A001790 -id:A001790 - cp's OEIS frontend

A127974 Numerators in expansion of (1-x)^(-2/3).

1, 2, 5, 40, 110, 308, 2618, 7480, 21505, 559130, 1621477, 4717024, 41273960, 120646960, 353323240, 3109244512, 9133405754, 26862958100, 711868389650, 2098138411600, 6189508314220, 54821359354520, 161972198092900

Offset: 0

Paul Barry, Feb 09 2007

Numerators of n!/A008544(n) are A127975.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A001790, A008544, A104133, A127975.

Numerator[CoefficientList[Series[(1 - x)^(-2/3), {x, 0, 50}], x]] (* G. C. Greubel, May 07 2018 *)

a(n) = denominator(n!/A008544(n)).
a(n) = denominator(n!/(Product_{k=0..n-1} (2+3*k))).
Conjecture: a(n-1) = the numerator of (1/(2*sqrt(3)*Pi)) * Integral_{x >= 0} 1/(1 + x^3)^n. - Peter Bala, Jun 11 2024

A180955 Triangle read by rows T(n,k) = numerators of A180955/A180956.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 5, 3, 1, 1, 35, 5, 3, 1, 1, 63, 35, 5, 3, 1, 1, 231, 63, 35, 5, 3, 1, 1, 429, 231, 63, 35, 5, 3, 1, 1, 6435, 429, 231, 63, 35, 5, 3, 1, 1, 12155, 6435, 429, 231, 63, 35, 5, 3, 1, 1, 46189, 12155, 6435, 429, 231, 63, 35, 5, 3, 1, 1, 88179, 46189, 12155, 6435, 429, 231, 63, 35, 5, 3, 1, 1

Offset: 0

Mats Granvik, Sep 28 2010

Consider the fractional triangle A180955/A180956, call it triangle A. Consider also a triangle defined by k=1: T(n,1)=1, k>1 and n>=k: T(n,k)= any random number, else 0, call it triangle B. Calculate the matrix inverse of triangle B, call it triangle C. Multiply C with A, call the result triangle D. Calculate the matrix inverse of D, call it triangle E. Then the first column in both matrix A and matrix E will have the fraction A001790/A046161 in the first column.

			Triangle starts:
      1;
      1,     1;
      3,     1,     1;
      5,     3,     1,    1;
     35,     5,     3,    1,   1;
     63,    35,     5,    3,   1,   1;
    231,    63,    35,    5,   3,   1,  1;
    429,   231,    63,   35,   5,   3,  1,  1;
   6435,   429,   231,   63,  35,   5,  3,  1,  1;
  12155,  6435,   429,  231,  63,  35,  5,  3,  1,  1;
  46189, 12155,  6435,  429, 231,  63, 35,  5,  3,  1,  1;
  88179, 46189, 12155, 6435, 429, 231, 63, 35,  5,  3,  1,  1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A001790, A046161, A180956.

A180955:= func< n,k | Numerator((n-k+1)*Catalan(n-k)/4^(n-k)) >;
[A180955(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 22 2024

A180955[n_, k_]:= Numerator[Binomial[2*(n-k), n-k]/4^(n-k)];
Table[A180955[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 22 2024 *)

def A180955(n,k): return numerator(binomial(2*(n-k), n-k)/4^(n-k))
flatten([[A180955(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 22 2024

From G. C. Greubel, Sep 22 2024: (Start)
T(n, k) = A001790(n-k) = numerator(binomial(2*(n-k), n-k)/4^(n-k)).
T(n, 0) = T(2*n, n) = A001790(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A001790(n+1) + Sum_{j=0..n+1} (-1)^(n+j)*A001790(j).
Sum_{k=0..floor(n/2)} T(n-k, k) = (1/2)*Sum_{j=0..n} (1+(-1)^(n+j))*A001790(j). (End)

Offset changed by G. C. Greubel, Sep 22 2024

A180956 Triangle read by rows T(n,k) = denominators of A180955/A180956.

