A060818 -id:A060818 - cp's OEIS frontend

A243758 a(n) = Product_{i=1..n} A234959(i).

1, 1, 1, 1, 1, 1, 6, 6, 6, 6, 6, 6, 36, 36, 36, 36, 36, 36, 216, 216, 216, 216, 216, 216, 1296, 1296, 1296, 1296, 1296, 1296, 7776, 7776, 7776, 7776, 7776, 7776, 279936, 279936, 279936, 279936, 279936, 279936, 1679616, 1679616, 1679616, 1679616, 1679616, 1679616, 10077696

Offset: 0

Tom Edgar, Jun 10 2014

This is the generalized factorial for A234959.

a(0) = 1 as it represents the empty product.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
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.

Cf. A060828, A060818, A234959, A242954, A243757.

a243758 n = a243758_list !! n
a243758_list = scanl (*) 1 a234959_list
-- Reinhard Zumkeller, Feb 09 2015

Table[Product[6^IntegerExponent[k, 6], {k, 1, n}], {n, 0, 20}] (* G. C. Greubel, Dec 24 2016 *)

valp(n,p)=my(s); while(n\=p, s+=n); s
a(n)=6^valp(n,6) \\ Charles R Greathouse IV, Oct 03 2016

S=[0]+[6^valuation(i,6) for i in [1..100]]
[prod(S[1:i+1]) for i in [0..99]]

a(n) = Product_{i=1..n} A234959(i).

a(n) = 6^(A054895(n)).

A375931 The product of the prime powers in the prime factorization of n that have an exponent that is equal to the maximum exponent in this factorization.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 4, 13, 14, 15, 16, 17, 9, 19, 4, 21, 22, 23, 8, 25, 26, 27, 4, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 8, 41, 42, 43, 4, 9, 46, 47, 16, 49, 25, 51, 4, 53, 27, 55, 8, 57, 58, 59, 4, 61, 62, 9, 64, 65, 66, 67, 4, 69, 70, 71

Offset: 1

Amiram Eldar, Sep 03 2024

Differs from A327526 at n = 12, 20, 28, 40, 44, 45, ... .
Each positive number appears in this sequence either once or infinitely many times:
1. If m is squarefree then the only solution to a(x) = m is x = m.
2. If m = s^k is a power of a squarefree number s with k >= 2, then x = m * i is a solution to a(x) = m for all numbers i that are k-free numbers (i.e., having exponents in their prime factorizations that are all less than k) that are coprime to m.

			180 = 2^2 * 3^2 * 5, and the maximum exponent in the prime factorization of 180 is 2, which is the exponent of its prime factors 2 and 3. Therefore a(180) = 2^2 * 3^2 = (2*3)^2 = 36.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A001221 (omega), A001222 (Omega), A005117, A051903, A060818, A072774, A104141, A261969, A327526, A356862, A362611, A375932.

a[n_] := Module[{f = FactorInteger[n], p, e, i, m}, p = f[[;; , 1]]; e = f[[;; , 2]]; m = Max[e]; i = Position[e, m] // Flatten; (Times @@ p[[i]])^m]; Array[a, 100]

a(n) = {my(f = factor(n), p = f[,1], e = f[,2], m); if(n == 1, 1, m = vecmax(e); prod(i = 1, #p, if(e[i] == m, p[i], 1))^m);}

If n = Product_{i} p_i^e_i (where p_i are distinct primes) then a(n) = Product_{i} p_i^(e_i * [e_i = max_{j} e_j]), where [] is the Iverson bracket.
a(n) = A261969(n)^A051903(n).
a(n) = n / A375932(n).
a(n) = n if and only if n is a power of a squarefree number (A072774).
A051903(a(n)) = A051903(n).
omega(a(n)) = A362611(n).
omega(a(n)) = 1 if and only if n is in A356862.
Omega(a(n)) = A051903(n) * A362611(n).
a(n!) = A060818(n) for n != 3.
Sum_{k=1..n} a(k) ~ c * n^2, where c = 3/Pi^2 = 0.303963... (A104141).

A118381 Largest 3-smooth number dividing n!.

