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 1-4 of 4 results.

A165424 a(1) = 1, a(2) = 6, a(n) = product of the previous terms for n >= 3.

Original entry on oeis.org

1, 6, 6, 36, 1296, 1679616, 2821109907456, 7958661109946400884391936, 63340286662973277706162286946811886609896461828096
Offset: 1

Views

Author

Jaroslav Krizek, Sep 17 2009

Keywords

Links

Crossrefs

See A173501 for another version. - N. J. A. Sloane, Feb 23 2010

Programs

  • Mathematica
    a[1]:= 1; a[2]:= 6; a[n_]:= Product[a[j], {j,1,n-1}]; Table[a[n],{n,1, 12}] (* G. C. Greubel, Oct 19 2018 *)
  • PARI
    {a(n) = if(n==1, 1, if(n==2, 6, prod(j=1,n-1, a(j))))};
for(n=1,10, print1(a(n), ", ")) \\ G. C. Greubel, Oct 19 2018

Formula

a(1) = 1, a(2) = 6, a(n) = Product_{i=1..n-1} a(i), n >= 3.
a(1) = 1, a(2) = 6, a(n) = A000400(2^(n-3)) = 6^(2^(n-3)), n >= 3.
a(1) = 1, a(2) = 6, a(3) = 6, a(n) = (a(n-1))^2, n >= 4.

A225158 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 6/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 5, 31, 1141, 1502761, 2555339110801, 7279526598745139799221281, 58396508924557918552199410007906486608310469119041, 3723292553725227196293782783863296586090351965218332181732394788182320381276998127547535467381368961
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence of fractions f(n) is A165424(n+1), hence sum(A165424(i+1)/a(i),i=1..n) = product(A165424(i+1)/a(i),i=1..n) = A165424(n+2)/A225165(n) = A173501(n+2)/A225165(n).

Examples

			f(n) = 6, 6/5, 36/31, 1296/1141, ...
6 + 6/5 = 6 * 6/5 = 36/5; 6 + 6/5 + 36/31 = 6 * 6/5 * 36/31 = 1296/155; ...

Crossrefs

Cf. A100441, A165424, A173501, A225165.

Programs

  • Maple
    b:=n->6^(2^(n-2)); # n > 1
b(1):=6;
p:=proc(n) option remember; p(n-1)*a(n-1); end;
p(1):=1;
a:=proc(n) option remember; b(n)-p(n); end;
a(1):=1;
seq(a(i),i=1..9);

Formula

a(n) = 6^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
a(n) = 6^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A241241 If x is in the sequence then so are x^2 and x(x+1)/2.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 9, 10, 16, 21, 36, 45, 55, 81, 100, 136, 231, 256, 441, 666, 1035, 1296, 1540, 2025, 3025, 3321, 5050, 6561, 9316, 10000, 18496, 26796, 32896, 53361, 65536, 97461, 194481, 222111, 443556, 536130, 840456, 1071225, 1186570, 1679616, 2051325
Offset: 1

Views

Author

Reinhard Zumkeller, Apr 17 2014

Keywords

Links

Crossrefs

Cf. A000290, A000217.
Subsequence of A005214; some subsequences: A001146, A007501, A011764, A176594, A173501, A050909.

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
a241241 n = a241241_list !! (n-1)
a241241_list = 0 : 1 : f (singleton 2) where
   f s = m : f (insert (a000290 m) $ insert (a000217 m) s')
         where (m, s') = deleteFindMin s
  • Mathematica
    Nest[Flatten[{#,#^2,(#(#+1))/2}]&,{0,1,2},5]//Union (* Harvey P. Dale, Aug 12 2016 *)

Extensions

Initial 0 and 1 prepended by Jon Perry, Apr 17 2014

A225165 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 6/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 5, 155, 176855, 265770796655, 679134511201261085170655, 4943777738415359153962876938905400001585992709055
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence s(n) of the sum resp. product of fractions f(n) is A165424(n+2), hence sum(A165424(i+1)/A225158(i),i=1..n) = product(A165424(i+1)/A225158(i),i=1..n) = A165424(n+2)/a(n) = A173501(n+2)/a(n).

Examples

			f(n) = 6, 6/5, 36/31, 1296/1141, ...
6 + 6/5 = 6 * 6/5 = 36/5; 6 + 6/5 + 36/31 = 6 * 6/5 * 36/31 = 1296/155; ...
s(n) = 1/b(n) = 6, 36/5, 1296/155, ...

Crossrefs

Cf. A076628, A165424, A173501, A225158.

Programs

  • Maple
    b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
b(1):=1/6;
a:=n->6^(2^(n-1))*b(n);
seq(a(i),i=1..8);

Formula

a(n) = 6^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/6.
Showing 1-4 of 4 results.

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