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.

Previous Showing 11-15 of 15 results.

A304721 Numbers m with A304720(m) = 1.

Original entry on oeis.org

2, 3, 5, 7, 9, 10, 11, 12, 13, 19, 21, 23, 26, 28, 30, 39, 41, 46, 50, 51, 53, 55, 57, 59, 77, 89, 93, 101, 113, 129, 149, 151, 153, 161, 165, 178, 185, 189, 201, 221, 237, 245, 246, 297, 364, 377, 489, 553, 581, 639
Offset: 1

Views

Author

Zhi-Wei Sun, May 17 2018

Keywords

Comments

Conjecture: The sequence only has 112 terms as listed in the b-file.
We have verified that there is no new term below 2*10^9.

Examples

			a(9) = 13 since 13 - (4^1 - 1) = 2*5 is squarefree,  13 - (4^0 - 0) = 2^2*3 is not squarefree, and 13 - (4^k -k ) < 0 for any integer k > 1.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=4^n-n;
    tab={};Do[r=0;k=0;Label[bb];If[f[k]>=m,Goto[aa]];If[SquareFreeQ[m-f[k]],r=r+1];If[r>1,Goto[cc]];k=k+1;Goto[bb];Label[aa];If[r==1,tab=Append[tab,m]];Label[cc],{m,1,640}];Print[tab]

A304122 Squarefree numbers of the form 2^k + 5^m, where k is a positive integer and m is a nonnegative integer.

Original entry on oeis.org

3, 5, 7, 13, 17, 21, 29, 33, 37, 41, 57, 65, 69, 89, 127, 129, 133, 141, 157, 253, 257, 281, 381, 517, 537, 627, 629, 633, 641, 689, 753, 881, 1049, 1137, 1149, 1649, 2049, 2053, 2073, 2173, 3127, 3129, 3133, 3157, 3189, 3253, 3637, 4097, 4101, 4121
Offset: 1

Views

Author

Zhi-Wei Sun, May 07 2018

Keywords

Comments

The conjecture in A304081 has the following equivalent version: Any even number greater than 4 can be written as the sum of a prime and a term of the current sequence, and also any odd number greater than 8 can be written as the sum of a prime and twice a term of the current sequence.

Examples

			a(1) = 3 since 3 = 2^1 + 5^0 is squarefree.
a(6) = 21 since 21 = 2^4 + 5^1 = 3*7 is squarefree.
		

Crossrefs

Programs

  • Mathematica
    V={}; Do[If[SquareFreeQ[2^k+5^m],V=Append[V,2^k+5^m]],{k,1,12},{m,0,5}];
    LL:=LL=Sort[DeleteDuplicates[V]];
    a[n_]:=a[n]=LL[[n]];
    Table[a[n],{n,1,50}]

A304943 Number of ways to write n as the sum of a positive tribonacci number (A000073) and a positive odd squarefree number.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 22 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 6, 10, 11, 13, 29, 76, 1332, 25249.

Examples

			a(2) = 1 with 2 = 1 + 1, where 1 = A000073(2) = A000073(3) is a positive tribonacci number, and 1 is also odd and squarefree.
a(29) = 1 since 29 = A000073(8) + 5 with 5 odd and squarefree.
a(76) = 1 since 76 = A000073(6) + 3*23 with 3*23 odd and squarefree.
a(1332) = 1 since 1332 = A000073(7) + 1319 with 1319 odd and squarefree.
a(25249) = 1 since 25249 = A000073(4) + 25247 with 25247 odd and squarefree.
		

Crossrefs

Programs

  • Mathematica
    f[0]=0;f[1]=0;f[2]=1;
    f[n_]:=f[n]=f[n-1]+f[n-2]+f[n-3];
    QQ[n_]:=QQ[n]=Mod[n,2]==1&&SquareFreeQ[n];
    tab={};Do[r=0;k=3;Label[bb];If[f[k]>=n,Goto[aa]];If[QQ[n-f[k]],r=r+1];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,100}];Print[tab]

A304945 Number of nonnegative integers k such that n - k*L(k) is positive and squarefree, where L(k) denotes the k-th Lucas number A000032(k).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 22 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 5, 9, 10, 24, 28, 64, 100, 104, 136, 153, 172, 176, 344, 496, 856, 928, 1036, 1084, 1216, 1860.

Examples

			 a(1) = 1 since 1 = 0*L(0) + 1 with 1 squarefree.
a(10) = 1 since 10 = 0*L(0) + 2*5 with 2*5 squarefree.
a(136) = 1 since 136 = 2*L(2) + 2*5*13 with 2*5*13 squarefree.
a(344) = 1 since 344 = 7*L(7) + 3*47 with 3*47 squarefree.
a(1036) = 1 since 1036 = 2*L(2) + 2*5*103 with 2*5*103 squarefree.
a(1860) = 1 since 1860 = 7*L(7) + 1657 with 1657 squarefree.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=n*LucasL[n];
    QQ[n_]:=QQ[n]=SquareFreeQ[n];
    tab={};Do[r=0;k=0;Label[bb];If[f[k]>=n,Goto[aa]];If[QQ[n-f[k]],r=r+1];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,100}];Print[tab]

A305232 Number of ordered ways to write 2*n+1 as p + binomial(2k,k) + 2*binomial(2m,m), where p is an odd prime, and k and m are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 27 2018

Keywords

Comments

The first value of n > 2 with a(n) = 0 is 15212443837. Neither 2*15212443837 + 1 = 30424887675 nor 2*15657981007 + 1 = 31315962015 can be written as the sum of a prime, a central binomial coefficient and twice a central binomial coefficient.

Examples

			a(3) = 1 since 2*3 + 1 = 7 = 3 + binomial(2*1,1) + 2*binomial(2*0,0) with 3 an odd prime.
a(368233372) = 1 since 2*368233372 + 1 = 736466745 = 735761311 + binomial(2*11,11) + 2*binomial(2*0,0) with 735761311 an odd prime.
a(5274658504) = 1 since 2*5274658504 + 1 = 10549317009 = 10549316083 + binomial(2*6,6) + 2*binomial(2*0,0) with 10549316083 an odd prime.
a(8722422187) = 1 since 2*8722422187 + 1 = 17444844375 = 17444844367 + binomial(2*2,2) + 2*binomial(2*0,0) with 17444844367 an odd prime.
a(10296844792) = 1 since 2*10296844792 + 1 = 20593689585 = 20593688659 + binomial(2*6,6) + 2*binomial(2*0,0) with 20593688659 an odd prime.
		

Crossrefs

Programs

  • Mathematica
    tab={};Do[r=0;k=0;Label[aa];k=k+1;If[Binomial[2k,k]>=2n+1`,Goto[cc]];m=0;Label[bb];If[2*Binomial[2m,m]>=2n+1-Binomial[2k,k],Goto[aa]]; If[PrimeQ[2n+1-Binomial[2k,k]-2*Binomial[2m,m]],r=r+1];m=m+1;Goto[bb];Label[cc];tab=Append[tab,r],{n,1,90}];Print[tab]
Previous Showing 11-15 of 15 results.