A249421 -id:A249421 - 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

A249151 Largest m such that m! divides the product of elements on row n of Pascal's triangle: a(n) = A055881(A001142(n)).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 6, 1, 2, 4, 10, 7, 12, 6, 4, 1, 16, 2, 18, 4, 6, 10, 22, 11, 4, 12, 2, 6, 28, 25, 30, 1, 10, 16, 6, 36, 36, 18, 12, 40, 40, 6, 42, 10, 23, 22, 46, 19, 6, 4, 16, 12, 52, 2, 10, 35, 18, 28, 58, 47, 60, 30, 63, 1, 12, 10, 66, 16, 22, 49, 70, 41, 72, 36, 4, 18, 10, 12, 78, 80, 2

Offset: 0

Antti Karttunen, Oct 25 2014

A000225 gives the positions of ones.

A006093 seems to give all such k, that a(k) = k.

			              Binomial coeff.   Their product  Largest k!
                 A007318          A001142(n)   which divides
Row 0                1                    1        1!
Row 1              1   1                  1        1!
Row 2            1   2   1                2        2!
Row 3          1   3   3   1              9        1!
Row 4        1   4   6   4   1           96        4! (96 = 4*24)
Row 5      1   5  10  10   5   1       2500        2! (2500 = 1250*2)
Row 6    1   6  15  20  15   6   1   162000        6! (162000 = 225*720)

Chai Wah Wu, Table of n, a(n) for n = 0..10000 (terms 0..4096 from Antti Karttunen)

One more than A249150.
Cf. A249423 (numbers k such that a(k) = k+1).
Cf. A249429 (numbers k such that a(k) > k).
Cf. A249433 (numbers k such that a(k) < k).
Cf. A249434 (numbers k such that a(k) >= k).
Cf. A249424 (numbers k such that a(k) = (k-1)/2).
Cf. A249428 (and the corresponding values, i.e. numbers n such that A249151(2n+1) = n).
Cf. A249425 (record positions).
Cf. A249427 (record values).
Cf. A001142, A006093, A000225, A007917, A055881, A187059, A249346, A249421, A249430, A249431, A249432.

A249151(n) = { my(uplim,padicvals,b); uplim = (n+3); padicvals = vector(uplim); for(k=0, n, b = binomial(n, k); for(i=1, uplim, padicvals[i] += valuation(b, prime(i)))); k = 1; while(k>0, for(i=1, uplim, if((padicvals[i] -= valuation(k, prime(i))) < 0, return(k-1))); k++); };
\\ Alternative implementation:
A001142(n) = prod(k=1, n, k^((k+k)-1-n));
A055881(n) = { my(i); i=2; while((0 == (n%i)), n = n/i; i++); return(i-1); }
A249151(n) = A055881(A001142(n));
for(n=0, 4096, write("b249151.txt", n, " ", A249151(n)));

from itertools import count
from collections import Counter
from math import comb
from sympy import factorint
def A249151(n):
    p = sum((Counter(factorint(comb(n,i))) for i in range(n+1)),start=Counter())
    for m in count(1):
        f = Counter(factorint(m))
        if not f<=p:
            return m-1
        p -= f # Chai Wah Wu, Aug 19 2025

(define (A249151 n) (A055881 (A001142 n)))

a(n) = A055881(A001142(n)).

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

A060175 Square array A(n,k) = exponent of the largest power of k-th prime which divides n, read by falling antidiagonals.

Original entry on oeis.org

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

Offset: 1

Henry Bottomley, Mar 14 2001

			a(12,1) = 2 since 4 = 2^2 = p_1^2 divides 12 but 8 = 2^3 does not.
a(12,2) = 1 since 3 = p_2 divides 12 but 9 = 3^2 does not.
See also examples in A249344, which is transpose of this array.
The top-left corner of the array:
n\k | 1  2  3  4  5  6  7  8
----+------------------------
1   | 0, 0, 0, 0, 0, 0, 0, 0,
2   | 1, 0, 0, 0, 0, 0, 0, 0,
3   | 0, 1, 0, 0, 0, 0, 0, 0,
4   | 2, 0, 0, 0, 0, 0, 0, 0,
5   | 0, 0, 1, 0, 0, 0, 0, 0,
6   | 1, 1, 0, 0, 0, 0, 0, 0,
7   | 0, 0, 0, 1, 0, 0, 0, 0,
8   | 3, 0, 0, 0, 0, 0, 0, 0,
9   | 0, 2, 0, 0, 0, 0, 0, 0,
10  | 1, 0, 1, 0, 0, 0, 0, 0,
11  | 0, 0, 0, 0, 1, 0, 0, 0,
12  | 2, 1, 0, 0, 0, 0, 0, 0,
...

