A084623 -id:A084623 - cp's OEIS frontend

A075101 Numerator of 2^n/n.

2, 2, 8, 4, 32, 32, 128, 32, 512, 512, 2048, 1024, 8192, 8192, 32768, 4096, 131072, 131072, 524288, 262144, 2097152, 2097152, 8388608, 2097152, 33554432, 33554432, 134217728, 67108864, 536870912, 536870912, 2147483648, 134217728, 8589934592, 8589934592, 34359738368, 17179869184, 137438953472

Offset: 1

Reinhard Zumkeller, Sep 01 2002

G. C. Greubel, Table of n, a(n) for n = 1..1000

Denominator is A000265(n).

Cf. A007814, A051179, A073113, A084623, A273893.

[Numerator(2^n/n): n in [1..50]]; // G. C. Greubel, Feb 28 2019

Maple
```
[seq(numer(2^n/n),n=1..50)];
```

f[n_]:=Numerator[2^n/n]; Array[f,50] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2011 *)

a(n) = numerator(2^n/n); \\ Michel Marcus, Mar 25 2018

a(n) = 2^(n - valuation(n, 2)) \\ Jianing Song, Oct 24 2018

from fractions import Fraction
def A075101(n):
    return (Fraction(2**n)/n).numerator # Chai Wah Wu, Mar 25 2018

[numerator(2^n/n) for n in (1..50)] # G. C. Greubel, Feb 28 2019

a(n) = 2^(n - A007814(n)).
a(n) = 2*A084623(n). - Paul Curtz, Jan 28 2013
a(n) = 2^A093048(n). - Paul Curtz, Jun 10 2016
From Peter Bala, Feb 25 2019: (Start)
a(n) = 2^n/gcd(n,2^n).
O.g.f.: F(2*x) - (1/2)*F((2*x)^2) - (1/4)*F((2*x)^4) - (1/8)*F((2*x)^8) - ..., where F(x) = x/(1 - x). Cf. A000265.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = F((x/2)) + F((x/2)^2) + 2*F((x/2)^4) + 4*F((x/2)^8) + 8*F((x/2)^16) + 16*F((x/2)^32) + .... (End)
Sum_{n>=1} 1/a(n) = Sum_{n>=1} 2^(2^(n-1)+n-1)/(2^(2^n) - 1) = Sum_{n>=1} A073113(n-1)/A051179(n) = 1.48247501707... - Amiram Eldar, Aug 14 2022

A247281 a(n) = 4^n -(-1)^n.

Original entry on oeis.org

0, 5, 15, 65, 255, 1025, 4095, 16385, 65535, 262145, 1048575, 4194305, 16777215, 67108865, 268435455, 1073741825, 4294967295, 17179869185, 68719476735, 274877906945, 1099511627775, 4398046511105, 17592186044415

Offset: 0

Paul Curtz, Sep 11 2014

OEIS Wiki, Autosequence
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. A015521, A037481, A084623.

LinearRecurrence[{3, 4}, {0, 5}, 23] (* Jean-François Alcover, May 22 2016 *)

concat(0, Vec(-5*x / ((x+1)*(4*x-1)) + O(x^100))) \\ Colin Barker, Sep 11 2014

vector(100,n,4^(n-1)+(-1)^n) \\ Derek Orr, Sep 11 2014

def A247281(n): return (1<<(n<<1))+(1 if n&1 else -1) # Chai Wah Wu, Jun 28 2023

a(n) = 5*A015521(n).
a(n+1) = 10*A037481(n) + 5.
a(n+1) = 4*a(n) + 5*(-1)^n.
a(n) = 3*a(n-1) + 4*a(n-2). - Colin Barker, Sep 11 2014
G.f.: -5*x / ((x+1)*(4*x-1)). - Colin Barker, Sep 11 2014

Typos in data fixed by Colin Barker, Sep 11 2014

A093048 a(n) = n minus exponent of 2 in n, with a(0) = 0.

