A053657 a(n) = Product_{p prime} p^{ Sum_{k>=0} floor[(n-1)/((p-1)p^k)]}.
1, 2, 24, 48, 5760, 11520, 2903040, 5806080, 1393459200, 2786918400, 367873228800, 735746457600, 24103053950976000, 48206107901952000, 578473294823424000, 1156946589646848000, 9440684171518279680000, 18881368343036559360000, 271211974879377138647040000
Offset: 1
A175669 Triangle of numerators of coefficients of the polynomial Q^(2)m(n) defined by the recursion Q^(2)_0(n)=1; for m>=1, Q^(2)_m(n) = Sum{i=1..n} i^2*Q^(2)_(m-1)(i). For m>=0, the denominator for all 3*m+1 terms of the m-th row is A202367(m+1).
1, 2, 3, 1, 0, 20, 96, 155, 90, 5, -6, 0, 280, 2772, 10518, 18711, 14385, 1323, -2863, -126, 360, 0, 2800, 47040, 323336, 1157760, 2238855, 2050020, 207158, -810600, -58505, 322740, 7956, -45360, 0, 12320, 314160, 3409472, 20401128, 72418826, 150057435, 154651321, 12413874, -101524412, -6408765, 82588957, 3394248, -37374084, -546480, 5443200, 0
Offset: 0
Comments
Consider sequence of sequences of polynomials {Q^(0)_m(x)}, {Q^(1)_m(x)},...,{Q^(r)_m(x)},..., such that in every sequence m=0,1,...
Sequence {Q^(r)m(x)} is defined by the recursion: Q^(r)_0(x)=1; for m>=1 and integer x=n, Q^(r)_m(n)=sum{i=1,...,n}i^rQ^(r)(m-1)(i). By the induction, we see that polynomial Q^(r)_m(x) has degree (r+1)*m. Note that Q^(0)_m(n) is C(n+m-1,m), Q^(1)_m(n)=S(n+m,n), where S(k,l) are Stirling numbers of the second kind. Thus Q^(r)_m(x) is an r-generalization of binomial coefficients and Stirling numbers of the second kind. Moreover, for every r, LCM of denominators of the coefficients of Q^(r)_m(x) generate sequences of factorial type which possess important arithmetic properties. For r=0, it is n!, for r=1, it is A053657, for r=2,3,4 we obtain A202367, A202368, A202369. Denote the general term of the sequence corresponding to a given r by n!^(r) and, for 0<=m<=n, denote C^(r)(n,m)=n!^(r)/(m!^(r)*(n-m)!^(r). Then, for the "r-Pascal triangle", we have the following conjectural regularity: if a prime p==1 mod r, then the ((p-1)/r)-th row contains two 1's and numbers multiple of p. Cf. triangles A202917, A202941.
Examples
The sequence of polynomials begins: Q^(2)_0=1, Q^(2)_1=(2*x^3+3*x^2+x)/6, Q^(2)_2=(20*x^6+96*x^5+155*x^4+90*x^3+5*x^2-6*x)/360, Q^(2)_3=(280*x^9+2772*x^8+10518*x^7+18711*x^6+14385*x^5+1323*x^4-2863*x^3 -126*x^2+360*x)/45360.
Formula
Q^(2)_n(1)=1.
A202917 For n >= 0, let n!^(1) = A053657(n+1) and, for 0 <= m <= n, C^(1)(n,m) = n!^(1)/(m!^(1)*(n-m)!^(1)). The sequence gives a triangle of numbers C^(1)(n,m) with rows of length n+1.
