A056606 -id:A056606 - 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

A002944 a(n) = LCM(1,2,...,n) / n.

Original entry on oeis.org

1, 1, 2, 3, 12, 10, 60, 105, 280, 252, 2520, 2310, 27720, 25740, 24024, 45045, 720720, 680680, 12252240, 11639628, 11085360, 10581480, 232792560, 223092870, 1070845776, 1029659400, 2974571600, 2868336900, 80313433200, 77636318760, 2329089562800

Offset: 1

N. J. A. Sloane

Equals LCM of all numbers of (n-1)-st row of Pascal's triangle [Montgomery-Breusch]. - J. Lowell, Apr 16 2014. Corrected by N. J. A. Sloane, Sep 04 2019

Williams proves that a(n+1) = A034386(n) for n=2, 11 and 23 only. This is trivially the case for n=0 and 1, too. - Michel Marcus, Apr 16 2020

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 = 1..500
Bakir Farhi, An identity involving the least common multiple of binomial coefficients and its application, arXiv:0906.2295 [math.NT], 2009.
Bakir Farhi, An identity involving the least common multiple of binomial coefficients and its application, Amer. Math. Monthly, 116 (2009), 836-839.
Peter L. Montgomery (proposer) and Robert Breusch (solver), LCM of Binomial Coefficients, Problem E2686, American Mathematical Monthly, Vol. 84 (1977), p. 820.
Peter L. Montgomery (proposer) and Robert Breusch (solver), LCM of Binomial Coefficients, Solution to Problem E2686, American Mathematical Monthly, Vol. 86 (1979), p. 131.
Ian S. Williams, On a problem of Kurt Mahler concerning binomial coefficents (sic), Bulletin of the Australian Mathematical Society, Volume 14, Issue 2, April 1976, pp. 299-302.
Index entries for sequences related to lcm's

Cf. A025527 and A025537.
Cf. A100561, A003418, A001142, A001405. - Franklin T. Adams-Watters, Jul 05 2009
Cf. A056606 (squarefree kernel).

a002944 n = a003418 n `div` n  -- Reinhard Zumkeller, Mar 16 2015

A003418 := n-> lcm(seq(i,i=1..n)); f:=n->A003418(n)/n;
BB:=n->sum(1/sqrt(k), k=1..n): a:=n->floor(denom(BB(n))/n): seq(a(n), n=1..29); # Zerinvary Lajos, Mar 29 2007

Table[Apply[LCM,Range[n]]/n,{n,1,30}]  (* Geoffrey Critzer, Feb 10 2013 *)

a(n) = lcm(vector(n, i, i))/n; \\ Michel Marcus, Apr 16 2014

a(n) = A003418(n) / n.
a(n) = LCM of C(n-1, 0), C(n-1, 1), ..., C(n-1, n-1). [Montgomery-Breusch] [Corrected by N. J. A. Sloane, Jun 11 2008]
Equally, a(n+1) = LCM_{k=0..n} binomial(n,k). - Franklin T. Adams-Watters, Jul 05 2009

More terms from Jud McCranie, Jan 17 2000

Edited by N. J. A. Sloane, Jun 11 2008 and Sep 04 2019

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

A256113 Table read by rows: T(1,1) = 1, for n > 1: row n = union of distinct prime factors occurring in terms of n-th row of Pascal's triangle, cf. A007318.

Original entry on oeis.org

1, 2, 3, 2, 3, 2, 5, 2, 3, 5, 3, 5, 7, 2, 5, 7, 2, 3, 7, 2, 3, 5, 7, 2, 3, 5, 7, 11, 2, 3, 5, 7, 11, 2, 3, 5, 11, 13, 2, 3, 7, 11, 13, 3, 5, 7, 11, 13, 2, 3, 5, 7, 11, 13, 2, 5, 7, 11, 13, 17, 2, 3, 5, 7, 11, 13, 17, 2, 3, 7, 11, 13, 17, 19, 2, 3, 5, 11, 13

Offset: 1

Reinhard Zumkeller, Mar 16 2015

			.  n |   T(n,k)   |                A001142(n) | A007318(n,0..n)
. ---+------------+---------------------------+-------------------------
.  1 | 1          |                         1 | 1  1
.  2 | 2          |                         2 | 1  2  1
.  3 | 3          |                         9 | 1  3  3   1
.  4 | 2 3        |                        96 | 1  4  6   4   1
.  5 | 2 5        |                      2500 | 1  5 10  10   5   1
.  6 | 2 3 5      |                    162000 | 1  6 15  20  15   6   1
.  7 | 3 5 7      |                  26471025 | 1  7 21  35  35  21   7   1
.  8 | 2 5 7      |               11014635520 | 1  8 28  56  70  56  28 ...
.  9 | 2 3 7      |            11759522374656 | 1  9 36  84 126 126  84 ...
. 10 | 2 3 5 7    |         32406091200000000 | 1 10 45 120 210 252 210 ...
. 11 | 2 3 5 7 11 |     231627686043080250000 | 1 11 55 165 330 462 462 ...
. 12 | 2 3 5 7 11 | 4311500661703860387840000 | 1 12 66 220 495 792 924 ...

Reinhard Zumkeller, b-file >>Rows n = 1..1000 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A027748, A001142, A004788 (row lengths), A056606 (row products).

a256113 n k = a256113_tabf !! (n-1) !! (n-1)
a256113_row n = a256113_tabf !! (n-1)
a256113_tabf = map a027748_row $ tail a001142_list

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

A329947 Integers k such that the radical of the cumulative product of k^k/k! equals its predecessor.

Original entry on oeis.org

1, 12, 30, 36, 40, 60, 70, 72, 96, 108, 112, 126, 150, 175, 180, 192, 198, 210, 224, 240, 270, 280, 306, 312, 324, 330, 336, 350, 351, 352, 378, 384, 396, 399, 400, 408, 418, 420, 432, 440, 441, 442, 448, 455, 456, 462, 475, 490, 494, 520, 539, 540, 546, 560

Offset: 1

Peter Luschny, Dec 21 2019

nonn

No prime numbers appear in this sequence.

			Consider the rows 11 and 12 of Pascal's triangle.
P11 = [1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1].
P12 = [1, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1].
lcm(P11) = 2310 and radical(2310) = 2310.
lcm(P12) = 27720 and radical(27720) = 2310.
Since radical(lcm(P11)) = radical(lcm(P12)) 12 is in this sequence.
Also: 1 is in this sequence because radical(lcm(P0)) = radical(lcm([1])) = radical(1) = 1 = radical(lcm([1, 1])) = radical(lcm(P1)).

Cf. A056606, A056077.

h := n -> mul(k^k/factorial(k), k=0..n):
rad := n -> mul(k, k = numtheory[factorset](n)):
g := proc(n) option remember; rad(h(n)) end:
isA329947 := n -> g(n) = g(n-1): select(isA329947, [$1..560]);

h[n_] := Product[k^k/k!, {k, 1, n}];
rad[n_] := Times @@ FactorInteger[n][[All, 1]];
g[n_] := g[n] = rad[h[n]];
isA329947[n_] := g[n] == g[n-1];
Select[Range[560], isA329947] (* Jean-François Alcover, Feb 28 2024, after Maple code *)

cp's OEIS Frontend

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

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A002944 a(n) = LCM(1,2,...,n) / 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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A256113 Table read by rows: T(1,1) = 1, for n > 1: row n = union of distinct prime factors occurring in terms of n-th row of Pascal's triangle, cf. A007318.

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

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A329947 Integers k such that the radical of the cumulative product of k^k/k! equals its predecessor.

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