Original entry on oeis.org

1, 2, 6, 24, 24, 144, 144, 1152, 10368, 20736, 20736, 248832, 248832, 497664, 1492992, 23887872, 23887872, 429981696, 429981696, 1719926784, 5159780352, 10319560704, 10319560704, 247669456896, 247669456896, 495338913792, 13374150672384, 53496602689536, 53496602689536

Offset: 1

Reinhard Zumkeller, May 16 2006

nonn

Amiram Eldar, Table of n, a(n) for n = 1..1861 (terms below 10^1000)
Index entries for sequences related to factorial numbers.

Cf. A003586, A000142, A011371, A054861, A060818, A060828.

s[n_, b_] := Sum[IntegerExponent[k, b], {k, 1, n}]; a[n_] := 2^s[n, 2] * 3^s[n, 3]; Array[a, 30] (* Amiram Eldar, Jan 29 2020 *)

a(n) = A060818(n) * A060828(n).

More terms from Amiram Eldar, Jan 29 2020

A124399 a(n) = 4^(n - bitcount(n)) where bitcount(n) = A000120(n).

Original entry on oeis.org

1, 1, 4, 4, 64, 64, 256, 256, 16384, 16384, 65536, 65536, 1048576, 1048576, 4194304, 4194304, 1073741824, 1073741824, 4294967296, 4294967296, 68719476736, 68719476736, 274877906944, 274877906944, 17592186044416, 17592186044416

Offset: 0

Wolfdieter Lang, Nov 10 2006

Numerators of one half of norm square of monic Legendre polynomials p_n(x).
The denominators of these polynomials are given by A069955.
The rationals N2(n) = 2*a(n)/A069955(n) give the minimal norm square for real monic polynomials. The norm square is defined as integral over the interval [-1,+1] of the square of the polynomials. Cf. the Courant-Hilbert reference.

			Rationals a(n)/A069955(n): [1, 1/3, 4/45, 4/175, 64/11025, 64/43659, 256/693693, ...].
Rationals N2(n): [2, 2/3, 8/45, 8/175, 128/11025, 128/43659, 512/693693,...].

Richard Courant and David Hilbert, Methoden der mathematischen Physik, Bd. I, 3, Auflage, Springer, 1993, pp. 73-74.

Wolfdieter Lang, Norm square, rationals and more.

Cf. A000120, A001790, A056982, A060818, A069955 (denominators of N2(n) as defined in the comments).

bitcount(n) = sum(digits(n, base=2))
a(n) = 4^(n - bitcount(n)) # Peter Luschny, Oct 01 2019

a[n_] := 4^(n - DigitCount[n, 2, 1]); Array[a, 25, 0] (* Amiram Eldar, Jul 25 2023 *)

a(n) = numerator((1/(2*n+1))*((2^n)/binomial(2*n,n))^2); \\ Michel Marcus, Aug 11 2019

a(n) = numerator(N2(n)/2) with N2(n)/2:=(1/(2*n+1))*((2^n)/binomial(2*n,n))^2.
N2(n)/2 = (1/(2*n+1))*(1/L(n))^2 with L(n)= A001790(n)/A060818(n), the leading coefficient of the Legendre polynomial P_n(x), in lowest terms.
Bisection: a(2*n)=a(2*n+1) = A056982(n), n>=0.

New name by Peter Luschny, Oct 01 2019

A336940 Number of odd divisors of n!.

Original entry on oeis.org

1, 1, 1, 2, 2, 4, 6, 12, 12, 20, 30, 60, 72, 144, 216, 336, 336, 672, 864, 1728, 2160, 3200, 4800, 9600, 10560, 14784, 22176, 28224, 35280, 70560, 86400, 172800, 172800, 245760, 368640, 497664, 559872, 1119744, 1679616, 2363904, 2626560, 5253120, 6451200, 12902400, 16128000

Offset: 0

Gus Wiseman, Aug 23 2020

nonn

			The a(1) = 1 through a(8) = 12 divisors:
  1  1  1  1  1   1   1    1
        3  3  3   3   3    3
              5   5   5    5
              15  9   7    7
                  15  9    9
                  45  15   15
                      21   21
                      35   35
                      45   45
                      63   63
                      105  105
                      315  315

Seiichi Manyama, Table of n, a(n) for n = 0..10000

A049606 gives the maximum among these divisors, with quotient A060818.
A337257 is the even version.
A000265 gives the maximum odd divisor of n.
A001227 counts odd divisors.
A183063 counts even divisors.
Cf. A000005, A001013, A001055, A006939, A113474, A124010, A253249.
Factorial numbers: A000142, A022559, A027423 (divisors), A048656, A071626, A076716 (factorizations), A325272, A325273, A325617, A336414, A336498.

