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 1-9 of 9 results.

A143484 "Fourth down, Extream [sic] between the two farthest Bells from it" in bell-ringing is a sequence of permutations p_1=(1,2,3,4), p_2=(1,2,4,3), .. which runs through all permutations of {1,2,3,4} with period 24; sequence gives number in position 1 of n-th permutation.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Aug 19 2008

Keywords

Comments

Start with (1,2,3,4), i.e. the first permutation of {1,2,3} followed by 4; then for each next permutation, transpose 4 one to the left; if at position 1, replace {1,2,3} recursively by the next permutation of these numbers. Thereafter, for each next permutation, transpose 4 to the right. And so on.

Examples

			The full list of the 24 permutations is as follows (the present sequence gives the first column):
1 2 3 4
1 2 4 3
1 4 2 3
4 1 2 3
4 1 3 2
1 4 3 2
1 3 4 2
1 3 2 4
3 1 2 4
3 1 4 2
3 4 1 2
4 3 1 2
4 3 2 1
3 4 2 1
3 2 4 1
3 2 1 4
2 3 1 4
2 3 4 1
2 4 3 1
4 2 3 1
4 2 1 3
2 4 1 3
2 1 4 3
2 1 3 4
		

Crossrefs

Programs

  • Maple
    ring:= proc(k::nonnegint) local p,i,left,l,nf,ini; if k<=1 then proc() [1$k] end else ini := proc() p:= ring(k-1); i:= k; left:= true; l:= p(); nf:= k! end; ini(); proc() local ll; ll:= [seq(l[t], t=1..(i-1)), k, seq(l[t], t=i..(k-1))]; if left then if i>1 then i:= i-1 else left:= false; l:=p() fi else if i bell(4)[modp(n-1,24)+1][1]: seq(a(n), n=1..121);
  • Mathematica
    LinearRecurrence[{2, -2, 1, 0, 0, -1, 2, -2, 1, 0, 0, -1, 2, -2, 1, 0, 0, -1, 2, -2, 1}  {1, 1, 1, 4, 4, 1, 1, 1, 3, 3, 3, 4, 4, 3, 3, 3, 2, 2, 2, 4, 4}, 105] (* Jean-François Alcover, Mar 14 2021 *)

Formula

Period 24.

A143490 "Fourth down, Extream [sic] between the two farthest Bells from it" in bell-ringing is a sequence of permutations p_1=(1,2,3,4), p_2=(1,2,4,3), .. which runs through all permutations of {1,2,3,4} with period 24; sequence gives position of bell 3 in n-th permutation.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Aug 19 2008

Keywords

Comments

Start with (1,2,3,4), i.e. the first permutation of {1,2,3} followed by 4; then for each next permutation, transpose 4 one to the left; if at position 1, replace {1,2,3} recursively by the next permutation of these numbers. Thereafter, for each next permutation, transpose 4 to the right. And so on.

Crossrefs

Programs

  • Maple
    ring:= proc(k::nonnegint) local p,i,left,l,nf,ini; if k<=1 then proc() [1$k] end else ini := proc() p:= ring(k-1); i:= k; left:= true; l:= p(); nf:= k! end; ini(); proc() local ll; ll:= [seq(l[t], t=1..(i-1)), k, seq(l[t], t=i..(k-1))]; if left then if i>1 then i:= i-1 else left:= false; l:=p() fi else if i indx (bell(4)[modp(n-1,24)+1], 3): seq (a(n), n=1..121);
  • Mathematica
    a[n_] := a[n] = If[n <= 13, {3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 1, 2, 2}[[n]], a[n-1] - a[n-12] + a[n-13]]; Array[a, 105] (* Jean-François Alcover, May 01 2019 *)
    LinearRecurrence[{1,0,0,0,0,0,0,0,0,0,0,-1,1},{3,4,4,4,3,3,2,2,1,1,1,2,2},120] (* Harvey P. Dale, Apr 28 2020 *)

Formula

Period 24.

A143486 "Fourth down, Extream [sic] between the two farthest Bells from it" in bell-ringing is a sequence of permutations p_1=(1,2,3,4), p_2=(1,2,4,3), .. which runs through all permutations of {1,2,3,4} with period 24; sequence gives number in position 3 of n-th permutation.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Aug 19 2008

Keywords

Comments

Start with (1,2,3,4), i.e. the first permutation of {1,2,3} followed by 4; then for each next permutation, transpose 4 one to the left; if at position 1, replace {1,2,3} recursively by the next permutation of these numbers. Thereafter, for each next permutation, transpose 4 to the right. And so on.

