A249343 -id:A249343 - cp's OEIS frontend

A001142 a(n) = Product_{k=1..n} k^(2k - 1 - n).

1, 1, 2, 9, 96, 2500, 162000, 26471025, 11014635520, 11759522374656, 32406091200000000, 231627686043080250000, 4311500661703860387840000, 209706417310526095716965894400, 26729809777664965932590782608648192

Offset: 0

N. J. A. Sloane

Absolute value of determinant of triangular matrix containing binomial coefficients.
These are also the products of consecutive horizontal rows of the Pascal triangle. - Jeremy Hehn (ROBO_HEN5000(AT)rose.net), Mar 29 2007
Limit_{n->oo} a(n)*a(n+2)/a(n+1)^2 = e, as follows from lim_{n->oo} (1 + 1/n)^n = e. - Harlan J. Brothers, Nov 26 2009
A000225 gives the positions of odd terms. - Antti Karttunen, Nov 02 2014
Product of all unreduced fractions h/k with 1 <= k <= h <= n. - Jonathan Sondow, Aug 06 2015
a(n) is a product of the binomial coefficients from the n-th row of the Pascal triangle, for n= 0, 1, 2, ... For n > 0, a(n) means the number of all maximum chains in the poset formed by the n-dimensional Boolean cube {0,1}^n and the relation "precedes by weight". This relation is defined over {0,1}^n as follows: for arbitrary vectors u, v of {0,1}^n we say that "u precedes by weight v" if wt(u) < wt(v) or if u = v, where wt(u) denotes the (Hamming) weight of u. For more details, see the sequence A051459. - Valentin Bakoev, May 17 2019

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..50
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Mohammad K. Azarian, On the Hyperfactorial Function, Hypertriangular Function, and the Discriminants of Certain Polynomials, International Journal of Pure and Applied Mathematics 36(2), 2007, pp. 251-257. MR2312537. Zbl 1133.11012.
Harlan J. Brothers, Finding e in Pascal's Triangle, Mathematics Magazine, 85 (2012), p. 51.
Harlan J. Brothers, Pascal's Triangle: The Hidden Stor-e, The Mathematical Gazette, 96 (2012), 145-148.
Jeffrey C. Lagarias and Harsh Mehta, Products of binomial coefficients and unreduced Farey fractions, arXiv:1409.4145 [math.NT], 2014.
Leroy Quet, Problem 1636, Mathematics Magazine, Dec. 2001, p. 403.

Cf. A000178, A002109, A007318, A000225, A056077, A249421, A187059 (2-adic valuation), A249343, A249345, A249346, A249347, A249151.

Cf. also A004788, A056606 (squarefree kernel), A256113.

List([0..15], n-> Product([0..n], k-> Binomial(n,k) )); # G. C. Greubel, May 23 2019

a001142 = product . a007318_row -- Reinhard Zumkeller, Mar 16 2015

[(&*[Binomial(n,k): k in [0..n]]): n in [0..15]]; // G. C. Greubel, May 23 2019

a:=n->mul(binomial(n,k), k=0..n): seq(a(n), n=0..14); # Zerinvary Lajos, Jan 22 2008

Table[Product[k^(2*k - 1 - n), {k, n}], {n, 0, 20}] (* Harlan J. Brothers, Nov 26 2009 *)
Table[Hyperfactorial[n]/BarnesG[n+2], {n, 0, 20}] (* Peter Luschny, Nov 29 2015 *)
Table[Product[(n - k + 1)^(n - 2 k + 1), {k, 1, n}], {n, 0, 20}] (* Harlan J. Brothers, Aug 26 2023 *)

a(n):= prod(binomial(n,k),k,0,n); n : 15; for i from 0 thru n do print (a(i)); /* Valentin Bakoev, May 17 2019 */

for(n=0,16,print(prod(m=1,n,binomial(n,m))))