Original entry on oeis.org

1, 2, 1, 8, 2, 1, 16, 8, 2, 1, 128, 16, 8, 2, 1, 256, 128, 16, 8, 2, 1, 1024, 256, 128, 16, 8, 2, 1, 2048, 1024, 256, 128, 16, 8, 2, 1, 32768, 2048, 1024, 256, 128, 16, 8, 2, 1, 65536, 32768, 2048, 1024, 256, 128, 16, 8, 2, 1, 262144, 65536, 32768, 2048, 1024, 256, 128, 16, 8, 2, 1

Offset: 0

Mats Granvik, Sep 28 2010

			Triangle starts:
       1;
       2,     1;
       8,     2,     1;
      16,     8,     2,    1;
     128,    16,     8,    2,    1;
     256,   128,    16,    8,    2,   1;
    1024,   256,   128,   16,    8,   2,   1;
    2048,  1024,   256,  128,   16,   8,   2,  1;
   32768,  2048,  1024,  256,  128,  16,   8,  2,  1;
   65536, 32768,  2048, 1024,  256, 128,  16,  8,  2,  1;
  262144, 65536, 32768, 2048, 1024, 256, 128, 16,  8,  2,  1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A001790, A046161, A180955.

A180956:= func< n,k | Denominator((n-k+1)*Catalan(n-k)/4^(n-k)) >;
[A180956(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 22 2024

A180956[n_, k_]:= Denominator[Binomial[2*(n-k), n-k]/4^(n-k)];
Table[A180956[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 23 2024 *)

def A180956(n,k): return denominator(binomial(2*(n-k), n-k)/4^(n-k))
flatten([[A180956(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 22 2024

From G. C. Greubel, Sep 23 2024: (Start)
T(n, k) = A046161(n-k) = denominator(binomial(2*(n-k), n-k)/4^(n-k)).
T(n, 0) = T(2*n, n) = A046161(n).
Sum_{k=0..n} T(n, k) = Sum_{j=0..n} A046161(j).
Sum_{k=0..n} (-1)^k*T(n, k) = A046161(n+1) + Sum_{j=0..n+1} (-1)^(n+j)*A046161(j).
Sum_{k=0..floor(n/2)} T(n-k, k) = floor(n/2) + (1/2)*Sum_{j=0..n} (1+(-1)^(n+j)) * A046161(j). (End)

Offset changed by G. C. Greubel, Sep 23 2024

A187791 Repeat n+1 times 2^A005187(n).

Original entry on oeis.org

1, 2, 2, 8, 8, 8, 16, 16, 16, 16, 128, 128, 128, 128, 128, 256, 256, 256, 256, 256, 256, 1024, 1024, 1024, 1024, 1024, 1024, 1024, 2048, 2048, 2048, 2048, 2048, 2048, 2048, 2048, 32768, 32768, 32768, 32768, 32768, 32768, 32768, 32768, 32768, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536, 65536

Offset: 0

Paul Curtz, Jan 06 2013

a(n) is the denominators of the antidiagonals of the Lorentz factor, which can be written A001790(n)/A046161(n), and its differences.
1,           1/2,     3/8,      5/16,    35/128,     63/256,...  the Lorentz gamma factor,
-1/2,       -1/8,   -1/16,    -5/128,    -7/256,   -21/1024, ...  -A098597(n)/A046161(n+1),from the Lorentz (beta) factor,
3/8,        1/16,   3/128,     3/256,    7/1024,     9/2048,...  A161200(n+2)/A046161(n+2),
-5/16,    -5/128,  -3/256,   -5/1024,   -5/2048,  -45/32768,...  A161202(n+3)/A046161(n+4),
35/128,    7/256,  7/1024,    5/2048,  35/32768,   35/65536, ...
-63/256, -21/1024, -9/2048, -45/32768, -35/65536, -63/262144, ...  .
Like 1/n and A164555(n)/A027642(n), the Lorentz factor is an autosequence of the second kind. The first column is the signed sequence.
The main diagonal is (-1)^n *A001790(n)/A061549(n).
The Lorentz factor is the differences of (0, followed by A001803(n)) / (1, followed by A046161(n)).
PiSK(n-2)=(0, 0, followed by A001803(n)) / (1, 1, followed by A046161(n)) is also an autosequence of second kind.
Remember that an autosequence of the second kind is a sequence whose inverse binomial transform is the sequence signed, with its main diagonal being the double of its first upper diagonal. - Paul Curtz, Oct 13 2013

			1,
2,   2,
8,   8,  8,
16, 16, 16, 16.

OEIS Wiki, Autosequence
Wikipedia, Lorentz Factor.

Cf. A003506.