Crossrefs

Programs

  • Maple
    ring:= proc(k::nonnegint) local p,i,left,l,nf,ini; if k<=1 then proc() [1$k] end else ini := proc() p:= ring(k-1); i:= k; left:= true; l:= p(); nf:= k! end; ini(); proc() local ll; ll:= [seq(l[t], t=1..(i-1)), k, seq(l[t], t=i..(k-1))]; if left then if i>1 then i:= i-1 else left:= false; l:=p() fi else if i bell(4)[modp(n-1,24)+1][3]: seq (a(n), n=1..121);
  • Mathematica
    LinearRecurrence[{2, -2, 1, 0, 0, -1, 2, -2, 1, 0, 0, -1, 2, -2, 1, 0, 0, -1, 2, -2, 1}, {3, 4, 2, 2, 3, 3, 4, 2, 2, 4, 1, 1, 2, 2, 4, 1, 1, 4, 3, 3, 1}, 105] (* Jean-François Alcover, Mar 14 2021 *)

Formula

Period 24.
From Chai Wah Wu, Jan 07 2020: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3) - a(n-6) + 2*a(n-7) - 2*a(n-8) + a(n-9) - a(n-12) + 2*a(n-13) - 2*a(n-14) + a(n-15) - a(n-18) + 2*a(n-19) - 2*a(n-20) + a(n-21) for n > 21.
G.f.: x*(-3*x^20 + 2*x^19 + x^18 - 4*x^17 + x^16 + 2*x^15 - 6*x^14 + 6*x^13 - 3*x^12 - 4*x^11 + 6*x^10 - 3*x^9 - 3*x^8 + 5*x^7 - 5*x^6 + x^5 + x^4 - 3*x^3 + 2*x - 3)/((x - 1)*(x^2 + 1)*(x^4 + 1)*(x^2 - x + 1)*(x^4 - x^2 + 1)*(x^8 - x^4 + 1)). (End)

A143485 "Fourth down, Extream [sic] between the two farthest Bells from it" in bell-ringing is a sequence of permutations p_1=(1,2,3,4), p_2=(1,2,4,3), .. which runs through all permutations of {1,2,3,4} with period 24; sequence gives number in position 2 of n-th permutation.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Aug 19 2008

Keywords

Comments

Start with (1,2,3,4), i.e. the first permutation of {1,2,3} followed by 4; then for each next permutation, transpose 4 one to the left; if at position 1, replace {1,2,3} recursively by the next permutation of these numbers. Thereafter, for each next permutation, transpose 4 to the right. And so on.

Crossrefs

Programs

  • Maple
    ring:= proc(k::nonnegint) local p,i,left,l,nf, ini; if k<=1 then proc() [1$k] end else ini := proc() p:= ring(k-1); i:= k; left:= true; l:= p(); nf:= k! end; ini(); proc() local ll; ll:= [seq(l[t], t=1..(i-1)), k, seq(l[t], t=i..(k-1))]; if left then if i>1 then i:= i-1 else left:= false; l:=p() fi else if i bell(4)[modp(n-1,24)+1][2]: seq (a(n), n=1..121);
  • Mathematica
    LinearRecurrence[{2, -2, 1, 0, 0, -1, 2, -2, 1, 0, 0, -1, 2, -2, 1, 0, 0, -1, 2, -2, 1}, {2, 2, 4, 1, 1, 4, 3, 3, 1, 1, 4, 3, 3, 4, 2, 2, 3, 3, 4, 2, 2}, 105] (* Jean-François Alcover, Mar 14 2021 *)

Formula

Period 24.
From Chai Wah Wu, Jan 15 2020: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3) - a(n-6) + 2*a(n-7) - 2*a(n-8) + a(n-9) - a(n-12) + 2*a(n-13) - 2*a(n-14) + a(n-15) - a(n-18) + 2*a(n-19) - 2*a(n-20) + a(n-21) for n > 21.
G.f.: x*(-x^20 + x^19 - 4*x^18 + 5*x^17 - 5*x^16 + 2*x^14 - 2*x^13 - 2*x^12 + 4*x^11 - 6*x^10 + 3*x^9 - x^8 - 2*x^7 + 2*x^6 - 5*x^4 + 5*x^3 - 4*x^2 + 2*x - 2)/((x - 1)*(x^2 + 1)*(x^4 + 1)*(x^2 - x + 1)*(x^4 - x^2 + 1)*(x^8 - x^4 + 1)). (End)

A143487 "Fourth down, Extream [sic] between the two farthest Bells from it" in bell-ringing is a sequence of permutations p_1=(1,2,3,4), p_2=(1,2,4,3), .. which runs through all permutations of {1,2,3,4} with period 24; sequence gives number in position 4 of n-th permutation.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Aug 19 2008

Keywords

Comments

Start with (1,2,3,4), i.e. the first permutation of {1,2,3} followed by 4; then for each next permutation, transpose 4 one to the left; if at position 1, replace {1,2,3} recursively by the next permutation of these numbers. Thereafter, for each next permutation, transpose 4 to the right. And so on.