1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 60, 10, 60, 1, 1, 1, 10, 10, 1, 1, 1, 126, 21, 1260, 21, 126, 1, 1, 1, 21, 21, 21, 21, 1, 1
Offset: 0
Comments
1) Note that A053657(n+1) is the LCM of the denominators of the coefficients of the polynomials Q^(1)n(x) which, for integer x=k, are defined by the recursion Q^(1)_0(x)=1, for n>=1, Q^(1)_n(x) = Sum{i=1..k} i*Q^(1)(n-1)(i). Also note that Q^(1)_n(k) = S(k+n,k), where the numbers S(l,m) are Stirling numbers of the second kind. The sequence of polynomials {Q^(1)_n(x)} includes the family of sequences of polynomials {{Q^(r)_n}}(r>=0) described in a comment at A175669. In particular, the LCM of the denominators of the coefficients of Q^(0)_n(x) is n!.
2) This triangle differs from triangle A186430 which is defined according to the theory of factorials over sets by Bhargava. Unfortunately, this theory does not have a conversion theorem. Therefore it is not known if there is a set A such that n!^(1) = n!_A in the Bhargava sense.
3) If p is an odd prime, then the (p-1)-th row contains two 1's and p-2 numbers that are multiples of p. For a conjectural generalization, see comment in A175669.
Examples
Triangle begins n/m.|..0.....1.....2.....3.....4.....5.....6.....7 ================================================== .0..|..1 .1..|..1.....1 .2..|..1.....6.....1 .3..|..1.....1 ... 1 .....1 .4..|..1....60....10......60.....1 .5..|..1.....1....10......10.....1.....1 .6..|..1...126....21....1260....21...126.....1 .7..|..1.....1....21......21....21....21.....1.....1 .8..|
Programs
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Mathematica
A053657[n_] := Product[p^Sum[Floor[(n-1)/((p-1) p^k)], {k, 0, n}], {p, Prime[Range[n]]}]; f1[n_] := A053657[n+1]; C1[n_, m_] := f1[n]/(f1[m] * f1[n-m]); Table[C1[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Nov 22 2016 *)
A202941 For n>=0, let n!^(2)=A202367(n+1) and, for 0<=m<=n, C^(2)(n,m)=n!^(2)/(m!^(2)*(n-m)!^(2)). The sequence gives triangle of numbers C^(2)(n,m) with rows of length n+1.
1, 1, 1, 1, 10, 1, 1, 21, 21, 1, 1, 20, 42, 20, 1, 1, 11, 22, 22, 11, 1, 1, 2730, 3003, 2860, 3003, 2730, 1, 1, 1, 273, 143, 143, 273, 1, 1
Offset: 0
Comments
Examples
Triangle begins n/m.|..0.....1.....2.....3.....4.....5.....6.....7 ================================================== .0..|..1 .1..|..1.....1 .2..|..1....10.....1 .3..|..1....21 ...21.....1 .4..|..1....20....42....20.....1 .5..|..1....11....22....22....11.....1 .6..|..1..2730..3003..2860..3003..2730.....1 .7..|..1.....1...273...143...143...273.....1.....1 .8..|
A203484 For n>=0, let n!^(3) = A202368(n+1) and, for 0<=m<=n, C^(3)(n,m) = n!^(3)/(m!^(3)*(n-m)!^(3)). The sequence gives triangle of numbers C^(3)(n,m) with rows of length n+1.
1, 1, 1, 1, 42, 1, 1, 5, 5, 1, 1, 1092, 130, 1092, 1, 1, 1, 26, 26, 1, 1, 1, 11970, 285, 62244, 285, 11970, 1, 1, 11, 3135, 627, 627, 3135, 11, 1
Offset: 0
Comments
Conjecture. If p is prime of the form 3*k+1, then the k-th row contains two 1's and k-1 numbers multiple of p; if p is prime of the form 3*k+2, then the (2*k+1)-th row contains two 1's and 2*k numbers multiple of p.