Original entry on oeis.org

0, 1, 1, 3, 2, 5, 5, 7, 5, 9, 9, 11, 10, 13, 13, 15, 12, 17, 17, 19, 18, 21, 21, 23, 21, 25, 25, 27, 26, 29, 29, 31, 27, 33, 33, 35, 34, 37, 37, 39, 37, 41, 41, 43, 42, 45, 45, 47, 44, 49, 49, 51, 50, 53, 53, 55, 53, 57, 57, 59, 58, 61, 61, 63, 58, 65, 65, 67, 66, 69

Offset: 0

Ralf Stephan, Mar 16 2004

nonn

			G.f. = x + x^2 + 3*x^3 + 2*x^4 +  5*x^5 + 5*x^6 + 7*x^7 + 5*x^8 + 9*x^9 + ... - _Michael Somos_, Jan 25 2020

R. J. Mathar, Table of n, a(n) for n = 0..1000

a(n) = n - A007814(n) = A093049(n) + 1, n > 0.
a(n) is the exponent of 2 in A002689(n-1), A014070(n), A060690(n), A075101(n).
See also A084623.

A093048 := proc(n)
    n-A007814(n) ;
end proc: # R. J. Mathar, Jul 24 2014

a[ n_] := If[ n == 0, n - IntegerExponent[n, 2]]; (* Michael Somos, Jan 25 2020 *)

a(n) = if(n<1, 0, if(n%2==0, a(n/2) + n/2 - 1, n))

a(n) = n - valuation(n, 2) \\ Jianing Song, Oct 24 2018

def A093048(n): return n-(~n& n-1).bit_length() if n else 0 # Chai Wah Wu, Jul 07 2022

Recurrence: a(2n) = a(n) + n - 1, a(2n+1) = 2n + 1.
G.f.: Sum_{k>=0} (t*(t^3 + t^2 + 1)/(1 - t^2)^2), with t = x^2^k.
a(n) = Sum_{k=1..n} sign(n mod 2^k). - Wesley Ivan Hurt, May 09 2021

A093049 n-1 minus exponent of 2 in n, a(0) = 0.

Original entry on oeis.org

0, 0, 0, 2, 1, 4, 4, 6, 4, 8, 8, 10, 9, 12, 12, 14, 11, 16, 16, 18, 17, 20, 20, 22, 20, 24, 24, 26, 25, 28, 28, 30, 26, 32, 32, 34, 33, 36, 36, 38, 36, 40, 40, 42, 41, 44, 44, 46, 43, 48, 48, 50, 49, 52, 52, 54, 52, 56, 56, 58, 57, 60, 60, 62, 57, 64, 64, 66, 65, 68

Offset: 0

Ralf Stephan, Mar 16 2004

nonn

			G.f. = 2*x^3 + x^4 + 4*x^5 + 4*x^6 + 6*x^7 + 4*x^8 + 8*x^9 + 8*x^10 + ... - _Michael Somos_, Jan 25 2020

a(n) = n - A007814(n) - 1 = A093048(n) - 1, n>0.

a(n) is the exponent of 2 in A001761(n+1), A002105(n), A002682(n-1), A006963(n), A036770(n-1), A059837(n), A084623(n), |A003707(n)|, |A011859(n)|.

a[ n_] := If[ n == 0, 0, n - 1 - IntegerExponent[n, 2]]; (* Michael Somos, Jan 25 2020 *)

a(n)=if(n<1,0,if(n%2==0,a(n/2)+n/2-1,n-1))

{a(n) = if( n, n - 1 - valuation(n, 2))}; /* Michael Somos, Jan 25 2020 */

def A093049(n): return n-1-(~n& n-1).bit_length() if n else 0 # Chai Wah Wu, Jul 07 2022

Recurrence: a(2n) = a(n) + n - 1, a(2n+1) = 2n.

G.f.: sum(k>=0, t^3(t+2)/(1-t^2)^2, t=x^2^k).

A206771 0 followed by the numerators of the reduced (A001803(n) + A001790(n)) / (2*A046161(n)).

