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 1-2 of 2 results.

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)).

A234959 Highest power of 6 dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 36, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 36, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6
Offset: 1

Views

Author

Tom Edgar, Jan 01 2014

Keywords

Comments

The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 6.

Examples

			Since 12 = 6 * 2, a(12) = 6. Likewise, since 6 does not divide 13, a(13) = 1.
		

Crossrefs

Programs

  • Haskell
    a234959 = f 1 where
       f y x = if m == 0 then f (y * 6) x' else y  where (x', m) = divMod x 6
    -- Reinhard Zumkeller, Feb 09 2015
    
  • Mathematica
    6^Table[IntegerExponent[n, 6], {n, 84}] (* Alonso del Arte, Jan 01 2014 *)
  • PARI
    a(n)=6^valuation(n,6) \\ Charles R Greathouse IV, Aug 05 2015
  • Sage
    n=200 #change n for more terms
    [6^(valuation(i,6)) for i in [1..n]]
    

Formula

a(n) = 6^(valuation(n,6)).
a(n) = 6^A122841(n). - Joerg Arndt, Jan 02 2014
G.f.: x/(1 - x) + 5 * Sum_{k>=1} 6^(k-1)*x^(6^k)/(1 - x^(6^k)). - Ilya Gutkovskiy, Jul 10 2019
Showing 1-2 of 2 results.