Crossrefs

Programs

  • Maple
    ring:= proc(k::nonnegint) local p,i,left,l,nf,ini; if k<=1 then proc() [1$k] end else ini := proc() p:= ring(k-1); i:= k; left:= true; l:= p(); nf:= k! end; ini(); proc() local ll; ll:= [seq(l[t], t=1..(i-1)), k, seq(l[t], t=i..(k-1))]; if left then if i>1 then i:= i-1 else left:= false; l:=p() fi else if i bell(4)[modp(n-1,24)+1][4]: seq (a(n), n=1..121);
  • Mathematica
    LinearRecurrence[{2, -2, 1, 0, 0, -1, 2, -2, 1, 0, 0, -1, 2, -2, 1, 0, 0, -1, 2, -2, 1}, {4, 3, 3, 3, 2, 2, 2, 4, 4, 2, 2, 2, 1, 1, 1, 4, 4, 1, 1, 1, 3}, 105] (* Jean-François Alcover, Mar 15 2021 *)

Formula

Period 24.
From Chai Wah Wu, Jan 15 2020: (Start)
a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3) - a(n-6) + 2*a(n-7) - 2*a(n-8) + a(n-9) - a(n-12) + 2*a(n-13) - 2*a(n-14) + a(n-15) - a(n-18) + 2*a(n-19) - 2*a(n-20) + a(n-21) for n > 21.
G.f.: x*(-4*x^20 + 5*x^19 - 5*x^18 + x^17 + 2*x^16 - 2*x^15 - 2*x^14 + 2*x^13 - 2*x^12 + x^11 - x^10 + x^9 - 3*x^8 + 3*x^7 - 3*x^6 - x^5 + x^4 + x^3 - 5*x^2 + 5*x - 4)/((x - 1)*(x^2 + 1)*(x^4 + 1)*(x^2 - x + 1)*(x^4 - x^2 + 1)*(x^8 - x^4 + 1)). (End)

A143489 "Fourth down, Extream [sic] between the two farthest Bells from it" in bell-ringing is a sequence of permutations p_1=(1,2,3,4), p_2=(1,2,4,3), .. which runs through all permutations of {1,2,3,4} with period 24; sequence gives position of bell 2 in n-th permutation.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Aug 19 2008

Keywords

Comments

Start with (1,2,3,4), i.e. the first permutation of {1,2,3} followed by 4; then for each next permutation, transpose 4 one to the left; if at position 1, replace {1,2,3} recursively by the next permutation of these numbers. Thereafter, for each next permutation, transpose 4 to the right. And so on.

Crossrefs

Programs

  • Maple
    ring:= proc(k::nonnegint) local p,i,left,l,nf, ini; if k<=1 then proc() [1$k] end else ini := proc() p:= ring(k-1); i:= k; left:= true; l:= p(); nf:= k! end; ini(); proc() local ll; ll:= [seq(l[t], t=1..(i-1)), k, seq(l[t], t=i..(k-1))]; if left then if i>1 then i:= i-1 else left:= false; l:=p() fi else if i indx (bell(4)[modp(n-1,24)+1], 2): seq (a(n), n=1..121);
  • Mathematica
    LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1}, {2, 2, 3, 3, 4, 4, 4, 3, 3, 4, 4, 4, 3, 3, 2, 2, 1, 1, 1, 2, 2}, 105] (* Jean-François Alcover, Mar 15 2021 *)

Formula

Period 24.
From Chai Wah Wu, Jan 15 2020: (Start)
a(n) = a(n-1) - a(n-12) + a(n-13) for n > 13.
G.f.: x*(-x^12 - x^9 + x^7 - x^4 - x^2 - 2)/(x^13 - x^12 + x - 1). (End)

A143488 "Fourth down, Extream [sic] between the two farthest Bells from it" in bell-ringing is a sequence of permutations p_1=(1,2,3,4), p_2=(1,2,4,3), .. which runs through all permutations of {1,2,3,4} with period 24; sequence gives position of bell 1 (the treble bell) in n-th permutation.

Original entry on oeis.org

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

Views

Author

Alois P. Heinz, Aug 19 2008

Keywords

Comments

Start with (1,2,3,4), i.e. the first permutation of {1,2,3} followed by 4; then for each next permutation, transpose 4 one to the left; if at position 1, replace {1,2,3} recursively by the next permutation of these numbers. Thereafter, for each next permutation, transpose 4 to the right. And so on.

