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.

A243759 Triangle T(m,k): exponent of the highest power of 3 dividing the binomial coefficient binomial(m,k).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 2, 1, 2, 2, 1, 2, 2, 0, 0, 0, 2, 1, 1, 2, 1, 1, 2, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 2, 2, 1, 2, 2
Offset: 0

Views

Author

Tom Edgar, Jun 10 2014

Keywords

Comments

T(m,k) is the number of 'carries' that occur when adding k and n-k in base 3 using the traditional addition algorithm.

Examples

			The triangle begins:
0,
0, 0,
0, 0, 0,
0, 1, 1, 0;
0, 0, 1, 0, 0;
0, 0, 0, 0, 0, 0;
0, 1, 1, 0, 1, 1, 0;
0, 0, 1, 0, 0, 1, 0, 0;
0, 0, 0, 0, 0, 0, 0, 0, 0;
0, 2, 2, 1, 2, 2, 1, 2, 2, 0;
		

Crossrefs

Programs

  • Maple
    A243759:= (m,k) -> padic[ordp](binomial(m,k),3);
    for m from 0 to 50 do
      seq(A243759(m,k),k=0..m)
    od;   # Robert Israel, Jun 15 2014
  • Mathematica
    T[m_, k_] := IntegerExponent[Binomial[m, k], 3];
    Table[T[m, k], {m, 0, 12}, {k, 0, m}] // Flatten (* Jean-François Alcover, Jun 05 2022 *)
  • Sage
    m=50
    T=[0]+[3^valuation(i, 3) for i in [1..m]]
    Table=[[prod(T[1:i+1])/(prod(T[1:j+1])*prod(T[1:i-j+1])) for j in [0..i]] for i in [0..m-1]]
    [log(Integer(x),base=3) for sublist in Table for x in sublist]
    
  • Scheme
    (define (A243759 n) (A007949 (A007318 n))) ;; Antti Karttunen, Oct 28 2014

Formula

T(m,k) = log_3(A242849(m,k)).
From Antti Karttunen, Oct 28 2014: (Start)
a(n) = A007949(A007318(n)).
a(n) * A083093(n) = 0 and a(n) + A083093(n) > 0 for all n.
(End)

Extensions

Name clarified by Antti Karttunen, Oct 28 2014

A254609 Triangle read by rows: T(n,k) = A243757(n)/(A243757(k)*A243757(n-k)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 5, 5, 1, 1, 1, 5, 5, 5, 1, 1, 1, 1, 1, 5, 5, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 5, 5, 1, 5, 5, 5, 5, 1, 1, 1, 5, 5, 5, 1, 1, 5, 5, 5, 1, 1, 1, 1, 1, 5, 5, 1, 1, 1, 5, 5
Offset: 0

Views

Author

Tom Edgar, Feb 02 2015

Keywords

Comments

These are the generalized binomial coefficients associated with A060904.
The exponent of T(n,k) is the number of 'carries' that occur when adding k and n-k in base 5 using the traditional addition algorithm.
If T(n,k) != 0 mod 5, then n dominates k in base 5.
A194459(n) = number of ones in row n. - Reinhard Zumkeller, Feb 04 2015

Examples

			The first five terms in A060904 are 1, 1, 1, 1, and 5 and so T(4,2) = 1*1*1*1/((1*1)*(1*1))=1 and T(5,3) = 5*1*1*1*1/((1*1*1)*(1*1))=5.
The triangle begins:
1
1, 1
1, 1, 1
1, 1, 1, 1
1, 1, 1, 1, 1
1, 5, 5, 5, 5, 1
1, 1, 5, 5, 5, 1, 1
1, 1, 1, 5, 5, 1, 1, 1
1, 1, 1, 1, 5, 1, 1, 1, 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1
1, 5, 5, 5, 5, 1, 5, 5, 5, 5, 1
1, 1, 5, 5, 5, 1, 1, 5, 5, 5, 1, 1
1, 1, 1, 5, 5, 1, 1, 1, 5, 5, 1, 1, 1
1, 1, 1, 1, 5, 1, 1, 1, 1, 5, 1, 1, 1, 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
		

Crossrefs

Programs

  • Haskell
    import Data.List (inits)
    a254609 n k = a254609_tabl !! n !! k
    a254609_row n = a254609_tabl !! n
    a254609_tabl = zipWith (map . div)
       a243757_list $ zipWith (zipWith (*)) xss $ map reverse xss
       where xss = tail $ inits a243757_list
    -- Reinhard Zumkeller, Feb 04 2015

Formula

T(n,k) = A243757(n)/(A243757(k)*A243757(n-k)).
T(n,k) = Product_{i=1..n} A060904(i)/(Product_{i=1..k} A060904(i)*Product_{i=1..n-k} A060904(i)).
T(n,k) = A060904(n)/n*(k/A060904(k)*T(n-1,k-1)+(n-k)/A060904(n-k)*T(n-1,k)).