Examples
Triangle begins n/m.|..0.....1.....2.....3.....4.....5.....6.....7 ================================================== .0..|..1 .1..|..1......1 .2..|..1.....42.....1 .3..|..1......5 ....5......1 .4..|..1...1092...130...1092.....1 .5..|..1......1....26.....26.....1......1 .6..|..1..11970...285..62244...285..11970....1 .7..|..1.....11..3135....627...627...3135...11.....1 .8..|
A178473 For n>=0, let n!^(4) = A202369(n+1) and, for 0<=m<=n, C^(4)(n,m) = n!^(4)/(m!^(4)*(n-m)!^(4)). The sequence gives triangle of numbers C^(4)(n,m) with rows of length n+1.
1, 1, 1, 1, 2, 1, 1, 273, 273, 1, 1, 68, 9282, 68, 1, 1, 55, 1870, 1870, 55, 1, 1, 546, 15015, 3740, 15015, 546, 1, 1, 29, 7917, 1595, 1595, 7917, 29, 1
Offset: 0
Comments
Conjecture. If p is prime of the form 4*k+1, then the k-th row contains two 1's and k-1 numbers multiple of p; if p is prime of the form 4*k+3, then the (2*k+1)-th row contains two 1's and 2*k numbers multiple of p.
Examples
Triangle begins n/m.|..0.....1.....2.....3.....4.....5.....6.....7 ================================================== .0..|..1 .1..|..1......1 .2..|..1......2......1 .3..|..1....273 ...273......1 .4..|..1.....68...9282.....68......1 .5..|..1.....55...1870...1870.....55......1 .6..|..1....546..15015...3740..15015....546....1 .7..|..1.....29...7917...1595...1595...7917...29.....1 .8..|
A202717 Triangle of numerators of coefficients of the polynomial Q^(3)m(n) defined by the recursion Q^(3)_0(n)=1; for m>=1, Q^(3)_m(n) = Sum{i=1...n} i^3*Q^(3)_(m-1)(i).
1, 1, 2, 1, 0, 0, 21, 132, 294, 252, 21, -56, 0, 8, 0, 35, 450, 2293, 5700, 6405, 770, -3661, -240, 2320, 40, -672, 0, 0, 9555, 207480, 1889316, 9216312, 25051026, 33229560, 3678948, -35339304, -2666157, 51171120, 2178176, -49878192, -792064, 24460800, 4160, -3714816, 0
Offset: 0
Comments
Examples
The sequence of polynomials begins Q^(3)_0=1, Q^(3)_1=(x^4+2*x^3+x^2)/4, Q^(3)_2=(21*x^8+132*x^7+294*x^6+252*x^5+21*x^4-56*x^3+8*x)/672, Q^(3)_3=(35*x^12+450*x^11+2293*x^10+5700*x^9+6405*x^8+770*x^7-3661*x^6-240*x^5+2320*x^4+40x^3-672*x^2)/13440.
Formula
Q^(3)_n(1)=1.
A202749 Triangle of numerators of coefficients of the polynomial Q^(4)m(n) defined by the recursion Q^(4)_0(n)=1; for m>=1,Q^(4)_m(n)=sum{i=1,...,n}i^4*Q^(4)(m-1)(i). For m>=0, the denominator for all 5*m+1 terms of the m-th row is A202369(m+1).
1, 6, 15, 10, 0, -1, 0, 36, 280, 795, 900, 88, -450, -20, 200, 1, -30, 0, 19656, 311220, 1991430, 6354075, 9367722, 1283100, -10854935, -1064700, 16237338, 615615, -16336320, -136500, 8189909, 8190, -1243800, 0
Offset: 0
Comments
See comment in A175669.
Examples
The sequence of polynomials begins Q^(3)_0=1, Q^(3)_1=(6*x^5+15*x^4+10*x^3-x)/30, Q^(3)_2=(36*x^10+280*x^9+795*x^8+900*x^7+88*x^6-450*x^5-20*x^4+200*x^3+x^2-30*x)/1800.
Formula
Q^(4)_n(1)=1.
Comments
Examples
References
Links
Crossrefs
Programs
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Mathematica
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Formula
Extensions