A001142(n) = prod(k=1, n, k^((k+k)-1-n)); \\ Antti Karttunen, Nov 02 2014

from math import factorial, prod
from fractions import Fraction
def A001142(n): return prod(Fraction((k+1)**k,factorial(k)) for k in range(1,n)) # Chai Wah Wu, Jul 15 2022

a = lambda n: prod(k^k/factorial(k) for k in (1..n))
[a(n) for n in range(20)] # Peter Luschny, Nov 29 2015

(define (A001142 n) (mul (lambda (k) (expt k (+ k k -1 (- n)))) 1 n))
(define (mul intfun lowlim uplim) (let multloop ((i lowlim) (res 1)) (cond ((> i uplim) res) (else (multloop (+ 1 i) (* res (intfun i)))))))
;; Antti Karttunen, Oct 28 2014

a(n) = C(n, 0)*C(n, 1)* ... *C(n, n).
From Harlan J. Brothers, Nov 26 2009: (Start)
a(n) = Product_{j=1..n-2} Product_{k=1..j} (1+1/k)^k, n >= 3.
a(1) = a(2) = 1, a(n) = a(n-1) * Product_{k=1..n-2} (1+1/k)^k. (End)
a(n) = hyperfactorial(n)/superfactorial(n) =  A002109(n)/A000178(n). - Peter Luschny, Jun 24 2012
a(n) ~ A^2 * exp(n^2/2 + n - 1/12) / (n^(n/2 + 1/3) * (2*Pi)^((n+1)/2)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Jul 10 2015
a(n) = Product_{i=1..n} Product_{j=1..i} (i/j). - Pedro Caceres, Apr 06 2019
a(n) = Product_{k=1..n} (n - k + 1)^(n - 2*k + 1). - Harlan J. Brothers, Aug 26 2023

More terms from James Sellers, May 01 2000

Better description from Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 30 2001

A187059 The exponent of highest power of 2 dividing the product of the elements of the n-th row of Pascal's triangle (A001142).

Original entry on oeis.org

0, 0, 1, 0, 5, 2, 4, 0, 17, 10, 12, 4, 18, 8, 11, 0, 49, 34, 36, 20, 42, 24, 27, 8, 58, 36, 39, 16, 47, 22, 26, 0, 129, 98, 100, 68, 106, 72, 75, 40, 122, 84, 87, 48, 95, 54, 58, 16, 162, 116, 119, 72, 127, 78, 82, 32, 147, 94, 98, 44, 108, 52, 57, 0, 321, 258, 260, 196, 266, 200, 203, 136, 282, 212, 215, 144, 223, 150, 154, 80, 322, 244, 247, 168, 255, 174, 178, 96, 275, 190, 194, 108, 204, 116, 121, 32, 418, 324, 327, 232, 335

Offset: 0

Bruce Reznick, Mar 05 2011

The exponent of the highest power of 2 which divides Product_{k=0..n} binomial(n, k). This can be computed using de Polignac's formula.

This is the function ord_2(Ḡ_n) extensively studied in Lagarias-Mehta (2014), and plotted in Fig. 1.1. - Antti Karttunen, Oct 22 2014

			For example, if n = 4, the power of 2 that divides 1*4*6*4*1 is 5.

I. Niven, H. S. Zuckerman, H. L. Montgomery, An Introduction to the Theory of Numbers, Wiley, 1991, pages 182, 183, 187 (Ex. 34).

Antti Karttunen and Paul Tek, Table of n, a(n) for n = 0..8191 (First 4096 terms from Karttunen)
J. Lagarias, Products of binomial coefficients and Farey fractions, Lecture in DIMACS Conference on Challenges of Identifying Integer Sequences, October 9, 2014; Part 1, Part 2.
Jeffrey C. Lagarias and Harsh Mehta, Products of binomial coefficients and unreduced Farey fractions, arXiv:1409.4145 [math.NT], 2014.