A243756 Triangle read by rows: T(n,k) = A242954(n)/(A242954(k) * A242954(n-k)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 4, 1, 1, 1, 4, 4, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 4, 1, 4, 4, 4, 1, 1, 1, 4, 4, 1, 1, 4, 4, 1, 1, 1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 4, 1, 4, 4, 4, 1, 4, 4, 4, 1
Offset: 0

Views

Author

Tom Edgar, Jun 09 2014

Keywords

Comments

The exponent of T(n,k) is the number of 'carries' that occur when adding k and n-k in base 4 using the traditional addition algorithm.
If T(n,k) != 0 mod 4, then n dominates k in base 4.

Examples

			The triangle begins:
1;
1, 1;
1, 1, 1;
1, 1, 1, 1;
1, 4, 4, 4, 1;
1, 1, 4, 4, 1, 1;
1, 1, 1, 4, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1, 1;
1, 4, 4, 4, 1, 4, 4, 4, 1;
1, 1, 4, 4, 1, 1, 4, 4, 1, 1;
1, 1, 1, 4, 1, 1, 1, 4, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
		

Crossrefs

Programs

  • Sage
    m=50
    T=[0]+[4^valuation(i, 4) for i in [1..m]]
    Table=[[prod(T[1:i+1])/(prod(T[1:j+1])*prod(T[1:i-j+1])) for j in [0..i]] for i in [0..m-1]]
    [x for sublist in Table for x in sublist]

Formula

T(n,k) = A242954(n)/(A242954(k) * A242954(n-k)).
T(n,k) = Product_{i=1..n} A234957(i)/(Product_{i=1..k} A234957(i)*Product_{i=1..n-k} A234957(i)).
T(n,k) = A234957(n)/n*(k/A234957(k)*T(n-1,k-1)+(n-k)/A234957(n-k)*T(n-1,k)).

A254730 Triangle read by rows: T(n,k) = A243758(n)/(A243758(k)*A243758(n-k)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 6, 6, 6, 6, 1, 1, 1, 6, 6, 6, 6, 1, 1, 1, 1, 1, 6, 6, 6, 1, 1, 1, 1, 1, 1, 1, 6, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 6, 6, 6, 6, 1, 6, 6, 6
Offset: 0

Views

Author

Tom Edgar, Feb 06 2015

Keywords

Comments

These are the generalized binomial coefficients associated with A234959.
The exponent of T(n,k) is the number of 'carries' that occur when adding k and n-k in base 6 using the traditional addition algorithm.
If T(n,k) != 0 mod 6, then n dominates k in base 6.

Examples

			The first six terms in A234959 are 1, 1, 1, 1, 1 and 6 and so T(4,2) = 1*1*1*1/((1*1)*(1*1))=1 and T(6,3) = 6*1*1*1*1*1/((1*1*1)*(1*1*1))=6.
The triangle begins:
1
1, 1
1, 1, 1
1, 1, 1, 1
1, 1, 1, 1, 1
1, 1, 1, 1, 1, 1
1, 6, 6, 6, 6, 6, 1
1, 1, 6, 6, 6, 6, 1, 1
1, 1, 1, 6, 6, 6, 1, 1, 1
1, 1, 1, 1, 6, 6, 1, 1, 1, 1
1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
1, 6, 6, 6, 6, 6, 1, 6, 6, 6, 6, 6, 1
1, 1, 6, 6, 6, 6, 1, 1, 6, 6, 6, 6, 1, 1
1, 1, 1, 6, 6, 6, 1, 1, 1, 6, 6, 6, 1, 1, 1
1, 1, 1, 1, 6, 6, 1, 1, 1, 1, 6, 6, 1, 1, 1, 1
1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
		

Crossrefs

Programs

  • Haskell
    import Data.List (inits)
    a254730 n k = a254730_tabl !! n !! k
    a254730_row n = a254730_tabl !! n
    a254730_tabl = zipWith (map . div)
       a243758_list $ zipWith (zipWith (*)) xss $ map reverse xss
       where xss = tail $ inits a243758_list
    -- Reinhard Zumkeller, Feb 09 2015
  • Sage
    P=[0]+[6^valuation(i,6) for i in [1..100]]
    [m for sublist in [[mul(P[1:n+1])/(mul(P[1:k+1])*mul(P[1:(n-k)+1])) for k in [0..n]] for n in [0..len(P)-1]] for m in sublist]
    

Formula

T(n,k) = A243758(n)/(A243758(k)*A243758(n-k)).
T(n,k) = Product_{i=1..n} A234959(i)/(Product_{i=1..k} A234959(i)*Product_{i=1..n-k} A234959(i)).
T(n,k) = A234959(n)/n*(k/A234959(k)*T(n-1,k-1)+(n-k)/A234959(n-k)*T(n-1,k)).
Showing 1-4 of 4 results.