Antti Karttunen, Table of n, a(n) for n = 1..22155; the first 210 antidiagonals

Transpose: A249344.
Column 1: A007814.
Column 2: A007949.
Column 3: A112765.
Column 4: A214411.
Row sums: A001222.
Cf. also A002260, A004736, A054841, A060176, A085604, A090622, A115627, A249421, A249422.

T[n_, k_] := IntegerExponent[n, Prime[k]];
Table[T[n-k+1, k], {n, 1, 15}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 18 2019 *)

a(n, k) = valuation(n, prime(k)); \\ Michel Marcus, Jun 24 2017

from sympy import prime
def a(n, k):
    p=prime(n)
    i=z=0
    while p**i<=k:
        if k%(p**i)==0: z=i
        i+=1
    return z
for n in range(1, 10): print([a(n - k + 1, k) for k in range(1, n + 1)]) # Indranil Ghosh, Jun 24 2017

(define (A060175 n) (A249344bi (A004736 n) (A002260 n)))
(define (A249344bi row col) (let ((p (A000040 row))) (let loop ((n col) (i 0)) (cond ((not (zero? (modulo n p))) i) (else (loop (/ n p) (+ i 1)))))))
;; Antti Karttunen, Oct 28 2014

A(n, k) = log(A060176(n, k))/log(A000040(k)) = k-th digit from right of A054841(n).

Erroneous example corrected and more terms computed by Antti Karttunen, Oct 28 2014

Name clarified by Antti Karttunen, Jan 16 2025

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)).

Original entry on oeis.org

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.

A249344 A(n,k) = exponent of the largest power of n-th prime which divides k, square array read by antidiagonals.

Original entry on oeis.org

0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 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), ... (transpose of array A060175).
A(n,k) is the (p_n)-adic valuation of k, where p_n is the n-th prime, A000040(n).
Each row is effectively a ruler function, s, with s(1) = 0. - Peter Munn, Apr 30 2022

			The top-left corner of the array:
  0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ...
  0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 0, 1, 0, ...
  0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, ...
  0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, ...
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, ...
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, ...
  ...
A(1,8) = 3, because 2^3 is the largest power of 2 (= p_1 = A000040(1)) that divides 8.
a(2,9) = 2, because 3^2 is the largest power of 3 (= p_2) that divides 9.
a(3,15) = 1, because 5^1 is the largest power of 5 (= p_3) that divides 15.

Transpose: A060175.
Row 1: A007814.
Row 2: A007949.
Row 3: A112765.
Row 4: A214411.
Cf. also A000040, A002260, A004736, A085604, A090622, A115627, A249421.
Completely additive sequences where more than one prime is mapped to 1, all other primes to 0: A065339, A083025, A087436, A169611.
Ruler functions, s, with s(1) = 0 that are not rows here: A122840, A122841, A235127, A244413.

A[n_, k_] := IntegerExponent[k, Prime[n]]; Table[A[k, n - k + 1], {n, 1, 15}, {k, 1, n}] // Flatten (* Amiram Eldar, Oct 01 2023 *)

a(n, k) = valuation(k, prime(n)); \\ Michel Marcus, Jun 24 2017

from sympy import prime
def a(n, k):
    p=prime(n)
    i=z=0
    while p**i<=k:
        if k%(p**i)==0: z=i
        i+=1
    return z
for n in range(1, 10): print([a(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Jun 24 2017

(define (A249344 n) (A249344bi (A002260 n) (A004736 n)))
(define (A249344bi row col) (let ((p (A000040 row))) (let loop ((n col) (i 0)) (cond ((not (zero? (modulo n p))) i) (else (loop (/ n p) (+ i 1)))))))

Row n, as a sequence, is completely additive with A(n, prime(n)) = 1, A(n, prime(m)) = 0 for m <> n. - Peter Munn, Apr 30 2022

Sum_{k=1..m} A(n,k) ~ (1/(prime(n)-1)) * m. - Amiram Eldar, Oct 01 2023

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.

cp's OEIS Frontend

A249422 Transpose of square array A249421.

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

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A249151 Largest m such that m! divides the product of elements on row n of Pascal's triangle: a(n) = A055881(A001142(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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A060175 Square array A(n,k) = exponent of the largest power of k-th prime which divides n, read by falling antidiagonals.

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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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