A001142 -id:A001142 - cp's OEIS frontend

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

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

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.

A249150 Number of trailing zeros in the factorial base representation of products of binomial coefficients: a(n) = A230403(A001142(n)).

Original entry on oeis.org

0, 0, 1, 0, 3, 1, 5, 0, 1, 3, 9, 6, 11, 5, 3, 0, 15, 1, 17, 3, 5, 9, 21, 10, 3, 11, 1, 5, 27, 24, 29, 0, 9, 15, 5, 35, 35, 17, 11, 39, 39, 5, 41, 9, 22, 21, 45, 18, 5, 3, 15, 11, 51, 1, 9, 34, 17, 27, 57, 46, 59, 29, 62, 0, 11, 9, 65, 15, 21, 48, 69, 40, 71, 35, 3, 17, 9, 11, 77, 79, 1

Offset: 0

Antti Karttunen, Oct 25 2014

nonn

a(n) = A249151(n)-1. Please see the comments and graph of that sequence.

One less than A249151.
Cf. A249423 (values k such that a(k) = k).
Cf. A249425 (record positions).
Cf. A249426 (record values).
Cf. A001142, A187059, A230403.

(define (A249150 n) (A230403 (A001142 n)))

a(n) = A230403(A001142(n)).

A056077 Indices n of terms of sequence A001142, Product_{k=0..n} binomial(n,k), that are divisible by all primes <= n.

Original entry on oeis.org

1, 2, 4, 6, 10, 11, 12, 16, 18, 22, 23, 28, 29, 30, 35, 36, 39, 40, 42, 44, 46, 47, 52, 55, 58, 59, 60, 62, 66, 69, 70, 71, 72, 78, 79, 82, 83, 88, 89, 95, 96, 100, 102, 104, 106, 107, 108, 111, 112, 119, 125, 126, 130, 131, 134, 136, 138, 139, 143, 148, 149, 150, 153

Offset: 1

Leroy Quet, Jul 26 2000

a(n) + 1 is either a prime or a "mutinous number" (A027854).

			11 is included because Product_{k=0..11} binomial(11, k) is divisible by 2, 3, 5, 7 and 11.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Hans Montanus and Ron Westdijk, Cellular Automation and Binomials, Green Blue Mathematics (2022), p. 69.

Cf. A001142, A056606, A027854, A002110.

isA056077 := proc(n) local radh; radh := proc(n) option remember;
mul(k, k = numtheory:-factorset(mul(k^k/factorial(k), k=0..n))) end;
type(radh(n)/radh(n-1), integer) end: # isA056077(0) = true.
select(isA056077, [$1..153]); # Peter Luschny, Dec 21 2019

With[{s = Select[Range@ 154, Function[n, (n/Apply[Power, Last@ #]) > #[[-1, 1]] &@ FactorInteger[n]]]}, -1 + Union[s, Prime@ Range@ PrimePi@ Max@ s]] (* Michael De Vlieger, Sep 23 2017 *)

Let h(m) = Product(PrimeDivisors(Product_{k=0..m} k^k/k!)). If h(m-1) divides h(m) then m is in this sequence. # Peter Luschny, Dec 21 2019

Extended by Ray Chandler, Nov 17 2008

A109873 a(n) = product of terms in row n of Pascal's triangle (A001142) divided by n^k, where n^k is the largest power of n dividing it.

Original entry on oeis.org

1, 1, 1, 6, 4, 125, 225, 336140, 2458624, 324060912, 8930250000, 835597712998125, 9001015156742400, 6600661714966989472803, 68987440762943255933340961, 28036608657071518646200652343750, 377177413291384771899817984000000

Offset: 1

Amarnath Murthy, Jul 10 2005

nonn

If p is a prime then a(p) = A001142(p)/(p^(p-1)).

			a(5) = 1*5*10*10*5*1/625= 4.

Cf. A001142, A109874.

A001142 := proc(n) local k ; mul(k^(2*k-1-n),k=1..n) ; end: A109873 := proc(n) local a; a := A001142(n) ; while a mod n = 0 and a > 1 do a := a/n ; od; RETURN(a) ; end: seq(A109873(n),n=1..20) ; # R. J. Mathar, Aug 15 2007

More terms from R. J. Mathar, Aug 15 2007

A109874 Largest exponent e such that n^e that divides A001142(n).