Original entry on oeis.org

0, 1, 1, 9, 5, 175, 189, 1617, 429, 57915, 60775, 508079, 264537, 8788507, 9100525, 75218625, 9694845, 5109183315, 5250613995, 43106892675, 22090789875, 723694276305, 740104577355, 6049284520695, 1543768261425, 201547523019375

Offset: 0

Paul Curtz, Jan 10 2013

We write the fractions a(n)/b(n) and higher order differences as a matrix:
0,         1,      1,    9/8,    5/4,...
1,         0,    1/8,    1/8, 15/128,... = (A001790(n)/A046161(n) +Lorbeta(n)) /2
-1,      1/8,      0, -1/128, -1/128,... = (Lorbeta(n+1) +A161200(n+1)/A046161(n+1)) / 2
9/8,    -1/8, -1/128,      0, 1/1024,...
-5/4, 15/128,  1/128, 1/1024,      0,...
Here, Lorbeta(0)=1 and Lorbeta(n) = -A098597(n-1)/A046161(n) for n>0 is the inverse of the Lorentz factor.
The first line with numerators a(n) and denominators b(n) is 0, 1, 1, 9/8, 5/4, 175/128, 189/128, 1617/1024, 429/256, 57915/32768, 60775/32768,... It is an autosequence: Its inverse binomial transform is the signed sequence.
a(n+1)/(2*n-1)= 1, 3, 1, 25, 21, 147, 33, 3861, 3575, 26741,... .
a(n+1)/A146535(n) = 9, 5, 35, 27, 539, 39, 4455,... .
A001790(n)/A046161(n) yields the coefficients of the Lorentz factor (or Lorentz gamma factor). With b for beta and g for gamma:
g = (1-b^2)^-1 = 1 + (b^2)/2 + 3*(b^4)/8 + 5*(b^6)/16 + ... .
b = (1-g^-2)^-1 = 1 - (g^-2)/2 - (g^-4)/8 - (g^-6)/16  - ... .
Are the denominators of the first subdiagonal 1, 1/8, -1/128, 1/1024,...  A061549(n) ?
a(n+1)/(A000108(n)*b(n)) = 1, 1, 9/16, 1/4, 25/256, 9/256, 49/4096, 1/256, 81/65536, 25/65536, 121/1048576,... = A191871(n+1)/ A084623(n+1)^2 ?

			From the first formula: a(1)=1*1, a(2)=1*1, a(3)=3*3, a(4)=1*5, a(5)=5*35, a(6)=3*63.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
OEIS Wiki, Autosequence
Wikipedia, Lorentz Factor.

Cf. A187791, A000108.

/* By definition: */ m:=25; R:=PowerSeriesRing(Rationals(), m); p:=Coefficients(R!(1/(1-x)^(1/2))); q:=Coefficients(R!((1-x)^(-3/2))); A001790:=[Numerator(p[i]): i in [1..m]]; A001803:=[Numerator(q[i]): i in [1..m]]; A046161:=[Denominator(Binomial(2*n,n)/4^n): n in [0..m-1]]; [0] cat [Numerator((A001803[n]+A001790[n])/(2*A046161[n])): n in [1..m]]; // Bruno Berselli, Mar 11 2013

A206771 := proc(n)
        A001790(n)+A001803(n) ;
        %/2/A046161(n) ;
        numer(%) ;
end proc: # R. J. Mathar, Jan 18 2013

max = 25; A001803 = CoefficientList[Series[(1 - x)^(-3/2), {x, 0, max}], x] // Numerator; A001790 = CoefficientList[Series[1/Sqrt[(1 - x)], {x, 0, max}], x] // Numerator; A046161 = Table[Binomial[2n, n]/4^n, {n, 0, max}] // Denominator; a[n_] := (A001803[[n]] + A001790[[n]])/(2*A046161[[n]]) // Numerator; a[0] = 0; Table[a[n], {n, 0, max}]
(* or (from 1st formula) : *) Table[ n*Numerator[4^(1-n)*Binomial[2n-2, n-1]]/2^IntegerExponent[n, 2], {n, 0, max}]
(* or (from 2nd formula) : *) Table[ Numerator[ CatalanNumber[n-1]/2^(2n-1)]*Numerator[n^2/2^n], {n, 0, max}] (* Jean-François Alcover, Jan 31 2013 *)

a(n) = A000265(n) * A001790(n-1).
a(n) = A098597(n-1) * A191871(n). See also A181318
a(n) = numerator of n*binomial(2n-2,n-1)/4^(n-1). - Nathaniel Johnston, Dec 16 2022

a(11)-a(25) from Jean-François Alcover, Jan 13 2013

A213268 Denominators of the Inverse semi-binomial transform of A001477(n) read downwards antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 4, 4, 1, 2, 2, 8, 2, 1, 1, 4, 1, 16, 16, 1, 2, 1, 8, 8, 32, 16, 1, 1, 4, 4, 16, 8, 64, 64, 1, 2, 2, 8, 4, 32, 32, 128, 16, 1, 1, 4, 2, 16, 16, 64, 8, 256, 256, 1, 2, 1, 8, 8, 32, 4, 128, 128, 512, 256, 1, 1, 4, 4, 16, 2, 64, 64, 256, 128, 1024, 1024