Table[Length[Select[Divisors[n!],OddQ]],{n,0,15}]

a(n) = sumdiv(n!, d, d%2); \\ Michel Marcus, Aug 24 2020

a(n) = numdiv(prod(k=1, n, k >> valuation(k, 2))); \\ Michel Marcus, Aug 27 2020

a(n) = A001227(n!).
a(n) = A000005(A049606(n)).
a(n) + A337257(n) = A027423(n) = A000005(n!).
From Seiichi Manyama, Aug 27 2020: (Start)
If p is odd prime, a(p) = 2 * a(p-1).
a(n) = A027423(n) / A113474(n) for n > 0. (End)

a(36)-a(44) from Seiichi Manyama, Aug 26 2020

A082093 a(n) is the least number m such that (m+n)!/m! = (m+1)(m+2)...*(m+n) divides lcm(1,...,m).

Original entry on oeis.org

5, 13, 19, 32, 73, 89, 140, 199, 294, 468, 1072, 1072, 1072, 2161, 2976, 32805, 32806, 65732, 65732, 262153, 262154, 524457, 524640, 4194464, 4194464, 8388640, 8388640, 33554432, 33554432, 67108992, 67109088, 2147483659, 2147484110, 4294967312, 4294967312, 17179869209, 17179869210

Offset: 1

Labos Elemer, Apr 10 2003

nonn

From David A. Corneth, Aug 30 2019: (Start)
As (m+1)*(m+2)*...*(m+n) is the product of n consecutive integers, it's divisible by n! and so a(n) >= 2^A011371(n) = A060818(n).
None of m+1..m+n are prime. (End)

			a(6)=89: lcm(1,...,89) = 718766754945489455304472257065075294400 is divisible by 625757605200 = 90*91*92*93*94*95 = (89+6)!/89! and the quotient is 1148634469597477063638686172.
For n=1 see A080765(1) = A082093(1).

David A. Corneth, Table of n, a(n) for n = 1..75

Cf. A011371, A060818, A080765.

k = 1; lc = 1; Do[While[lc = LCM[lc, k]; Mod[lc, (n + k)!/k! ] != 0, k++ ]; Print[{n, k}], {n, 0, 50}] (* Robert G. Wilson v, Apr 12 2006 *)

a(16)-a(19) from Robert G. Wilson v, Apr 12 2006
a(20)-a(23) from Vaclav Kotesovec, Aug 30 2019
a(24)-a(37) from David A. Corneth, Aug 30 2019

A135354 a(0)=1, a(n) = largest divisor of n! that is coprime to a(n-1).

Original entry on oeis.org

1, 1, 2, 3, 8, 15, 16, 315, 128, 2835, 256, 155925, 1024, 6081075, 2048, 638512875, 32768, 10854718875, 65536, 1856156927625, 262144, 194896477400625, 524288, 49308808782358125, 4194304, 3698160658676859375, 8388608, 1298054391195577640625, 33554432, 263505041412702261046875, 67108864

Offset: 0

Leroy Quet, Dec 07 2007

Robert Israel, Table of n, a(n) for n = 0..504

Cf. A060818, A049606.

f:= proc(n,a)
  local P,R,i;
  P:= select(t -> isprime(t) and igcd(t,a)=1, [2,seq(i,i=3..n,2)]);
  R:= map(proc(p) local k; add(floor(n/p^k), k=1 ..ilog[p](n)) end proc, P);
  mul(P[i]^R[i],i=1..nops(P));
end proc:
R:= 1: r:= 1: for i from 1 to 50 do r:= f(i,r); R:= R,r od:
R; # Robert Israel, Jul 21 2024

a = {1}; For[n = 1, n < 25, n++, AppendTo[a, Select[Divisors[n! ], GCD[a[[ -1]], # ] == 1 &][[ -1]]]]; a (* Stefan Steinerberger, Dec 10 2007 *)
ldnf[{n_,a_}]:={n+1,Max[Select[Divisors[(n+1)!],CoprimeQ[#,a]&]]}; Transpose[ NestList[ldnf,{0,1},30]][[2]] (* Harvey P. Dale, Jan 21 2016 *)

a(2n) = the largest power of 2 that divides (2n)!. a(2n+1) = the largest odd divisor of (2n+1)! = (2n+1)!/a(2n).

