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.

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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]

A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))*5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))*5^m squarefree.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 06 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 7.
This has been verified for n up to 2*10^10.
See also A303821, A303934, A303949, A304031 and A304122 for related information, and A304034 for a similar conjecture.
The author would like to offer 2500 US dollars as the prize to the first proof of the conjecture, and 250 US dollars as the prize to the first explicit counterexample. - Zhi-Wei Sun, May 08 2018

Examples

			a(6) = 1 since 6 = 3 + 2^1 + 5^0 with 3 an odd prime and 2^1 + 5^0 = 3 squarefree.
a(15) = 1 since 15 = 5 + 2^3 + 2*5^0 with 5 an odd prime and 2^3 + 2*5^0 = 2*5 squarefree.
a(35) = 1 since 35 = 29 + 2^2 + 2*5^0 with 29 an odd prime and 2^2 + 2*5^0 = 2*3 squarefree.
a(91) = 1 since 91 = 17 + 2^6 + 2*5^1 with 17 an odd prime and 2^6 + 2*5^1 = 2*37 squarefree.
a(9574899) = 1 since 9574899 = 9050609 + 2^19 + 2*5^0 with 9050609 an odd prime and 2^19 + 2*5^0 = 2*5*13*37*109 squarefree.
a(6447154629) = 2 since 6447154629 = 6447121859 + 2^15 + 2*5^0 with 6447121859 prime and 2^15 + 2*5^0 = 2*5*29*113 squarefree, and 6447154629 = 5958840611 + 2^15 + 2*5^12 with 5958840611 prime and 2^15 + 2*5^12 = 2*17*41*433*809 squarefree.
		

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=n>2&&PrimeQ[n];
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*5^m]&&PQ[n-2^k-(1+Mod[n,2])*5^m],r=r+1],{k,0,Log[2,n]},{m,0,If[2^k==n,-1,Log[5,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A304034 Number of ways to write n as p + 2^k + (1+(n mod 2))*3^m with p prime, where k and m are positive integers with 2^k + (1+(n mod 2))*3^m squarefree.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 06 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 11.
This has been verified for n up to 10^10.
See also A304081 for a similar conjecture.

Examples

			a(8) = 1 since 8 = 3 + 2^1 + 3^1 with 3 prime and 2^1 + 3^1 = 5 squarefree.
a(13) = 1 since 13 = 3 + 2^2 + 2*3^1 with 3 prime and 2^2 + 2*3^1 = 2*5 squarefree.
a(19) = 1 since 19 = 5 + 2^3 + 2*3^1 with 5 prime and 2^3 + 2*3^1 = 2*7 squarefree.
a(23) = 1 since 23 = 13 + 2^2 + 2*3^1 with 13 prime and 2^2 + 2*3 = 2*5 squarefree.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*3^m]&&PrimeQ[n-2^k-(1+Mod[n,2])*3^m],r=r+1],{k,1,Log[2,n]},{m,1,If[2^k==n,-1,Log[3,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A303702 Number of ways to write 2*n as p + 2^k + 3^m, where p is a prime, and k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 5, 7, 6, 6, 9, 9, 5, 8, 9, 6, 9, 11, 8, 10, 11, 7, 12, 15, 8, 10, 12, 7, 10, 9, 8, 12, 11, 5, 12, 16, 7, 13, 17, 8, 10, 15, 10, 13, 14, 10, 12, 17, 7, 12, 18, 11, 13, 17, 10, 13, 20, 11, 14, 17, 8, 10, 16, 7, 10
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 29 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. In other words, any even number greater than 2 can be written as the sum of a prime, a power of 2 and a power of 3.
It has been verified that a(n) > 0 for all n = 2..3*10^9.
a(n) > 0 for n <= 10^11. - Jud McCranie, Jun 25 2023
a(n) > 0 for n < 10^12. - Jud McCranie, Jul 11 2023
a(n) > 0 for n <= 4*10^12. - Jud McCranie, Aug 17 2023

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 3^0 with 2 prime.
a(3) = 2 since 2*3 = 2 + 2^0 + 3^1 = 3 + 2^1 + 3^0 with 2 and 3 prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[PrimeQ[2n-2^x-3^y],r=r+1],{x,0,Log[2,2n-1]},{y,0,Log[3,2n-2^x]}];tab=Append[tab,r],{n,1,65}];Print[tab]

A303821 Number of ways to write 2*n as p + 2^x + 5^y, where p is a prime, and x and y are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 3, 4, 4, 5, 3, 6, 5, 5, 6, 6, 4, 7, 6, 7, 7, 10, 4, 9, 10, 6, 10, 8, 5, 8, 6, 7, 7, 9, 5, 8, 11, 6, 10, 11, 6, 11, 8, 6, 8, 11, 4, 9, 9, 7, 6, 11, 6, 7, 11, 7, 10, 11, 5, 11, 9, 6, 7, 6, 6, 5, 12, 7, 10, 15, 8, 15, 10, 11, 13, 11, 7, 9, 8, 9, 12, 14
Offset: 1

Views

Author

Zhi-Wei Sun, May 01 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. Moreover, for any integer n > 4, we can write 2*n as p + 2^x + 5^y, where p is an odd prime, and x and y are positive integers.
This has been verified for n up to 10^10.
See also A303934 and A304081 for further refinements, and A303932 and A304034 for similar conjectures.