Offset: 0

Paul Curtz, Jun 08 2012

Starting from any sequence a(k) in the first row, define the array T(n,k) of the inverse semi-binomial transform  by T(0,k) = a(k), T(n,k) = T(n-1,k+1) -T(n-1,k)/2, n>=1.
Here, where the first row is the nonnegative integers, the array is
0        1      2      3    4      5      6      7     8   =A001477(n)
1       3/2     2    5/2    3    7/2      4    9/2     5   =A026741(n+2)/A000034(n)
1       5/4   3/2    7/4    2    9/4    5/2   11/4     3   =A060819(n+4)/A176895(n)
3/4     7/8     1    9/8  5/4   11/8    3/2   13/8   7/4   =A106609(n+6)/A205383(n+6)
1/2    9/16   5/8  11/16  3/4  13/16    7/8  15/16     1   =A106617(n+8)/TBD
5/16  11/32   3/8  13/32 7/16  15/32    1/2  17/32  9/16
3/16  13/64  7/32  15/64  1/4  17/64   9/32  19/64  5/16
7/64 15/128   1/8 17/128 9/64 19/128   5/32 21/128 11/64
1/16 17/256 9/128 19/256 5/64 21/256 11/128 23/256  3/32.
The first column contains 0, followed by fractions A000265/A084623, that is Oresme numbers n/2^n multiplied by 2 (see A209308).

			The array of denominators starts:
  1   1   1   1   1   1   1   1   1   1   1 ...
  1   2   1   2   1   2   1   2   1   2   1 ...
  1   4   2   4   1   4   2   4   1   4   2 ...
  4   8   1   8   4   8   2   8   4   8   1 ...
  2  16   8  16   4  16   8  16   1  16   8 ...
16  32   8  32  16  32   2  32  16  32   8 ...
16  64  32  64   4  64  32  64  16  64  32 ...
64 128   8 128  64 128  32 128  64 128  16 ...
16 256 128 256  64 256 128 256  32 256 128 ...
256 512 128 512 256 512  64 512 256 512 128 ...
All entries are powers of 2.

A213268frac := proc(n,k)
        if n = 0 then
                return k ;
        else
                return procname(n-1,k+1)-procname(n-1,k)/2 ;
        end if;
end proc:
A213268 := proc(n,k)
        denom(A213268frac(n,k)) ;
end proc: # R. J. Mathar, Jun 30 2012

T[0, k_] := k; T[n_, k_] := T[n, k] = T[n-1, k+1] - T[n-1, k]/2; Table[T[n-k, k] // Denominator, {n, 0, 11}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Sep 12 2014 *)

A301278 Numerator of variance of n-th row of Pascal's triangle.

Original entry on oeis.org

0, 0, 1, 4, 47, 244, 1186, 1384, 25147, 112028, 98374, 1067720, 1531401, 39249768, 166656772, 88008656, 2961699667, 12412521388, 51854046982, 108006842264, 448816369361, 3721813363288, 15401045060572, 15904199160592, 131178778841711, 1080387930269464, 4443100381114156, 9124976352166288

Offset: 0

N. J. A. Sloane, Mar 18 2018

Variance here is the sample variance unbiased estimator. For population variance, see A301631.

			The first few variances are 0, 0, 1/3, 4/3, 47/10, 244/15, 1186/21, 1384/7, 25147/36, 112028/45, 98374/11, 1067720/33, 1531401/13, 39249768/91, 166656772/105, 88008656/15, 2961699667/136, 12412521388/153, 51854046982/171, 108006842264/95, 448816369361/105, ...