Original entry on oeis.org

1, 2, 2, 4, 4, 6, 5, 7, 8, 10, 9, 12, 11, 12, 12, 16, 14, 18, 16, 18, 20, 22, 19, 22, 24, 22, 23, 28, 26, 30, 25, 30, 32, 30, 36, 36, 36, 36, 40, 40, 36, 42, 40, 39, 44, 46, 40, 45, 44, 46, 48, 52, 45, 50, 49, 54, 56, 58, 54

Offset: 2

Amarnath Murthy, Jul 10 2005

nonn

a(n) = n-1, if n is a prime. If n is composite, a(n) >= 2.
Conjectures: (1) If n is even and n = 2^r*m, m odd and >1, then a(n)= n-r-1. (2) If n = 2^r then a(n) = n-3. (3) If n is odd and composite then a(n) = n-2.
a(n) is the highest exponent e such that n^e divides Product_{k=0..n} binomial(n, k). - Joerg Arndt, Jun 04 2022

Cf. A001142, A109873.

A001142 := proc(n) local k ; mul(k^(2*k-1-n),k=1..n) ; end: A109874 := proc(n) local a,k; a := A001142(n) ; k := 0 ; while a mod n = 0 and a > 1 do a := a/n ; k := k+1 ; od; RETURN(k) ; end: seq(A109874(n),n=2..60) ; # R. J. Mathar, Aug 15 2007

a[n_] := IntegerExponent[Product[Binomial[n, k], {k, 0, n}], n];
Table[a[n], {n, 2, 60}] (* Jean-François Alcover, Apr 02 2024 *)

for(n=2,60,print1(valuation(prod(k=0,n,binomial(n,k)),n),", ")); \\ Joerg Arndt, Jun 04 2022

Corrected and extended by R. J. Mathar, Aug 15 2007

Name corrected by Joerg Arndt, Jun 04 2022

A219268 Logarithmic derivative of A001142, where A001142(n) = product{k=1..n} k^k/k!.

Original entry on oeis.org

1, 3, 22, 347, 11986, 956334, 184142134, 87903876147, 105736320973732, 323943204887363938, 2547547949361933790328, 51735228018482706470521574, 2726127372514537039881847535054, 374214400937086673452020875815709240, 134262616041282033840675468757467513112522

Offset: 1

Paul D. Hanna, Nov 16 2012

nonn

A001142(n) = hyperfactorial(n)/superfactorial(n) = A002109(n)/A000178(n).

			L.g.f.: L(x) = x + 3*x^2/2 + 22*x^3/3 + 347*x^4/4 + 11986*x^5/5 + 956334*x^6/6 +...
where
exp(L(x)) = 1 + x + 2*x^2 + 9*x^3 + 96*x^4 + 2500*x^5 + 162000*x^6 + 26471025*x^7 + 11014635520*x^8 +...+ A001142(n)*x^n +...

Cf. A001142, A219266, A219268.

nmax=15; Rest[CoefficientList[Series[Log[Sum[Product[j^j/j!,{j,1,k}]*x^k,{k,0,nmax}]],{x,0,nmax}],x] * Range[0,nmax]] (* Vaclav Kotesovec, Jul 10 2015 *)

{a(n)=n*polcoeff(log(sum(k=0,n+1,prod(j=0,k,j^j/j!)*x^k)+x*O(x^n)),n)}
for(n=1,21,print1(a(n),", "))

a(n) ~ A^2 * exp(n^2/2 + n - 1/12) / (n^(n/2 - 2/3) * (2*Pi)^((n+1)/2)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Jul 10 2015

A056609 a(n) = rad(n!)/rad(A001142(n)) where rad(n) is the squarefree kernel of n, A007947(n).

