + C(1) + C(1) = 0^2 + 0^2 + C(1) + C(2).
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    c[n_]:=c[n]=Binomial[2n,n]/(n+1);
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;k=1;Label[bb];If[c[k]>n,Goto[aa]];Do[If[QQ[n-c[k]-c[j]],Do[If[SQ[n-c[k]-c[j]-x^2],r=r+1],{x,0,Sqrt[(n-c[k]-c[j])/2]}]],{j,1,k}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]

A303637 Number of ways to write n as x^2 + y^2 + 2^z + 5*2^w, where x,y,z,w are nonnegative integers with x <= y.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 2, 2, 2, 4, 3, 4, 5, 5, 5, 4, 4, 5, 4, 5, 9, 8, 6, 6, 9, 7, 6, 8, 8, 10, 8, 4, 8, 5, 7, 9, 12, 9, 6, 10, 9, 11, 10, 8, 16, 12, 8, 9, 12, 9, 11, 12, 11, 9, 10, 12, 14, 10, 10
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 27 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 5.
This has been verified for n up to 5*10^9. Note that 321256731 cannot be written as x^2 + (2*y)^2 + 2^z + 5*2^w with x,y,z,w nonnegative integers.
In contrast, Crocker proved in 2008 that there are infinitely many positive integers not representable as the sum of two squares and at most two powers of 2.
570143 cannot be written as x^2 + y^2 + 2^z + 3*2^w with x,y,z,w nonnegative integers, and 2284095 cannot be written as x^2 + y^2 + 2^z + 7*2^w with x,y,z,w nonnegative integers.
Jiao-Min Lin (a student at Nanjing University) found a counterexample to the conjecture: a(18836421387) = 0. - Zhi-Wei Sun, Jul 21 2022

Examples

			a(6) = 1 with 6 = 0^2 + 0^2 + 2^0 + 5*2^0.
a(8) = 2 with 8 = 1^2 + 1^2 + 2^0 + 5*2^0 = 0^2 + 1^2 + 2^1 + 5*2^0.
a(9) = 2 with 9 = 1^2 + 1^2 + 2^1 + 5*2^0 = 0^2 + 0^2 + 2^2 + 5*2^0.
a(10) = 2 with 10 = 0^2 + 2^2 + 2^0 + 5*2^0 = 0^2 + 1^2 + 2^2 + 5*2^0.
		

References

  • R. C. Crocker, On the sum of two squares and two powers of k, Colloq. Math. 112(2008), 235-267.

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[n-5*2^k-2^m],Do[If[SQ[n-5*2^k-2^m-x^2],r=r+1],{x,0,Sqrt[(n-5*2^k-2^m)/2]}]],{k,0,Log[2,n/5]},{m,0,If[n/5==2^k,-1,Log[2,n-5*2^k]]}];tab=Append[tab,r],{n,1,60}];Print[tab]

A303601 Number of ways to write n as a*(a+1)/2 + b*(b+1)/2 + Bell(k) + Bell(m) with 0 <= a <= b and 0 < k <= m, where Bell(k) denotes the k-th Bell number A000110(k).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 26 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any integer n > 1 can be expressed as the sum of two triangular numbers and two Bell numbers.
This has been verified for all n = 2..7*10^8. Note that 111277 cannot be written as the sum of two squares and two Bell numbers.
As log(Bell(n)) is asymptotically equivalent to n*log(n), Bell numbers eventually grow faster than any exponential function.
See also A303389, A303540, A303543 and A303637 for similar conjectures.

Examples

			a(2) = 1 with 2 = 0*(0+1)/2 + 0*(0+1)/2 + Bell(1) + Bell(1).
a(3) = 2 with 3 =  0*(0+1)/2 + 1*(1+1)/2 + Bell(1) + Bell(1) = 0*(0+1)/2 + 0*(0+1)/2 + Bell(1) + Bell(2).
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
    b[n_]:=b[n]=BellB[n];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;k=1;Label[bb];If[b[k]>n,Goto[aa]];Do[If[QQ[4(n-b[k]-b[j])+1],Do[If[TQ[n-b[k]-b[j]-x(x+1)/2],r=r+1],{x,0,(Sqrt[4(n-b[k]-b[j])+1]-1)/2}]],{j,1,k}];k=k+1;Goto[bb];Label[aa];
    tab=Append[tab,r],{n,1,70}];Print[tab]

A303639 Number of ways to write n as a^2 + b^2 + binomial(2*c+1,c) + binomial(2*d+1,d), where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 2, 1, 2, 3, 3, 3, 3, 4, 2, 2, 2, 3, 4, 4, 5, 2, 4, 1, 2, 3, 3, 5, 3, 5, 1, 3, 1, 1, 6, 3, 8, 3, 6, 2, 4, 4, 2, 7, 5, 6, 2, 5, 2, 4, 5, 4, 8, 4, 7, 2, 4, 1, 3, 6, 4, 7, 3, 5, 2, 4, 2, 4, 9, 5, 6, 2, 6, 4, 5, 4, 7, 5, 2
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 27 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This is similar to the author's conjecture in A303540.
It has been verified that a(n) > 0 for all n = 2..6*10^8.

Examples

			a(9) = 1 with 9 = 1^2 + 2^2 + binomial(2*0+1,0) + binomial(2*1+1,1).
a(2530) = 1 with 2530 = 0^2 + 49^2 + binomial(2*1+1,1) + binomial(2*4+1,4).
a(3258) = 1 with 3258 = 22^2 + 52^2 + binomial(2*3+1,3) + binomial(2*3+1,3).
a(5300) = 1 with 5300 = 10^2 + 59^2 + binomial(2*1+1,1) + binomial(2*6+1,6).
a(13453) = 1 with 13453 = 51^2 + 104^2 + binomial(2*0+1,0) + binomial(2*3+1,3).
a(20964) = 1 with 20964 = 13^2 + 138^2 + binomial(2*3+1,3) + binomial(2*6+1,6).
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    c[n_]:=c[n]=Binomial[2n+1,n];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;k=0;Label[bb];If[c[k]>n,Goto[aa]];Do[If[QQ[n-c[k]-c[j]],Do[If[SQ[n-c[k]-c[j]-x^2],r=r+1],{x,0,Sqrt[(n-c[k]-c[j])/2]}]],{j,0,k}];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,80}];Print[tab]
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