Row sums of triangular table A065040.
Row 1 of array A249421.
Cf. A000295 (a(2^k-2)), A000337 (a(2^k)), A005803 (a(2^k-3)), A036799 (a(2^k+1)), A109363 (a(2^k-4)).
Cf. A000178, A000788, A001142, A002109, A007318, A007814, A174605, A249152, A249150, A249151, A249154, A249343, A249345, A249346, A249347.

a187059 = a007814 . a001142  -- Reinhard Zumkeller, Mar 16 2015

a[n_] := Sum[IntegerExponent[Binomial[n, k], 2], {k, 0, n}]; Array[a, 100, 0]
A187059[n_] := Sum[#*((#+1)*2^k - n - 1) & [Floor[n/2^k]], {k, Floor[Log2[n]]}];
Array[A187059, 100, 0] (* Paolo Xausa, Feb 11 2025 *)
2*Accumulate[#] - Range[Length[#]]*# & [DigitCount[Range[0, 99], 2, 1]] (* Paolo Xausa, Feb 11 2025 *)

a(n)=sum(k=0,n,valuation(binomial(n,k),2))

\\ Much faster version, based on code for A065040 by Charles R Greathouse IV which if reduced even further gives the formula a(n) = 2*A000788(n) - A249154(n):
A065040(m,k) = (hammingweight(k)+hammingweight(m-k)-hammingweight(m));
A187059(n) = sum(k=0, n, A065040(n, k));
for(n=0, 4095, write("b187059.txt", n, " ", A187059(n)));
\\ Antti Karttunen, Oct 25 2014

def A187059(n): return (n+1)*n.bit_count()+sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1)) # Chai Wah Wu, Nov 11 2024

a(2^k-1) = 0 (19th century); a(2^k) = (k-1)*2^k+1 for k >= 1. (Use de Polignac.)
a(n) = Sum_{i=0..n} A065040(n,i) [where the entries of triangular table A065040(m,k) give the exponent of the maximal power of 2 dividing binomial coefficient A007318(m,k)].
a(n) = A007814(A001142(n)). - Jason Kimberley, Nov 02 2011
a(n) = A249152(n) - A174605(n). [Exponent of 2 in the n-th hyperfactorial minus exponent of 2 in the n-th superfactorial. Cf. for example Lagarias & Mehta paper or Peter Luschny's formula for A001142.] - Antti Karttunen, Oct 25 2014
a(n) = 2*A000788(n) - A249154(n). - Antti Karttunen, Nov 02 2014
a(n) = Sum_{i=1..n} (2*i-n-1)*v_2(i), where v_2(i) = A007814(i) is the exponent of the highest power of 2 dividing i. - Ridouane Oudra, Jun 02 2022
a(n) = Sum_{k=1..floor(log_2(n))} t*((t+1)*2^k - n - 1), where t = floor(n/(2^k)). - Paolo Xausa, Feb 11 2025, derived from Ridouane Oudra's formula above.

Name clarified by Antti Karttunen, Oct 22 2014

A249421 A(n,k) = exponent of the largest power of n-th prime which divides the product of the elements on row (k-1) of Pascal's triangle; a square array read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 5, 2, 0, 0, 0, 2, 1, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 4, 4, 0, 0, 0, 0, 0, 17, 2, 3, 0, 0, 0, 0, 0, 0, 10, 0, 2, 0, 0, 0, 0, 0, 0, 0, 12, 14, 1, 6, 0, 0, 0, 0, 0, 0, 0, 4, 10, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 18, 6, 8, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 8, 13, 6, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 8, 4, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Antti Karttunen, Oct 28 2014

Square array A(n,k), where n = row, k = column, read by antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ....

A(n,k) is A000040(n)-adic valuation of A001142(k-1).