Chai Wah Wu, Table of n, a(n) for n = 0..1659
Simon Demers, Taylor's Law Holds for Finite OEIS Integer Sequences and Binomial Coefficients, American Statistician, online: 19 Jan 2018.

Mean and variance of n-th row of Pascal's triangle: A084623/A000265, A301278/A301279, A054650, A301280.

M:=70;
m := n -> 2^n/(n+1);
m1:=[seq(m(n),n=0..M)]; # A084623/A000265
v := n -> (1/n) * add((binomial(n,i) - m(n))^2, i=0..n );
v1:= [0, 0, seq(v(n),n=2..60)]; # A301278/A301279

a(n) = if(n==0, 0, numerator(binomial(2*n,n)/n - 4^n/(n*(n+1)))); \\ Altug Alkan, Mar 25 2018

from fractions import Fraction
from sympy import binomial
def A301278(n):
    return (Fraction(int(binomial(2*n,n)))/n - Fraction(4**n)/(n*(n+1))).numerator if n > 0 else 0 # Chai Wah Wu, Mar 23 2018

a(0) = 0; a(n) = numerator of binomial(2n,n)/n - 4^n/(n*(n+1)) for n >= 1. - Chai Wah Wu, Mar 23 2018

A301279 Denominator of variance of n-th row of Pascal's triangle.

Original entry on oeis.org

1, 1, 3, 3, 10, 15, 21, 7, 36, 45, 11, 33, 13, 91, 105, 15, 136, 153, 171, 95, 105, 231, 253, 69, 150, 325, 351, 189, 203, 435, 155, 31, 528, 51, 595, 315, 111, 703, 741, 195, 410, 861, 903, 473, 495, 1035, 1081, 141, 588, 1225, 255, 663, 689, 1431, 1485

Offset: 0

N. J. A. Sloane, Mar 18 2018

Variance here is sample variance unbiased estimator. For population variance, the denominator is A191871(n+1) = A000265(n+1)^2. - Chai Wah Wu, Mar 25 2018

			The first few variances are 0, 0, 1/3, 4/3, 47/10, 244/15, 1186/21, 1384/7, 25147/36, 112028/45, 98374/11, 1067720/33, 1531401/13, 39249768/91, 166656772/105, 88008656/15, 2961699667/136, 12412521388/153, 51854046982/171, 108006842264/95, 448816369361/105, ...

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Simon Demers, Taylor's Law Holds for Finite OEIS Integer Sequences and Binomial Coefficients, American Statistician, online: 19 Jan 2018.

Mean and variance of n-th row of Pascal's triangle: A084623/A000265, A301278/A301279, A054650, A301280.

M:=70;
m := n -> 2^n/(n+1);
m1:=[seq(m(n),n=0..M)]; # A084623/A000265
v := n -> (1/n) * add((binomial(n,i) - m(n))^2, i=0..n );
v1:= [0, 0, seq(v(n),n=2..60)]; # A301278/A301279

a(n) = if(n==0, 1, denominator(binomial(2*n,n)/n - 4^n/(n*(n+1)))); \\ Altug Alkan, Mar 25 2018

from fractions import Fraction
from sympy import binomial
def A301279(n):
    return (Fraction(int(binomial(2*n,n)))/n - Fraction(4**n)/(n*(n+1))).denominator if n > 0 else 1 # Chai Wah Wu, Mar 23 2018

a(0) = 1; a(n) = denominator of binomial(2n,n)/n - 4^n/(n*(n+1)) for n >= 1. - Chai Wah Wu, Mar 23 2018

A301280 Nearest integer to variance of n-th row of Pascal's triangle.

Original entry on oeis.org

0, 0, 0, 1, 5, 16, 56, 198, 699, 2490, 8943, 32355, 117800, 431316, 1587207, 5867244, 21777203, 81127591, 303240041, 1136914129, 4274441613, 16111746161, 60873695892, 230495640009, 874525192278, 3324270554675, 12658405644200, 48280298159610

Offset: 0

N. J. A. Sloane, Mar 18 2018

nonn

			The first few variances are 0, 0, 1/3, 4/3, 47/10, 244/15, 1186/21, 1384/7, 25147/36, 112028/45, 98374/11, 1067720/33, 1531401/13, 39249768/91, 166656772/105, 88008656/15, 2961699667/136, 12412521388/153, 51854046982/171, 108006842264/95, 448816369361/105, ...