Examples

			The full list of the 24 permutations is as follows (the present sequence gives position of bell 1):
1 2 3 4
1 2 4 3
1 4 2 3
4 1 2 3
4 1 3 2
1 4 3 2
1 3 4 2
1 3 2 4
3 1 2 4
3 1 4 2
3 4 1 2
4 3 1 2
4 3 2 1
3 4 2 1
3 2 4 1
3 2 1 4
2 3 1 4
2 3 4 1
2 4 3 1
4 2 3 1
4 2 1 3
2 4 1 3
2 1 4 3
2 1 3 4
		

Crossrefs

Programs

  • Maple
    ring:= proc(k::nonnegint) local p,i,left,l,nf, ini; if k<=1 then proc() [1$k] end else ini:= proc() p:= ring(k-1); i:= k; left:= true; l:= p(); nf:= k! end; ini(); proc() local ll; ll:= [seq(l[t], t=1..(i-1)), k, seq(l[t], t=i..(k-1))]; if left then if i>1 then i:= i-1 else left:= false; l:=p() fi else if i indx(bell(4)[modp(n-1,24)+1], 1): seq(a(n), n=1..121);
  • Mathematica
    LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1}, {1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 3, 4, 4, 4, 3, 3, 4, 4, 4, 3}, 105] (* Jean-François Alcover, Mar 15 2021 *)

Formula

Period 24.
From Chai Wah Wu, Jan 15 2020: (Start)
a(n) = a(n-1) - a(n-12) + a(n-13) for n > 13.
G.f.: x*(-2*x^12 - x^10 - x^8 + x^5 - x^3 - 1)/(x^13 - x^12 + x - 1). (End)

A219388 Basic quantic arrangement for the 1 to 120 planetary electrons and elementary periods (circles I to XX) distributed by energy levels.

Original entry on oeis.org

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

Views

Author

Paul Curtz, Nov 19 2012

Keywords

Comments

In his last publication (see the reference, leaflet 5 (number 12)), Janet (1849-1932) wrote: Arrangement quantique de base,des électrons planétaires,(1 à 120),et des périodes élémentaires (cercles I à XX) disposées par niveaux d'énergie. This corresponds to 1,2,3,4,4,3,2,1 terms (*) of 8 to 1. In the example.
Hence for the periodic table,the distribution
120 119
88 87 118....113
56 55 86.....81 112....103
38 37 54.....49 80.....71 102....89
20 19 36.....31 48.....39 70....57
12 11 18.....13 30.....21
4 3 10......5
2 1.
From right to left this is the ADOMAH periodic table (2006) in A217927(n).
A rotation of 180 degrees leads to the (B) periodic table in A217927(n).
Note that the rotation of the example gives
1
2 2
3 3 3
4 4 4 4
5 5 5 5
6 6 6
7 7
8.
Hence the (B) table in A217927.
(*) See A090281.

Examples

			8
7 7
6 6 6
5 5 5 5
4 4 4 4
3 3 3
2 2
1
The 20 numbers are presented each one in a circle.
The columns are what is now quantum numbers l=0,1,2,3.
		

References

  • Charles Janet, Concordance de l'arrangement quantique, de base, des électrons planétaires des atomes, avec la classification scalariforme, hélicoïdale, des éléments chimiques, Novembre 1930, N6, Beauvais, 55 pages, 6 leaflets.

A382910 a(n) = A003266(n)^2.

Original entry on oeis.org

1, 1, 1, 4, 36, 900, 57600, 9734400, 4292870400, 4962558182400, 15011738501760000, 118907980672440960000, 2465675887223735746560000, 133859078241489389944995840000, 19025256931384645503492313743360000, 7079298104168226591849489943904256000000, 6896432754839457130755425769163265163264000000
Offset: 0

Views

Author

Edwin Hermann, Apr 08 2025

Keywords

Comments

For n>=3 number of valid symmetrical change ringing methods on n bells with the shortest number of rows per lead where the treble plain hunts out to the back. See Wikipedia and the Polster and Ross link for an explanation of bell ringing terminology.

Crossrefs

Programs

  • Maple
    a:= proc(n) a(n):= `if`(n=0, 1, a(n-1)*(<<0|1>, <1|1>>^n)[1, 2]^2) end:
    seq(a(n), n=0..16);  # Alois P. Heinz, Apr 14 2025
  • Mathematica
    k = 1; {1, 1}~Join~Reap[Do[k *= Fibonacci[n]; Sow[k^2], {n, 16}] ][[-1, 1]] (* Michael De Vlieger, Apr 14 2025 *)

Formula

a(n) = Product_{j=1..n} Fibonacci(j)^2.
a(0) = 1; a(n) = a(n-1)*A007598(n). - Hugo Pfoertner, Apr 13 2025
a(n) ~ c^2 * phi^(n*(n+1)) / 5^n where phi is the golden ratio (A001622) and c = A062073. - Amiram Eldar, Aug 18 2025
Showing 1-9 of 9 results.