			The top left corner of the array:
0, 0, 1, 0, 5, 2, 4, 0, 17, 10, 12,  4, 18,  8, 11,  0, 49, 34, 36, 20, 42,
0, 0, 0, 2, 1, 0, 4, 2,  0, 14, 10,  6, 13,  8,  3, 12,  6,  0, 28, 20, 12,
0, 0, 0, 0, 0, 4, 3, 2,  1,  0,  8,  6,  4,  2,  0, 12,  9,  6,  3,  0, 16,
0, 0, 0, 0, 0, 0, 0, 6,  5,  4,  3,  2,  1,  0, 12, 10,  8,  6,  4,  2,  0,
0, 0, 0, 0, 0, 0, 0, 0,  0,  0,  0, 10,  9,  8,  7,  6,  5,  4,  3,  2,  1,
0, 0, 0, 0, 0, 0, 0, 0,  0,  0,  0,  0,  0, 12, 11, 10,  9,  8,  7,  6,  5,
0, 0, 0, 0, 0, 0, 0, 0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 16, 15, 14, 13,
0, 0, 0, 0, 0, 0, 0, 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0, 18, 17,
...

Jeffrey C. Lagarias, Harsh Mehta, Products of binomial coefficients and unreduced Farey fractions, arXiv:1409.4145 [math.NT], 2014.

Transpose: A249422.
Row 1: A187059, Row 2: A249343, Row 3: A249345, Row 4 A249347. (Cf. also A249346).
Cf. A000040, A001142, A007318, A249344, A249151.

(define (A249421 n) (A249421bi (A002260 n) (A004736 n)))
(define (A249421bi row col) (A249344bi row (A001142 (- col 1)))) ;; Code for A249344bi given in A249344.

A(n, k) = A249344(n, A001142(k-1)).

A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 3, 0, 4, 2, 0, 2, 1, 0, 12, 6, 0, 8, 4, 0, 4, 2, 0, 24, 21, 18, 19, 14, 9, 14, 7, 0, 28, 20, 12, 20, 13, 6, 12, 6, 0, 32, 19, 6, 21, 12, 3, 10, 5, 0, 48, 42, 36, 38, 28, 18, 28, 14, 0, 50, 37, 24, 36, 24, 12, 22, 11, 0, 52, 32, 12, 34, 20, 6, 16, 8, 0

Offset: 0

Antti Karttunen, Nov 04 2014

nonn

Number of zeros on row n of A095143 (Pascal's triangle reduced modulo 9).

This should have a formula. See for example A062296, A006047 and A048967.

			Row 9 of Pascal's triangle is {1, 9, 36, 84, 126, 126, 84, 36, 9, 1}. The terms 9, 36, and 126 are the only multiples of nine, and each of them occurs two times on that row, thus a(9) = 2*3 = 6.
Row 10 of Pascal's triangle is {1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1}. The terms 45 (= 9*5) and 252 (= 9*28) are the only multiples of nine, and the former occurs twice, while the latter is alone at the center, thus a(10) = 2+1 = 3.

Antti Karttunen, Table of n, a(n) for n = 0..6561
James G. Huard, Blair K. Spearman and Kenneth S. Williams, Pascal's triangle (mod 9), Acta Arithmetica (1997), Volume: 78, Issue: 4, page 331-349.

Cf. A007318, A048277, A048967, A062296, A095143, A249343, A249723, A249731, A249732, A051382, A249719, A249720, A006047.

Total/@Table[If[Mod[Binomial[n,k],9]==0,1,0],{n,0,80},{k,0,n}] (* Harvey P. Dale, Feb 12 2020 *)

A249733(n) = { my(c=0); for(k=0,n\2,if(!(binomial(n,k)%9),c += (if(k<(n/2),2,1)))); return(c); } \\ Unoptimized.
for(n=0, 6561, write("b249733.txt", n, " ", A249733(n)));

import re
from gmpy2 import digits
def A249733(n):
    s = digits(n,3)
    n1 = s.count('1')
    n2 = s.count('2')
    n01 = s.count('10')
    n02 = s.count('20')
    n11 = len(re.findall('(?=11)',s))
    n12 = s.count('21')
    return n+1-(((3*(n01+1)+(n02<<2)+n12<<2)+3*n11)*(3**n2<Chai Wah Wu, Jul 24 2025

For all n >= 0, the following holds:
a(n) <= A048277(n).
a(n) <= A062296(n).
a(2*A249719(n)) > 0 and a((2*A249719(n))-1) > 0.
a(n) is odd if and only if n is one of the terms of A249720.