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 2, 3, 5, 1, 1, 1, 7, 5, 2, 1, 3, 1, 5, 7, 11, 1, 1, 5, 13, 3, 7, 1, 1, 1, 2, 11, 17, 7, 1, 1, 19, 13, 1, 1, 7, 1, 11, 1, 23, 1, 1, 7, 5, 17, 13, 1, 3, 11, 1, 19, 29, 1, 1, 1, 31, 1, 2, 13, 11, 1, 17, 23, 1, 1, 1, 1, 37, 5, 19, 11, 13, 1, 1, 3, 41, 1, 1, 17, 43, 29, 11, 1, 1, 13

Offset: 1

Labos Elemer, Aug 07 2000

nonn

The previous name, which does not match the data as observed by Luc Rousseau, was: Quotient of squarefree kernels of A002944(n) and A001405.

a(n) is the unique prime p not greater than n missing in the prime factorization of A001142(n), if such a prime exists; a(n) is 1 otherwise. - Luc Rousseau, Jan 01 2019

			From _Luc Rousseau_, Jan 02 2019: (Start)
In Pascal's triangle,
- row n=3 (1 3 3 1) contains no number with prime factor 2, so a(3) = 2;
- row n=4 (1 4 6 4 1) contains, for all p prime <= 4, a multiple of p, so a(4) = 1;
- row n=5 (1 5 10 10 5 1) contains no number with prime factor 3, so a(5) = 3;
etc.
(End)

Luc Rousseau, Table of n, a(n) for n = 1..1000 (first 90 terms from Labos Elemer)

Cf. A002944, A001405, A003418, A002110, A001142, A034386, A056606.

L[n_] := Table[Binomial[n, k], {k, 1, Floor[n/2]}]
c[n_] := Complement[Prime /@ Range[PrimePi[n]], First /@ FactorInteger[Times @@ L[n]]]
a[n_] := Module[{x = c[n]}, If[x == {}, 1, First[x]]]
Table[a[n], {n, 1, 100}]
(* Luc Rousseau, Jan 01 2019 *)

rad(n) = factorback(factorint(n)[, 1]); \\ A007947
b(n) = prod(m=1, n, binomial(n, m)); \\ A001142
a(n) = rad(n!)/rad(b(n)); \\ Michel Marcus, Jan 02 2019

a(n) = A034386(n) / A056606(n). - Sean A. Irvine, Apr 24 2022

Definition and example changed by Luc Rousseau, Jan 02 2019

A092593 a(n) is the smallest number k > 1 for which A001142(k)/A002944(k+1)^n is an integer.

Original entry on oeis.org

2, 3, 9, 9, 15, 15, 38, 45, 45, 45, 61, 61, 225, 225, 225, 225, 225, 225, 225, 225, 225, 225, 635, 635, 1545, 1545, 1545, 1545, 2137, 2137, 2137, 2137, 2137, 2137, 2137, 2137, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660, 2660

Offset: 1

Labos Elemer, Mar 10 2004

nonn

a(62) > 12500. - Robert Israel, Jan 24 2019

			n=4, A001142(9) = 1*9*36*...*9*1 = 11759522374656,
A002944(10) = lcm(1,2,...,10)/10=252 and A001142(9) = 2916*(252^4) = 11759522374656,
so a(4)=9, the smallest relevant number.

Cf. A001142, A002944, A092593.

A001142:= proc(n) option remember; procname(n-1)*n^(n-1)/(n-1)! end proc:
A001142(0):= 1:
A002944:= proc(n) option remember; ilcm(n,procname(n-1)*(n-1))/n end proc:
A002944(1):= 1:
f:= proc(n) option remember; local k;
for k from procname(n-1) do
   if type(A001142(k)/A002944(k+1)^n, integer) then return k fi
od
end proc:
f(1):= 2:
map(f, [$1..61]); # Robert Israel, Jan 23 2019

Table[fla=1;Do[s1=Apply[Times, Table[Binomial[n, j], {j, 0, n}]]; s2=Apply[LCM, Table[Binomial[n, j], {j, 0, n}]]; If[IntegerQ[s1/(s2^k)]&&!Equal[n, 1]&&Equal[fla, 1], Print[{n, k}];fla=0], {n, 1, 230}], {k, 1, 25}]

Corrected and extended by Robert Israel, Jan 23 2019

cp's OEIS Frontend

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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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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A249150 Number of trailing zeros in the factorial base representation of products of binomial coefficients: a(n) = A230403(A001142(n)).

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A056077 Indices n of terms of sequence A001142, Product_{k=0..n} binomial(n,k), that are divisible by all primes <= n.

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A109873 a(n) = product of terms in row n of Pascal's triangle (A001142) divided by n^k, where n^k is the largest power of n dividing it.

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A109874 Largest exponent e such that n^e that divides A001142(n).

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