More terms from Stefan Steinerberger, Dec 10 2007

More terms from Robert Israel, Jul 21 2024

A227772 Sequence based on factorial representation converging to 1 in 2-adic numbers, and 0 in p-adic numbers for any other p.

Original entry on oeis.org

0, 1, 3, 9, 105, 225, 945, 36225, 76545, 2253825, 9511425, 89345025, 1526349825, 26434433025, 287969306625, 12057038618625, 179439357722625, 5870438207258625, 37882306735898625, 1984203913277210625, 11715811945983770625, 982443713208463130625, 15594453174317362970625

Offset: 1

Franklin T. Adams-Watters, Jul 30 2013

nonn

This is an example to show how a sequence can be constructed to converge to an arbitrary p-adic number chosen independently for each p.

			5! = 2^3 * 3 * 5. Solving for m == 1 (mod 2^3), 0 (mod 3) and 0 (mod 5), we get m == 105 (mod 120), so we take a(5) = 105.
The factorial base representation is ...114111.

Jinyuan Wang, Table of n, a(n) for n = 1..450

Cf. A000142, A049606, A060818, A007623.

a(n)=lift(chinese(Mod(1,denominator(polcoeff(pollegendre(n), n))),Mod(0,denominator(2^n/n!)))) /* Ralf Stephan, Aug 01 2013 */

Solve for a(n) == 1 (mod A060818(n)) and a(n) == 0 (mod A049606), taking the least nonnegative residue.

More terms from Ralf Stephan, Aug 01 2013

More terms from Jinyuan Wang, Jan 16 2021

A360603 Triangle read by rows. T(n, k) = A360604(n, k) * A001187(k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 4, 0, 8, 6, 12, 38, 0, 64, 32, 48, 152, 728, 0, 1024, 320, 320, 760, 3640, 26704, 0, 32768, 6144, 3840, 6080, 21840, 160224, 1866256, 0, 2097152, 229376, 86016, 85120, 203840, 1121568, 13063792, 251548592

Offset: 0

Peter Luschny, Feb 20 2023

			Triangle T(n, k) starts:
[0] 1;
[1] 0,       1;
[2] 0,       1,      1;
[3] 0,       2,      2,     4;
[4] 0,       8,      6,    12,    38;
[5] 0,      64,     32,    48,   152,    728;
[6] 0,    1024,    320,   320,   760,   3640,   26704;
[7] 0,   32768,   6144,  3840,  6080,  21840,  160224,  1866256;
[8] 0, 2097152, 229376, 86016, 85120, 203840, 1121568, 13063792, 251548592.

F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973.

Algorithms for Competitive Programming, Counting labeled graphs, cp-algorithms.
Albert Nijenhuis and Herbert S Wilf, The enumeration of connected graphs and linked diagrams, Journal of Combinatorial Theory, Vol. 27(3), 1979, Pages 356-359.
Eric Weisstein's World of Mathematics, Connected Graph.
Eric Weisstein's World of Mathematics, Labeled Graph.

Cf. A006125 Graphs on n labeled nodes, T(n+1, 1) and Sum_{k=0..n} T(n, k).
Cf. A054592 Disconnected labeled graphs with n nodes, Sum_{k=0..n-1} T(n, k).
Cf. A001187 Connected labeled graphs with n nodes, T(n, n).
Cf. A123903 Isolated nodes in all simple labeled graphs on n nodes, T(n+2, 2).
Cf. A053549 Labeled rooted connected graphs, T(n+1, n).
Cf. A275462 Leaves in all simple labeled connected graphs on n nodes T(n+2, n).
Cf. A060818 gcd_{k=0..n} T(n, k) = gcd(n!, 2^n).
Cf. A143543 Labeled graphs on n nodes with k connected components.
Cf. A095340 Total number of nodes in all labeled graphs on n nodes.
Cf. A360604, A360860 (accumulation triangle).