A249345 The exponent of the highest power of 5 dividing the product of the elements on the n-th row of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 0, 4, 3, 2, 1, 0, 8, 6, 4, 2, 0, 12, 9, 6, 3, 0, 16, 12, 8, 4, 0, 44, 38, 32, 26, 20, 43, 36, 29, 22, 15, 42, 34, 26, 18, 10, 41, 32, 23, 14, 5, 40, 30, 20, 10, 0, 88, 76, 64, 52, 40, 82, 69, 56, 43, 30, 76, 62, 48, 34, 20, 70, 55, 40, 25, 10, 64, 48, 32, 16, 0

Offset: 0

Antti Karttunen, Oct 28 2014

nonn

Paolo Xausa, Table of n, a(n) for n = 0..10000 (terms 0..3124 from Antti Karttunen).
Jeffrey C. Lagarias and Harsh Mehta, Products of binomial coefficients and unreduced Farey fractions, arXiv:1409.4145 [math.NT], 2014.

Row 3 of array A249421.

Cf. A001142, A007318, A112765, A187059, A249343, A249347.

A249345[n_] := Sum[#*((#+1)*5^k - n - 1) & [Floor[n/5^k]], {k, Floor[Log[5, n]]}];
Array[A249345, 100, 0] (* Paolo Xausa, Feb 11 2025 *)

allocatemem(234567890);
A249345(n) = sum(k=0, n, valuation(binomial(n, k), 5));
for(n=0, 3124, write("b249345.txt", n, " ", A249345(n)));

(define (A249345 n) (A112765 (A001142 n)))

(define (A249345 n) (add (lambda (n) (A112765 (A007318 n))) (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) = A112765(A001142(n)).
a(n) = Sum_{k=0..n} A112765(binomial(n,k)).
a(n) = Sum_{i=1..n} (2*i-n-1)*v_5(i), where v_5(i) = A112765(i) is the exponent of the highest power of 5 dividing i. - Ridouane Oudra, Jun 02 2022
a(n) = Sum_{k=1..floor(log_5(n))} t*((t+1)*5^k - n - 1), where t = floor(n/(5^k)). - Paolo Xausa, Feb 11 2025, derived from Ridouane Oudra's formula above.

A249347 The exponent of the highest power of 7 dividing the product of the elements on the n-th row of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 6, 5, 4, 3, 2, 1, 0, 12, 10, 8, 6, 4, 2, 0, 18, 15, 12, 9, 6, 3, 0, 24, 20, 16, 12, 8, 4, 0, 30, 25, 20, 15, 10, 5, 0, 36, 30, 24, 18, 12, 6, 0, 90, 82, 74, 66, 58, 50, 42, 89, 80, 71, 62, 53, 44, 35, 88, 78, 68, 58, 48, 38, 28, 87, 76, 65, 54, 43, 32, 21, 86, 74, 62, 50, 38, 26, 14, 85, 72, 59, 46, 33, 20, 7, 84, 70, 56, 42, 28, 14, 0

Offset: 0

Antti Karttunen, Oct 28 2014

nonn

Paolo Xausa, Table of n, a(n) for n = 0..10000 (terms 0..2400 from Antti Karttunen).
Jeffrey C. Lagarias and Harsh Mehta, Products of binomial coefficients and unreduced Farey fractions, arXiv:1409.4145 [math.NT], 2014.

Row 4 of array A249421.

Cf. A001142, A007318, A214411, A187059, A249343, A249345.