Robert Israel, Table of n, a(n) for n = 0..1667
Simon Demers, Taylor's Law Holds for Finite OEIS Integer Sequences and Binomial Coefficients, American Statistician, Volume 72, 2018 - Issue 4.

Mean and variance of n-th row of Pascal's triangle: A084623/A000265, A301278/A301279, A054650.

M:=70;
m := n -> 2^n/(n+1);
m1:=[seq(m(n),n=0..M)]; # A084623/A000265
v := n -> (1/n) * add((binomial(n,i) - m(n))^2, i=0..n );
v1:= [0, 0, seq(v(n),n=2..60)]; # A301278/A301279 and A301280
# Alternative:
f:= n -> round((binomial(2*n,n)-4^n/(n+1))/n): f(0):=0:
map(f, [$0..60]); # Robert Israel, Jul 18 2019

From Robert Israel, Jul 18 2019: (Start)
The variance is binomial(2*n,n)/n - 4^n/(n*(n+1)).
a(n) ~ 4^n/(sqrt(Pi)*n^(3/2)). (End)

A339392 Numerators of the probability that when a stick is broken up at n-1 points independently and uniformly chosen at random along its length there exist 3 of the n pieces that can form a triangle.

Original entry on oeis.org

0, 0, 1, 4, 23, 53, 87, 593, 5807, 415267, 8758459, 274431867, 12856077691, 905435186299, 481691519113703, 77763074616922439, 3824113551749834107, 1437016892446437662971, 165559472503434318118655, 146602912901791088694069887, 200050146291129782743679367167

Offset: 1

Amiram Eldar, Dec 04 2020

For the corresponding probability that any triple of pieces can form a triangle, see A001791. The probabilities for these two cases were found by Kong et al. (2013).

			Fractions begin with 0, 0, 1/4, 4/7, 23/28, 53/56, 87/88, 593/594, 5807/5808, 415267/415272, 8758459/8758464, 274431867/274431872, ...
For n = 1 or 2 the number of pieces is less than 3, so the probability is 0.
For n = 3, the stick is being broken into 3 pieces and the probability that they can form a triangle is 1/4, the solution to the classical broken stick problem (see, e.g., Gardner, 2001).

Amiram Eldar, Table of n, a(n) for n = 1..100
Alexander Bogomolny, Stick Broken Into Three Pieces (Trilinear Coordinates), Interactive Mathematics Miscellany and Puzzles, Cut the Knot website.
P. A. Crowdmath, The Broken Stick Project, arXiv:1805.06512 [math.HO], 2018.
Martin Gardner, Probability and Ambiguity, The Colossal Book of Mathematics, W. W. Norton, New York, 2001, chapter 21, pp. 273-285.
Lingyi Kong, Luvsandondov Lkhamsuren, Abigail Turner, Aananya Uppal and A. J. Hildebrand, Random Points, Broken Sticks, and Triangles, Project Report, Illinois Geometry Lab, 2013.

Cf. A000045, A001791, A084623, A234951, A243398, A339393 (denominators).

f = Table[k/(Fibonacci[k + 2] - 1), {k, 2, 20}]; Numerator[1 - FoldList[Times, 1, f]]

a(n) = numerator(1 - Product_{k=2..n} k/(Fibonacci(k+2)-1)).

Lim_{n->oo} a(n)/A339393(n) = 1.

cp's OEIS Frontend

A075101 Numerator of 2^n/n.

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A247281 a(n) = 4^n -(-1)^n.

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A093048 a(n) = n minus exponent of 2 in n, with a(0) = 0.

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A093049 n-1 minus exponent of 2 in n, a(0) = 0.

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Mathematica

PARI

PARI

Python

Formula

A206771 0 followed by the numerators of the reduced (A001803(n) + A001790(n)) / (2*A046161(n)).

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Magma

Maple

Mathematica

Formula

Extensions

A213268 Denominators of the Inverse semi-binomial transform of A001477(n) read downwards antidiagonals.

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Maple

Mathematica

A301278 Numerator of variance of n-th row of Pascal's triangle.

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