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 1-7 of 7 results.

A304523 Number of ordered ways to write n as the sum of a Lucas number (A000032) and a positive odd squarefree number.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 13 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 11, 13, 27, 29, 67, 139, 193, 247, 851.
It has been verified that a(n) > 0 for all n = 2..5*10^9.
See also A304331, A304333 and A304522 for similar conjectures.

Examples

			a(3) = 1 since 3 = A000032(0) + 1 with 1 odd and squarefree.
a(27) = 1 since 27 = A000032(3) + 23 with 23 odd and squarefree.
a(29) = 1 since 29 = A000032(6) + 11 with 11 odd and squarefree.
a(67) = 1 since 67 = A000032(0) + 5*13 with 5*13 odd and squarefree.
a(247) = 1 since 247 = A000032(6) + 229 with 229 odd and squarefree.
a(851) = 1 since 851 = A000032(0) + 3*283 with 3*283 odd and squarefree.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=LucasL[n];
    QQ[n_]:=QQ[n]=n>0&&Mod[n,2]==1&&SquareFreeQ[n];
    tab={};Do[r=0;k=0;Label[bb];If[k>0&&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,1,90}];Print[tab]

A304689 Number of nonnegative integers k such that n - F(k)*F(k+1) is positive and squarefree, where F(k) denotes the k-th Fibonacci number A000045(k).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 17 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 10, 90, 66690, 67452, 22756020.

Examples

			a(1) = 1 with 1 - F(0)*F(1) = 1 squarefree.
a(10) = 1 with 10 - F(0)*F(1) = 2*5 squarefree.
a(90) = 1 with 90 - F(1)*F(2) = 89 squarefree.
a(66690) = 1 with 66690 - F(10)*F(11) = 66690 - 55*89 = 5*17*727 squarefree.
a(67452) = 1 with 67452 - F(1)*F(2) = 37*1823 squarefree.
a(22756020) = 1 with 22756020 - F(2)*F(3) = 2*11378009 squarefree.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=Fibonacci[n]*Fibonacci[n+1];
    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]

A304720 Number of nonnegative integers k such that n - (4^k - k) is positive and squarefree.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 17 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This has been verified for n up to 2*10^10.
See A304721 for the values of n with a(n) = 1.
See A281192 for N such that none of N - 1 or N + 1 is squarefree: then n = N + 2 is such that n - 1 and n - 3 are not squarefree, i.e., one cannot take k = 0 or k = 1 in the present definition, and k > 1 is required to satisfy the conjecture. - M. F. Hasler, May 23 2018

Examples

			a(2) = 1 with 2 - (4^0 - 0) = 1 squarefree.
a(178) = 1 with 178 - (4^0 - 0) = 3*59 squarefree.
a(245) = 1 with 245 - (4^2 - 2) = 3*7*11 squarefree.
a(9196727) = 1 with 9196727 - (4^6 - 6) = 19*211*2293 squarefree.
a(16130577) = 1 with 16130577 - (4^9 - 9) = 2*7934221 squarefree.
a(38029402) = 1 with 38029402 - (4^1 - 1) = 1153*32983 squarefree.
a(180196927) = 1 with 180196927 - (4^11 - 11) = 2*139*227*2789 squarefree.
		

Crossrefs

Programs

  • Mathematica
    f[n_]:=f[n]=4^n-n;
    tab={};Do[r=0;k=0;Label[bb];If[f[k]>=n,Goto[aa]];If[SquareFreeQ[n-f[k]],r=r+1];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,100}];Print[tab]

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]

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]

A305424 Permutation of natural numbers: a(n) = A305422(2*n-1).

Original entry on oeis.org

1, 2, 4, 3, 6, 7, 11, 8, 16, 13, 5, 22, 19, 12, 14, 25, 50, 29, 31, 28, 37, 38, 24, 41, 9, 32, 26, 47, 44, 55, 59, 10, 20, 61, 21, 118, 67, 88, 110, 53, 69, 18, 64, 73, 94, 87, 43, 52, 91, 100, 58, 97, 56, 15, 103, 62, 82, 109, 115, 48, 23, 74, 76, 49, 98, 117, 113, 152, 131, 46, 148, 137, 143, 164, 218, 27, 96, 227, 145, 230, 89, 182, 200
Offset: 1

Views

Author

Antti Karttunen, Jun 08 2018

Keywords

Comments

Odd bisection of A305422 and A305425.

Crossrefs

Cf. A305423 (inverse).
Cf. also A064216.

Programs

  • PARI
    A091225(n) = polisirreducible(Pol(binary(n))*Mod(1, 2));
    A305419(n) = if(n<3,1, my(k=n-1); while(k>1 && !A091225(k),k--); (k));
    A305422(n) = { my(f = subst(lift(factor(Pol(binary(n))*Mod(1, 2))),x,2)); for(i=1,#f~,f[i,1] = Pol(binary(A305419(f[i,1])))); fromdigits(Vec(factorback(f))%2,2); };
    A305424(n) = A305422(n+n-1);

Formula

a(n) = A305422(2*n-1).

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]
Showing 1-7 of 7 results.