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime.
a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime.
a(5616) = 2 since 2*5616 = 9059 + 2^11 + 5^3 = 10979 + 2^7 + 5^3 with 9059 and 10979 both prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303660 Number of ways to write 2*n+1 as p + 3^k + 5^m, where p is a prime, and k and m are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 28 2018

Keywords

Comments

Note that a(21323543) = 0, i.e., the odd number 2*21323543 + 1 = 42647087 cannot be written as the sum of a prime, a power of 3 and a power of 5.

Examples

			a(2) = 1 since 2*2+1 = 3 + 3^0 + 5^0 with 3 prime.
a(3) = 2 since 2*3+1 = 3 + 3^1 + 5^0 = 5 + 3^0 + 5^0 with 3 and 5 prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[PrimeQ[2n+1-5^x-3^y],r=r+1],{x,0,Log[5,2n]},{y,0,Log[3,2n+1-5^x]}];tab=Append[tab,r],{n,1,70}];Print[tab]

A303932 Number of ways to write 2*n as p + 2^k + 3^m, where p is a prime with 11 a quadratic residue modulo p, and k and m are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 02 2018

Keywords

Comments

Conjecture 1. a(n) > 0 for all n > 1, i.e., any even number greater than two can be written as the sum of a power 2, a power of 3 and a prime p with 11 a quadratic residue modulo p.
Conjecture 2. For any integer n > 2, we can write 2*n as p + 2^k + 3^m, where p is a prime with 11 a quadratic nonresidue modulo p, and k and m are nonnegative integers.
We have verified Conjectures 1 and 2 for n up to 5*10^8.

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 3^0 with 11 a quadratic residue modulo the prime 2.
a(3) = 1 since 2*3 = 2 + 2^0 + 3^1 with 11 a quadratic residue modulo the prime 2.
a(10) = 1 since 2*10 = 7 + 2^2 + 3^2 with 11 a quadratic residue modulo the prime 7.
a(14) = 1 since 2*14 = 19 + 2^3 + 3^0 with 11 a quadratic residue modulo the prime 19.
a(17) = 1 since 2*17 = 5 + 2^1 + 3^3 with 11 a quadratic residue modulo the prime 5.
a(38) = 1 since 2*38 = 37 + 2^1 + 3^3 with 11 a quadratic residue modulo the prime 37.
		

Crossrefs

Programs

  • Mathematica
    PQ[n_]:=PQ[n]=n==2||(n>2&&PrimeQ[n]&&JacobiSymbol[11,n]==1);
    tab={};Do[r=0;Do[If[PQ[2n-2^k-3^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[3,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303934 Number of ways to write 2*n as p + 2^k + 5^m with p prime and 2^k + 5^m squarefree, where k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 3, 2, 2, 3, 3, 4, 3, 5, 4, 4, 3, 4, 5, 7, 4, 7, 4, 8, 7, 6, 7, 6, 5, 5, 5, 7, 5, 8, 5, 5, 8, 6, 9, 9, 6, 8, 6, 6, 7, 8, 4, 7, 8, 7, 3, 10, 6, 7, 8, 7, 7, 9, 5, 8, 7, 6, 5, 5, 6, 3, 11, 7, 9, 12, 8, 12, 10, 11, 11, 9, 7, 9, 7, 8, 8, 11, 7, 11, 8, 9, 15, 11, 8, 9, 8, 9
Offset: 1

Views

Author

Zhi-Wei Sun, May 03 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This has been verified for all n = 2..10^10.
Note that a(n) <= A303821(n).

Examples

			a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime and 2^0 + 5^0 squarefree.
a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime and 2^1 + 5^0 squarefree.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;Do[If[SquareFreeQ[2^k+5^m]&&PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A304031 Number of ways to write 2*n as p + 2^k + 5^m with p prime and 2^k + 5^m a product of at most three distinct primes, where k and m are nonnegative integers.

Original entry on oeis.org

0, 1, 1, 3, 3, 2, 2, 3, 3, 4, 3, 5, 4, 4, 3, 4, 5, 7, 4, 7, 4, 8, 7, 6, 7, 6, 5, 5, 5, 7, 5, 8, 5, 5, 8, 6, 9, 9, 6, 8, 6, 6, 7, 8, 4, 7, 8, 7, 3, 10, 6, 7, 8, 7, 7, 9, 5, 8, 7, 6, 5, 5, 6, 3, 11, 7, 9, 12, 8, 12, 10, 11, 11, 9, 7, 9, 7, 8, 8, 11, 7, 11, 8, 9, 15, 11, 8, 9, 8, 9
Offset: 1

Views

Author

Zhi-Wei Sun, May 04 2018

Keywords

Comments

a(n) > 0 for all 1 < n <= 10^10 with the only exception n = 3114603841, and 2*3114603841 = 6219442049 + 2^3 + 5^10 with 6219442049 prime and 2^3 + 5^10 = 3*17*419*457 squarefree.
Note that a(n) <= A303934(n) <= A303821(n).

Examples

			a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 = 2^1 + 5^0  prime.
a(7) = 2 since 2*7 = 7 + 2^1 + 5^1 with 7 = 2^1 + 5^1 prime, and 2*7 = 11 + 2^1 + 5^0 with 11 and 2^1 + 5^0 both prime.
a(42908) = 2 since 2*42908 = 85751 + 2^6 + 5^0 with 85751 prime and 2^6 + 5^0 = 5*13, and 2*42908 = 69431 + 2^14 + 5^0 with 69431 prime and 2^14 + 5^0 = 5*29*113.
		

Crossrefs

Programs

  • Mathematica
    qq[n_]:=qq[n]=SquareFreeQ[n]&&Length[FactorInteger[n]]<=3;
    tab={};Do[r=0;Do[If[qq[2^k+5^m]&&PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,90}];Print[tab]
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