A249347[n_] := Sum[#*((#+1)*7^k - n - 1) & [Floor[n/7^k]], {k, Floor[Log[7, n]]}];
Array[A249347, 100, 0] (* Paolo Xausa, Feb 08 2025 *)

allocatemem(234567890);
A249347(n) = sum(k=0, n, valuation(binomial(n, k), 7));
for(n=0, 2400, write("b249347.txt", n, " ", A249347(n)));

(define (A249347 n) (A214411 (A001142 n)))

(define (A249347 n) (add (lambda (n) (A214411 (A007318 n))) (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) = A214411(A001142(n)).
a(n) = Sum_{k=0..n} A214411(binomial(n,k)).
a(n) = Sum_{i=1..n} (2*i-n-1)*v_7(i), where v_7(i) = A214411(i) is the exponent of the highest power of 7 dividing i. - Ridouane Oudra, Jun 03 2022
a(n) = Sum_{k=1..floor(log_7(n))} t*((t+1)*7^k - n - 1), where t = floor(n/(7^k)). - Paolo Xausa, Feb 09 2025, derived from Ridouane Oudra's formula above.

A249346 The exponent of the highest power of 6 dividing the product of the elements on the n-th row of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 4, 0, 0, 10, 10, 4, 13, 8, 3, 0, 6, 0, 28, 20, 12, 24, 15, 6, 20, 10, 0, 16, 47, 22, 26, 0, 30, 48, 33, 18, 73, 56, 39, 40, 42, 24, 47, 28, 9, 54, 57, 16, 62, 40, 18, 46, 23, 0, 82, 32, 84, 94, 87, 44, 92, 52, 36, 0, 102, 72, 107, 76, 45, 82, 50, 18, 128, 94, 60, 100, 65, 30, 72, 36, 0

Offset: 0

Antti Karttunen, Oct 31 2014

Sounds good with MIDI player set to FX-7.

Antti Karttunen, Table of n, a(n) for n = 0..7775

Minimum of terms A187059(n) and A249343(n).

Cf. A001142, A122841, A249151.

a249346 = a122841 . a001142  -- Reinhard Zumkeller, Mar 16 2015

IntegerExponent[#,6]&/@Times@@@Table[Binomial[n,k],{n,0,80},{k,0,n}] (* Harvey P. Dale, Nov 21 2023 *)

A249346(n) = { my(b, s2, s3); s2 = 0; s3 = 0; for(k=0, n, b = binomial(n, k); s2 += valuation(b, 2); s3 += valuation(b, 3)); min(s2,s3); };
for(n=0, 7775, write("b249346.txt", n, " ", A249346(n)));

(define (A249346 n) (min (A187059 n) (A249343 n)))

(define (A249346 n) (A122841 (A001142 n)))

a(n) = min(A187059(n), A249343(n)).
a(n) = A122841(A001142(n)).
Other identities:
a(n) = 0 when A249151(n) < 3.

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

Tom Edgar, Jun 10 2014

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

			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;

Robert Israel, Table of n, a(n) for n = 0..10010
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: A249343.

Cf. A007318, A007949, A083093, A065040, A242849, A060828, A038500.

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

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 *)

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]

(define (A243759 n) (A007949 (A007318 n))) ;; Antti Karttunen, Oct 28 2014

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)

Name clarified by Antti Karttunen, Oct 28 2014

cp's OEIS Frontend

A001142 a(n) = Product_{k=1..n} k^(2k - 1 - n).

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A187059 The exponent of highest power of 2 dividing the product of the elements of the n-th row of Pascal's triangle (A001142).

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A249421 A(n,k) = exponent of the largest power of n-th prime which divides the product of the elements on row (k-1) of Pascal's triangle; a square array read by antidiagonals.

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A249733 Number of (not necessarily distinct) multiples of 9 on row n of Pascal's triangle.

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A249345 The exponent of the highest power of 5 dividing the product of the elements on the n-th row of Pascal's triangle.

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A249347 The exponent of the highest power of 7 dividing the product of the elements on the n-th row of Pascal's triangle.

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