cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A202367 LCM of denominators of the coefficients of polynomials 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).

Original entry on oeis.org

1, 6, 360, 45360, 5443200, 359251200, 5884534656000, 35307207936000, 144053408378880000, 1034591578977116160000, 3414152210624483328000000, 471153005066178699264000000, 15434972445968014187888640000000, 92609834675808085127331840000000, 161141112335906068121557401600000000
Offset: 1

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Author

Keywords

Comments

See comment in A175669.

Crossrefs

Formula

Conjecture: a(n) = Product_{primes p} p^(Sum_{k>=0} floor((n-1)/(ceiling((p-1)/2)*p^k))).
If the conjecture is true, then, for n >= 0, A007814(a(n+1)) = A007814(n!) + n.

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.

Original entry on oeis.org

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

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Keywords

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..|
		

Crossrefs

Programs

  • 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 *)

Formula

A007814(C^(1)(n,m)) = A007814(C(n,m)).

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.

Original entry on oeis.org

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

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Comments

Conjecture. If p is an odd prime, then the ((p-1)/2)-th row contains two 1's and (p-3)/2 numbers multiple of p.
See also comments in A175669 and A202917.

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..|
		

Crossrefs

Formula

If conjectural formula in A202367 is true, then A007814(C^(2)(n,m)) =A007814(C(n,m)).

A202368 LCM of denominators of the coefficients of polynomials 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).

Original entry on oeis.org

1, 4, 672, 13440, 58705920, 234823680, 11243357798400, 494707743129600, 4321766843980185600, 86435336879603712000, 450155234468976132096000
Offset: 1

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Author

Keywords

Comments

See comment in A175669.

Crossrefs

A202369 LCM of denominators of the coefficients of polynomials 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).

Original entry on oeis.org

1, 30, 1800, 14742000, 30073680000, 49621572000000, 812801349360000000, 707137173943200000000, 2885119669688256000000000, 49833835369821036293760000000000, 6742517925536786210545728000000000000
Offset: 1

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Author

Keywords

Comments

See comment in A175669.

Crossrefs

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.

Original entry on oeis.org

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

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Author

Keywords

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..|
		

Crossrefs

Formula

Conjecture. A007814(C^(3)(n,m)) = A007814(C(n,m)).

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.

Original entry on oeis.org

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

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Keywords

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..|
		

Crossrefs

Formula

Conjecture. A007814(C^(4)(n,m)) = A007814(C(n,m)).

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).

Original entry on oeis.org

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

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Keywords

Comments

For m>=0, the denominator for all 4*m+1 terms of the m-th row is A202368(m+1).
See comment to A175669.

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.
		

Crossrefs

Formula

Q^(3)_n(1)=1.

A376117 Irregular triangle of numerator polynomial coefficients of C({1..n},x), T(n,k) for n >= 0 and k >= A000217(n).

Original entry on oeis.org

1, 1, -2, -1, -6, 0, 10, 16, 4, -11, -17, -12, -5, -1, -24, 84, -60, 30, -144, -48, 104, 186, 268, -12, -240, -436, -348, -46, 262, 444, 391, 199, -23, -166, -207, -172, -109, -55, -21, -6, -1, 120, -1200, 4560, -7740, 5064, -2472, 9768, -19152, 35004, -39408
Offset: 0

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Author

John Tyler Rascoe, Sep 10 2024

Keywords

Examples

			For row n = 2, C({1,2},x) = (-2*x^3 - x^4)/(1 + x + 2*x^2 - x^3 - x^4).
Triangle begins
  k=0  1  2  3   4   5   6   7   8   9  10   11   12   13  14  15
n=0 1;
n=1 .  1;
n=2 .  .  . -2, -1;
n=3 .  .  .  .   .   .  -6,  0, 10, 16,  4, -11, -17, -12, -5, -1;
		

Crossrefs

Programs

  • PARI
    C_x(s)={my( g=if(#s <1, 1, sum(i=1, #s, C_x(s[^i]) * x^(s[i]) )/(1-sum(i=1, #s, x^(s[i]))))); return(g)}
    A376117_row(n)={my(t=n*(n+1)/2, c=C_x([1..n]), d=poldegree(numerator(c))-t, z=vector(d+1)); for(k=0,d,z[k+1]=polcoeff(numerator(c),k+t)); z}

Formula

C({s},x) = Sum_{i in {s}} (C({s}-{i},x)*x^i)/(1 - Sum_{i in {s}} (x^i)) with C({},x) = 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).

Original entry on oeis.org

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

Views

Author

Keywords

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.
		

Crossrefs

Formula

Q^(4)_n(1)=1.
Showing 1-10 of 10 results.