Flatten[Table[Denominator[Binomial[2n, n]/4^n], {n, 0, 19}, {n + 1}]] (* Alonso del Arte, Jan 07 2013 *)
(* Checking with the antidiagonals *) diff = Table[ Differences[ CoefficientList[ Series[1/Sqrt[1 - x], {x, 0, 9}], x], n], {n, 0, 9}]; Table[ diff[[n-k+1,k]] // Denominator,{n,0,10},{k,1,n}] // Flatten (* Jean-François Alcover, Jan 07 2013 *)
Flatten[Table[2^IntegerExponent[(2*n)!, 2], {n, 0, 19}, {n + 1}]]; (* Jean-François Alcover, Mar 27 2013, after A005187 *)

Repeat A046161(n) n+1 times. Triangle.

New definition by M. F. Hasler

A224270 Absolute values of the numerators of the third column of ( 0 followed by (interleave 0 , A001803(n))/A060818(n) ) and its successive differences.

Original entry on oeis.org

1, 1, 5, 11, 95, 203, 861, 1815, 30459, 63635, 264979, 550069, 4555915, 9412543, 38816525, 79898895, 2627302995, 5392044675, 22104436695, 45256266825, 370241638305, 756514878405, 3088866211275, 6300861570705, 102746354288175, 209286947903319

Offset: 0

Paul Curtz, Apr 02 2013

The array is
  0,          0,      1,     0,   3/2,     0,  15/8, 0,...
  0,          1,     -1,   3/2,  -3/2,  15/8, -15/8,...
  1,         -2,    5/2,    -3,  27/8, -15/4,...
  -3,       9/2,  -11/2,  51/8, -57/8,...
  15/2,     -10,   95/8, -27/2,...
  -35/2,  175/8, -203/8,...
  315/8, -189/4,...
  -693/8,...
Note A001803 in the first column and a variant of A206771(n) in the second column.
Now consider a(n)/A046161(n) and its differences:
  1,            1/2,     5/8,  11/16,  95/128, 203/256, 861/1024,...
  -1/2,         1/8,    1/16,  7/128,  13/256, 49/1024,...   =b(n)/A046161(n)
  5/8,        -1/16,  -1/128, -1/256, -3/1024,...
  -11/16,     7/128,   1/256, 1/1024,...
  95/128,   -13/256, -3/1024,...
  -203/256, 49/1024,...
  861/1024,...
This an autosequence of second kind. The first column is the signed sequence.
(Its companion, the corresponding autosequence of first kind, is 0, 1, 1, 9/8, 5/4,... in A206771).
Main diagonal: 1, 1/8, -1/128,... = A002596(n)/A061549(n) ?
b(n) = a(n+1) - A171977*a(n). Also for two successive rows (with shifted A171977).

			a(n)=numerators of 0+1=1, 0+1/2=1/2, 1/4+3/8=5/8, 3/8+5/16=11/16, 15/32+35/128=95/128,... .

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Cf. A001790, A001803, A002596, A046161, A061549, A098597, A171977, A206771.

nmax = 25; t1 = Table[ Numerator[ (2*n+1)*(Binomial[2*n, n]/4^n)] / Denominator[ Binomial[2*n, n]/4^n], {n, 0, Ceiling[nmax/2]}]; t2 = Join[{0}, Table[ If[ OddQ[n], 0, t1[[n/2]] ], {n, 1, nmax+2}] ]; t3 = Table[ Differences[t2, n], {n, 0, nmax}]; t3[[All, 3]] // Numerator // Abs (* Jean-François Alcover, Apr 02 2013 *)

Numerators of (0, 0 followed by A001803(n)/(4*A046161(n))) + A001790(n)/A046161(n).

More terms from Jean-François Alcover, Apr 02 2013

A269949 Triangle read by rows, T(n,k) = denominator(binomial(-1/2, n-k))*binomial(n-1/2, k-1/2), for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 5, 15, 5, 1, 35, 35, 35, 7, 1, 63, 315, 105, 63, 9, 1, 231, 693, 1155, 231, 99, 11, 1, 429, 3003, 3003, 3003, 429, 143, 13, 1, 6435, 6435, 15015, 9009, 6435, 715, 195, 15, 1, 12155, 109395, 36465, 51051, 21879, 12155, 1105, 255, 17, 1

Offset: 0

Peter Luschny, Apr 07 2016

Numerators of "gravitational descendent fields" presented on p. 28 of the Zhou reference. See also p. 31. - Tom Copeland, Feb 13 2017

			Triangle starts:
[  1]
[  1,   1]
[  3,   3,    1]
[  5,  15,    5,   1]
[ 35,  35,   35,   7,  1]
[ 63, 315,  105,  63,  9,  1]
[231, 693, 1155, 231, 99, 11, 1]

J. Zhou, Quantum deformation theory of the Airy curve and the mirror symmetry of a point, arXiv preprint arXiv:1405.5296 [math.AG], 2014.