T := (n, k) -> 2^binomial(n-k, 2)*binomial(n-1, k-1)*A001187(k):
for n from 0 to 9 do seq(T(n, k), k = 0..n) od;
# Based on the recursion:
Trow := proc(n) option remember; if n = 0 then return [1] fi;
seq(2^binomial(n-k, 2) * binomial(n-1, k-1) * Trow(k)[k+1], k = 1..n-1);
2^(n*(n-1)/2) - add(j, j = [%]); [0, %%, %] end:
seq(print(Trow(n)), n = 0..8);

A001187[n_] := A001187[n] = 2^((n - 1)*n/2) - Sum[Binomial[n - 1, k]*2^((k - n + 1)*(k - n + 2)/2)*A001187[k + 1], {k, 0, n - 2}];
T[n_, k_] := 2^Binomial[n - k, 2]*Binomial[n - 1, k - 1]*A001187[k];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 02 2023, after Peter Luschny in A001187 *)

from math import comb as binomial
from functools import cache
@cache
def A360603Row(n: int) -> list[int]:
    if n == 0: return [1]
    s = [2 ** (((k - n + 1) * (k - n)) // 2) * binomial(n - 1, k - 1) * A360603Row(k)[k] for k in range(1, n)]
    b = 2 ** (((n - 1) * n) // 2) - sum(s)
    return [0] + s + [b]

T(n, k) = 2^binomial(n-k, 2)*binomial(n-1, k-1) * A001187(k).
Recursion over the rows of the triangle: Set row(0) = [1] where [.] denotes a 0-based list. Assume now all rows(j) for j < n computed, next compute r = [2^binomial(n-k, 2) * binomial(n-1, k-1) * row(k)[k] for k = 1..n-1] and s = 2^(n*(n-1)/2) - Sum(r). Then row(n) = [0] & r & [s], where '&' denotes the concatenation of lists. (See the Python program for an implementation.)
T(n, n) = A001187(n) (connected labeled graphs).
T(n-1, n) = A053549(n-1) for n >= 1 (labeled rooted connected graphs).
T(n, 1) = Sum_{k>=0} T(n-1, k) = A006125(n-1) for n >= 1 (all labeled graphs).
Sum_{k=0..n-1} T(n, k) = A054592(n) for n >= 1 (disconnected labeled graphs).
See additional formulas in the cross-references.

A371357 a(n) = 2^(n - HammingWeight(n)) * binomial(2*n, n).

Original entry on oeis.org

1, 2, 12, 40, 560, 2016, 14784, 54912, 1647360, 6223360, 47297536, 180590592, 2769055744, 10650214400, 82158796800, 317680680960, 19696202219520, 76467608616960, 594748067020800, 2316387208396800, 36135640450990080, 141101072237199360, 1103153837490831360

Offset: 0

Peter Luschny, Mar 23 2024

nonn

Paolo Xausa, Table of n, a(n) for n = 0..1000

Cf. A060818, A000984.

HammingWeight := n -> add(convert(n, base, 2)):
a := n -> 2^(n - HammingWeight(n))*binomial(2*n, n):
seq(a(n), n = 0..22);

Array[2^(# - DigitCount[#, 2, 1]) * Binomial[2*#, #] &, 25, 0] (* Paolo Xausa, Mar 31 2024 *)

a(n) = A060818(n) * A000984(n).

cp's OEIS Frontend

A243758 a(n) = Product_{i=1..n} A234959(i).

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A375931 The product of the prime powers in the prime factorization of n that have an exponent that is equal to the maximum exponent in this factorization.

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A118381 Largest 3-smooth number dividing n!.

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A124399 a(n) = 4^(n - bitcount(n)) where bitcount(n) = A000120(n).

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A336940 Number of odd divisors of n!.

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A082093 a(n) is the least number m such that (m+n)!/m! = (m+1)*(m+2)*...*(m+n) divides lcm(1,...,m).

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A135354 a(0)=1, a(n) = largest divisor of n! that is coprime to a(n-1).

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A082093 a(n) is the least number m such that (m+n)!/m! = (m+1)(m+2)...*(m+n) divides lcm(1,...,m).