A243759 -id:A243759 - cp's OEIS frontend

A249343 The exponent of the highest power of 3 dividing the product of the elements on the n-th row of Pascal's triangle (A001142(n)).

0, 0, 0, 2, 1, 0, 4, 2, 0, 14, 10, 6, 13, 8, 3, 12, 6, 0, 28, 20, 12, 24, 15, 6, 20, 10, 0, 68, 55, 42, 58, 44, 30, 48, 33, 18, 73, 56, 39, 60, 42, 24, 47, 28, 9, 78, 57, 36, 62, 40, 18, 46, 23, 0, 136, 110, 84, 114, 87, 60, 92, 64, 36, 132, 102, 72, 107, 76, 45, 82, 50, 18, 128, 94, 60, 100, 65, 30, 72, 36, 0

Offset: 0

Antti Karttunen, Oct 28 2014

nonn

Paolo Xausa, Table of n, a(n) for n = 0..10000 (terms 0..6560 from Antti Karttunen).
Jeffrey C. Lagarias and Harsh Mehta, Products of binomial coefficients and unreduced Farey fractions, arXiv:1409.4145 [math.NT], 2014.

Row sums of A243759.
Row 2 of array A249421.
Cf. A001142, A007949, A187059, A249345, A249347.

a249343 = a007949 . a001142  -- Reinhard Zumkeller, Mar 16 2015

A249343[n_] := Sum[#*((#+1)*3^k - n - 1) & [Floor[n/3^k]], {k, Floor[Log[3, n]]}];
Array[A249343, 100, 0] (* Paolo Xausa, Feb 11 2025 *)

allocatemem(234567890);
A249343(n) = sum(k=0, n, valuation(binomial(n, k), 3));
for(n=0, 6560, write("b249343.txt", n, " ", A249343(n)));

(define (A249343 n) (add A243759 (A000217 n) (A000096 n)))
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (+ 1 i) (+ res (intfun i)))))))

a(n) = A007949(A001142(n)).
a(n) = Sum_{k=0..n} A243759(n,k).
a(n) = Sum_{i=1..n} (2*i-n-1)*v_3(i), where v_3(i) = A007949(i) is the exponent of the highest power of 3 dividing i. - Ridouane Oudra, Jun 02 2022
a(n) = Sum_{k=1..floor(log_3(n))} t*((t+1)*3^k - n - 1), where t = floor(n/(3^k)). - Paolo Xausa, Feb 11 2025, derived from Ridouane Oudra's formula above.

A065040 Triangle read by rows: T(m,k) = exponent of the highest power of 2 dividing the binomial coefficient binomial(m,k).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 1, 2, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 2, 3, 1, 3, 2, 3, 0, 0, 0, 2, 2, 1, 1, 2, 2, 0, 0, 0, 1, 0, 3, 1, 2, 1, 3, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 2, 1, 2, 0, 3, 2, 3, 0, 2, 1, 2, 0, 0, 0, 1, 1, 0, 0, 2, 2, 0, 0, 1, 1, 0, 0

Offset: 0

Claude Lenormand (hlne.lenormand(AT)voono.net), Nov 05 2001

T(m,k) is the number of 'carries' that occur when adding k and m-k in base 2 using the traditional addition algorithm. - Tom Edgar, Jun 10 2014

			Triangle begins:
[0]
[0, 0]
[0, 1, 0]
[0, 0, 0, 0]
[0, 2, 1, 2, 0]
[0, 0, 1, 1, 0, 0]
[0, 1, 0, 2, 0, 1, 0]
[0, 0, 0, 0, 0, 0, 0, 0]
[0, 3, 2, 3, 1, 3, 2, 3, 0]
[0, 0, 2, 2, 1, 1, 2, 2, 0, 0]
[0, 1, 0, 3, 1, 2, 1, 3, 0, 1, 0]
... - _N. J. A. Sloane_, Aug 21 2021

Antti Karttunen, Table of n, a(n) for n = 0..10010; the first 141 antidiagonals, flattened
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.

Row sums: A187059.

Cf. A007318, A007814, A001511, A000120, A047999, A049606, A000680, A048881, A011371, A005187, A000265, A001316, A001317, A243759.

A065040 := (n, k) -> padic[ordp](binomial(n, k), 2):
seq(seq(A065040(n,k), k=0..n), n=0..13); # Peter Luschny, Aug 15 2017

T[m_, k_] := IntegerExponent[Binomial[m, k], 2]; Table[T[m, k], {m, 0, 13}, {k, 0, m}] // Flatten (* Jean-François Alcover, Oct 06 2016 *)

T(m,k)=hammingweight(k)+hammingweight(m-k)-hammingweight(m)
for(m=0,9,for(k=0,m,print1(T(m,k)", "))) \\ Charles R Greathouse IV, Mar 26 2013

As an array f(i,j) = f(j,i) = T(i+j,j) read by antidiagonals: f(0,j) = 0, f(1,j) = A007814(j+1), f(i,j) = Sum_{k=0..i-1} (f(1,j+k) - f(1,k)). [corrected by Kevin Ryde, Oct 07 2021]
The n-th term a(n) is equal to the binomial coefficient binomial(m,k), where m = floor((1+sqrt(8*n+1))/2) - 1 and k = n - m(m+1)/2. Also a(n) = g(m) - g(k) - g(m-k), where g(x) = Sum_{i=1..floor(log_2(x))} floor(x/2^i), m = floor((1+sqrt(8*n+1))/2) - 1, k = n - m(m+1)/2. - Hieronymus Fischer, May 05 2007
T(m,k) <= log_2 m, for m > 0. - Charles R Greathouse IV, Mar 26 2013
T(m,k) = log_2(A082907(m,k)). - Tom Edgar, Jun 10 2014
From Antti Karttunen, Oct 28 2014: (Start)
a(n) = A007814(A007318(n)).
a(n) * A047999(n) = 0 and a(n) + A047999(n) > 0 for all n.
(End)

Name clarified by Antti Karttunen, Oct 28 2014

cp's OEIS Frontend

A249343 The exponent of the highest power of 3 dividing the product of the elements on the n-th row of Pascal's triangle (A001142(n)).

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