Cf. A001790 (col. 0), A001803 (col. 1), A161199 (col. 2), A161201 (col. 3).

Cf. A269950.

Table[Denominator[Binomial[-1/2, n - k]] Binomial[n - 1/2, k - 1/2], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 13 2017 *)

A269949 = lambda n,k: binomial(-1/2,n-k).denom()*binomial(n-1/2,k-1/2)
for n in range(8): print([A269949(n,k) for k in (0..n)])

A322756 Denominator of expected payoff in the "Guessing Card Colors" game with a 2n-card deck, using an optimal strategy.

Original entry on oeis.org

2, 6, 10, 70, 126, 462, 858, 12870, 24310, 92378, 176358, 1352078, 2600150, 10029150, 19389690, 601080390, 1166803110, 4537567650, 8836315950, 68923264410, 134564468610, 526024740930, 1029178840950, 16123801841550, 31602651609438, 123979633237026

Offset: 1

N. J. A. Sloane, Dec 27 2018

A 2n-card playing deck is shuffled and then revealed one-by-one to a player who guesses the color (red or black) of each card prior to its being revealed. The player earns one dollar for each card whose color he guesses correctly; there is no penalty for being wrong.

			3/2, 17/6, 41/10, 373/70, 823/126, 3565/462, 7625/858, 129293/12870, 272171/24310, 1139735/92378, 2376047/176358, ...

Thane Plambeck and others, Posting to Math Fun Mailing List, Dec 26 2018.

Michael Andreoli (proposer), Guessing Card Colors, Problem #630, College Mathematics Journal Vol. 30, No. 3 (May, 1999), pp. 234-235. Solution by John Henry Steelman.

Cf. A322755.

This is twice A001790.

a(n) = denominator(n - 1/2 + 2^(2*n-1)/binomial(2*n,n)); \\ Michel Marcus, Dec 28 2018

from fractions import Fraction
from math import comb
def A322756(n): return (n-Fraction(1,2)+Fraction(1<<(m:=n<<1)-1,comb(m,n))).denominator # Chai Wah Wu, Feb 12 2023

The optimal payoff is n - 1/2 + 2^(2n-1)/binomial(2n,n).

A344402 a(n) = denominator(R(n,3)), where R(n,d) = (Product_{j prime to d} Pochhammer(j/d, n)) / n!.

Original entry on oeis.org

1, 9, 81, 2187, 19683, 177147, 4782969, 43046721, 387420489, 31381059609, 282429536481, 2541865828329, 68630377364883, 617673396283947, 5559060566555523, 150094635296999121, 1350851717672992089, 12157665459056928801, 984770902183611232881, 8862938119652501095929

Offset: 0

Peter Luschny, May 20 2021

nonn

R(n, 1) = A000012 / A000012.
R(n, 2) = A001790 / A046161.
R(n, 3) = A273194 / (this sequence).

coprimes := n -> select(j -> igcd(j, n) = 1, {$1..n}):
R := (n, d) -> mul(pochhammer(j/d, n), j in coprimes(d)) / n!:
seq(denom(R(n, 3)), n = 0..16);

A094083 Numerators of ratio of sides of n-th triple of rectangles of unit area sum around a triangle.

Original entry on oeis.org

1, 1, 1, 4, 9, 64, 25, 256, 1225, 16384, 3969, 65536, 53361, 1048576, 184041, 4194304, 41409225, 1073741824, 147744025, 4294967296, 2133423721, 68719476736, 7775536041, 274877906944, 457028729521, 17592186044416, 1690195005625

Offset: 1

Peter J. C. Moses, Apr 30 2004

Page 13 of the link shows the type of configuration. When n is odd, the numerators 1,1,9,25,1225,3969,.. are A038534 and (A001790)^2, and the denominators 1,4,64,256,16384,65536,.. are A056982, A038533/2, and (A046161)^2. When n is even, the numerators 1,4,64,256,16384,65536,.. are A056982, A038533/2, and (A046161)^2, and the denominators 3,27,675,3675,297675,1440747,.. are 3*(A001803)^2. The limit of a(n+1)/a(n) as n(odd) tends to infinity = Pi^2/12, A072691. The limit of a(n+2)/a(n) as n tends to infinity = 1. a(n), for large odd n, tends to 2/(Pi*n). a(n), for large even n, tends to Pi/(6*n). The expansion of 2*x*EllipticK(x)/Pi gives the odd fractions. The expansion of 1/3*x*HypergeometricPFQ({1,1,1},{3/2,3/2},x) gives the even fractions.

