A191871 -id:A191871 - cp's OEIS frontend

A061049 Numerator of 1/64 - 1/n^2.

0, 17, 9, 57, 5, 105, 33, 161, 3, 225, 65, 297, 21, 377, 105, 465, 1, 561, 153, 665, 45, 777, 209, 897, 15, 1025, 273, 1161, 77, 1305, 345, 1457, 3, 1617, 425, 1785, 117, 1961, 513, 2145, 35, 2337, 609, 2537, 165, 2745, 713, 2961, 3, 3185, 825

Offset: 8

N. J. A. Sloane, May 26 2001

The 495 initial terms are the same as A000265(n * (n+16)), n > 0. - Simon Strandgaard, Oct 30 2021

G. C. Greubel, Table of n, a(n) for n = 8..1000

Cf. A061035-A061050.

Cf. also A191871.

Table[ Numerator[1/64 - 1/n^2], {n, 8, 58}] (* Jean-François Alcover, May 31 2013 *)

for(n=8,50, print1(numerator(1/64 - 1/n^2), ", ")) \\ G. C. Greubel, Jul 07 2017

A228564 Largest odd divisor of n^2 + 1.

Original entry on oeis.org

1, 1, 5, 5, 17, 13, 37, 25, 65, 41, 101, 61, 145, 85, 197, 113, 257, 145, 325, 181, 401, 221, 485, 265, 577, 313, 677, 365, 785, 421, 901, 481, 1025, 545, 1157, 613, 1297, 685, 1445, 761, 1601, 841, 1765, 925, 1937, 1013, 2117, 1105, 2305, 1201, 2501, 1301, 2705

Offset: 0

Jeremy Gardiner, Aug 25 2013

From Lamine Ngom, Jan 04 2023: (Start)
For n>2, a(n) = hypotenuse c of the primitive Pythagorean triple (a, b, c) such that n*a = b + c.
Terms that appear twice (1, 5, 145, 4901, ...) are the positive terms of A076218. Equivalently, the products of two consecutive terms of A001653, or one more than the squares of A001542.
These duplicated terms appear at indices i and j (i>j) such that (i^2-1)/2 = j^2 (A001541). In addition, they are hypotenuse in two primitive Pythagorean triples: (i, j^2, a(i)) and (2*j, j^2-1, a(i)). (End)

			A002522(3) = 3^2+1 = 10 => a(3) = 10/2 = 5.

Jeremy Gardiner, Table of n, a(n) for n = 0..1000
Wikipedia, Quasi-polynomial.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A000265, A002522, A164900, A191871.

for n = 0 to 45 : t=n^2+1
x: if not t mod 2 then t=t/2 : goto x
print str$(t);", "; : next n
print
end

List([0..60],n->NumeratorRat((n^2+1)/(n+1))); # Muniru A Asiru, Feb 20 2019

[(n^2+1)*(3+(-1)^n)/4: n in [0..60]]; // Bruno Berselli, Aug 26 2013

[Denominator(2*n^2/(n^2+1)): n in [0..60]]; // Vincenzo Librandi, Aug 19 2014

lod:= t -> t/2^padic:-ordp(t,2):
seq(lod(n^2+1),n=0..60); # Robert Israel, Aug 19 2014

Table[(n^2 + 1) (3 + (-1)^n)/4, {n, 0, 60}] (* Bruno Berselli, Aug 26 2013 *)

a(n)=if(n<2,n>0,m=n\4;[4*a(2*m)-3,2*a(2*m)+4*m-1,4*a(2*m)+16*m+1,2*a(2*m)+12*m+3][(n%4)+1]) \\ Ralf Stephan, Aug 26 2013

a(n)=(n^2+1)/2^valuation(n^2+1,2) \\ Ralf Stephan, Aug 26 2013

[(n^2+1)*(3+(-1)^n)/4 for n in (0..60)] # G. C. Greubel, Feb 20 2019

a(n) = A000265(A002522(n)).
From Ralf Stephan, Aug 26 2013: (Start)
a(4n) = 4*a(2n) - 3.
a(4n+1) = 2*a(2n) + 4*n - 1.
a(4n+2) = 4*a(2n) + 16*n + 1.
a(4n+3) = 2*a(2n) + 12*n + 3. (End)
From Bruno Berselli, Aug 26 2013: (Start)
G.f.: (1 + x + 2*x^2 + 2*x^3 + 5*x^4 + x^5) / (1-x^2)^3.
a(n) = 3*a(n-2) -3*a(n-4) +a(n-6).
a(n) = (n^2+1)*(3+(-1)^n)/4. (End)
From Peter Bala, Feb 14 2019: (Start)
a(n) = numerator((n^2 + 1)/(n + 1)).
a(n) is a quasi-polynomial in n: a(2*n) = 4*n^2 + 1; a(2*n + 1) = 2*n^2 + 2*n + 1. (End)
From Amiram Eldar, Aug 09 2022: (Start)
a(n) = numerator((n^2+1)/2).
Sum_{n>=0} 1/a(n) = (1 + coth(Pi/2)*Pi/2 + tanh(Pi/2)*Pi)/2. (End)
E.g.f.: ((2 + x + 2*x^2)*cosh(x) + (1 + x)^2*sinh(x))/2. - Stefano Spezia, Aug 04 2025

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

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

A301631 Numerator of population variance of n-th row of Pascal's triangle.

Original entry on oeis.org

0, 0, 2, 1, 94, 122, 2372, 173, 50294, 56014, 983740, 266930, 18376812, 19624884, 333313544, 5500541, 5923399334, 6206260694, 103708093964, 27001710566, 1795265477444, 1860906681644, 30802090121144, 1988024895074, 524715115366844, 540193965134732, 8886200762228312

Offset: 0

N. J. A. Sloane and Chai Wah Wu, Mar 24 2018

Denominator of population variance of n-th row of Pascal's triangle is A191871(n+1) = A000265(n+1)^2.

			The first few population variances are 0, 0, 2/9, 1, 94/25, 122/9, 2372/49, 173, 50294/81, 56014/25, 983740/121, 266930/9, 18376812/169, 19624884/49, 333313544/225, 5500541, 5923399334/289, ...

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.

Cf. A000108, A075101, A084623, A000265, A191871 (denominators), A301278, A301279, A054650, A301280.

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

from fractions import Fraction
from sympy import binomial
def A301631(n):
    return (Fraction(int(binomial(2*n,n)))/(n+1) - Fraction(4**n)/(n+1)**2).numerator

a(n) = numerator of binomial(2n,n)/(n+1) - 4^n/(n+1)^2.

a(n) = A000108(n)*A000265(n+1)^2 - A075101(n+1)^2/4.

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A061049 Numerator of 1/64 - 1/n^2.

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A228564 Largest odd divisor of n^2 + 1.

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

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

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A301631 Numerator of population variance of n-th row of Pascal's triangle.

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