			a(5) = a(5-2)*((5-2)/(5-1))^2 = 1/4*(3/4)^2 = 9/64

Paul Yiu, EuclideanGeometry Notes

Cf. A005386, A092936, A094084.

a[n_]:=If[OddQ[n], ((n/2-1)!)^2/(Pi*((n/2-1/2)!)^2), Pi*((n/2-1)!)^2/(12*((n/2-1/2)!)^2)] a[n_]:=If[OddQ[n], (2^(1-n)*(n-2)!!^2)/((n-1)/2)!^2, (2^(n-2)*((n-2)/2)!^2)/(3*(n-1)!!^2)] a[n_]:=((12+Pi^2+E^(I*n*Pi)*(Pi^2-12))*((n/2-1)!)^2)/(24*Pi*((n/2-1/2)!)^2) (CoefficientList[Series[(I*x*(6+Sqrt[3]*Pi)-2*x*Sqrt[3]*Log[x+Sqrt[x^2-1]])/(6*Sqrt[x^2-1]), {x, 0, 20}], x])^2

a(n)=a(n-2)*((n-2)/(n-1))^2, a(1)=1, a(2)=1/3. a(n)=((n/2-1)!)^2/(Pi*((n/2-1/2)!)^2) for n odd. a(n)=(2^(1-n)*(n-2)!!^2)/((n-1)/2)!^2 for n odd. a(n)=Pi*((n/2-1)!)^2/(12*((n/2-1/2)!)^2) for n even. a(n)=(2^(n-2)*((n-2)/2)!^2)/(3*(n-1)!!^2) for n even.

A162444 Denominators of the BG1[ -5,n] coefficients of the BG1 matrix.

Original entry on oeis.org

1, 1, 3, 5, 35, 9, 231, 143, 6435, 12155, 3553, 88179, 96577, 1300075, 5014575, 102051, 100180065, 116680311, 2268783825, 210388475, 6892326441, 67282234305, 17534158031, 39583801575, 8061900920775, 169906729083

Offset: 1

Johannes W. Meijer, Jul 06 2009

For the numerators of the BG1[ -5,n] coefficients see A162443.

We observe that BG1[ -3,n] = (-1)*A002595(n-1)/A055786(n-1), i.e. they equal the inverted coefficients of the series expansion of arcsin(x), and that BG1[ -1,n] = A046161(n-1)/A001790(n-1), i.e. they equal the inverted coefficients of the series expansion of 1/sqrt(1-x).

			The first few formulas for the BG1[1-2*m,n] matrix coefficients are:
BG1[ -1,n] = (1)*4^(n-1)*(n-1)!^2/(2*n-2)!
BG1[ -3,n] = (1-2*n)*4^(n-1)*(n-1)!^2/(2*n-2)!
BG1[ -5,n] = (1-8*n+12*n^2)*4^(n-1)*(n-1)!^2/(2*n-2)!
BG1[ -7,n] = (1-2*n+60*n^2-120*n^3)*4^(n-1)*(n-1)!^2/(2*n-2)!

A162443 are the numerators of the BG1[ -5, n] matrix coefficients.
The BG1[ -3, n] equal A002595(n-1)/A055786(n-1) for n =>1.
The BG1[ -1, n] equal A046161(n-1)/A001790(n-1) for n =>1.

a(n) = denom(BG1[ -5,n]) and A162443(n) = numer(BG1[ -5,n]) with BG1[ -5,n] = 4^(n-1)*(1-8*n+12*n^2)*(n-1)!^2/ (2*n-2)!.

cp's OEIS Frontend

A127974 Numerators in expansion of (1-x)^(-2/3).

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A180955 Triangle read by rows T(n,k) = numerators of A180955/A180956.

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A180956 Triangle read by rows T(n,k) = denominators of A180955/A180956.

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A187791 Repeat n+1 times 2^A005187(n).

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A224270 Absolute values of the numerators of the third column of ( 0 followed by (interleave 0 , A001803(n))/A060818(n) ) and its successive differences.

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A269949 Triangle read by rows, T(n,k) = denominator(binomial(-1/2, n-k))*binomial(n-1/2, k-1/2), for n>=0 and 0<=k<=n.

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A322756 Denominator of expected payoff in the "Guessing Card Colors" game with a 2n-card deck, using an optimal strategy.

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