A100044 -id:A100044 - cp's OEIS frontend

A019670 Decimal expansion of Pi/3.

1, 0, 4, 7, 1, 9, 7, 5, 5, 1, 1, 9, 6, 5, 9, 7, 7, 4, 6, 1, 5, 4, 2, 1, 4, 4, 6, 1, 0, 9, 3, 1, 6, 7, 6, 2, 8, 0, 6, 5, 7, 2, 3, 1, 3, 3, 1, 2, 5, 0, 3, 5, 2, 7, 3, 6, 5, 8, 3, 1, 4, 8, 6, 4, 1, 0, 2, 6, 0, 5, 4, 6, 8, 7, 6, 2, 0, 6, 9, 6, 6, 6, 2, 0, 9, 3, 4, 4, 9, 4, 1, 7, 8, 0, 7, 0, 5, 6, 8

Offset: 1

N. J. A. Sloane, Dec 11 1996

With an offset of zero, also the decimal expansion of Pi/30 ~ 0.104719... which is the average arithmetic area  of the 0-winding sectors enclosed by a closed Brownian planar path, of a given length t, according to Desbois, p. 1. - Jonathan Vos Post, Jan 23 2011
Polar angle (or apex angle) of the cone that subtends exactly one quarter of the full solid angle. See comments in A238238. - Stanislav Sykora, Jun 07 2014
60 degrees in radians. - M. F. Hasler, Jul 08 2016
Volume of a quarter sphere of radius 1. - Omar E. Pol, Aug 17 2019
Also smallest positive zero of Sum_{k>=1} cos(k*x)/k = -log(2*|sin(x/2)|). Proof of this identity: Sum_{k>=1} cos(k*x)/k = Re(Sum_{k>=1} exp(k*x*i)/k) = Re(-log(1-exp(x*i))) = -log(2*|sin(x/2)|), x != 2*m*Pi, where i = sqrt(-1). - Jianing Song, Nov 09 2019
The area of a circle circumscribing a unit-area regular dodecagon. - Amiram Eldar, Nov 05 2020

			Pi/3 = 1.04719755119659774615421446109316762806572313312503527365831486...
From _Peter Bala_, Nov 16 2016: (Start)
Case n = 1. Pi/3 = 18 * Sum_{k >= 0} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ).
Using the methods of Borwein et al. we can find the following asymptotic expansion for the tails of this series: for N divisible by 6 there holds Sum_{k >= N/6} (-1)^(k+1)( 1/((6*k - 5)*(6*k + 1)*(6*k + 7)) + 1/((6*k - 1)*(6*k + 5)*(6*k + 11)) ) ~ 1/N^3 + 6/N^5 + 1671/N ^7 - 241604/N^9 + ..., where the sequence [1, 0, 6, 0, 1671, 0, -241604, 0, ...] is the sequence of coefficients in the expansion of ((1/18)*cosh(2*x)/cosh(3*x)) * sinh(3*x)^2 = x^2/2! + 6*x^4/4! + 1671*x^6/6! - 241604*x^8/8! + .... Cf. A024235, A278080 and A278195. (End)

Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.3, p. 489.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
Kunle Adegoke, Infinite arctangent sums involving Fibonacci and Lucas numbers, arXiv:1603.08097 [math.NT], 2016.
J. M. Borwein, P. B. Borwein, and K. Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Jean Desbois and Stéphane Ouvry, Algebraic and arithmetic area for m planar Brownian paths, arXiv:1101.4135 [math-ph], Jan 21, 2011.
Index entries for transcendental numbers.

Cf. A000796 (Pi), A019669 (Pi/2), A019692 (2*Pi), A122952 (3*Pi), A003881(Pi/4), A137914, A238238, A002388, A091925, A093954, A016910, A019694, A136017, A352324.

Integral_{x=0..oo} 1/(1+x^m) dx: A013661 (m=2), A248897 (m=3), A093954 (m=4), A352324 (m=5), this sequence (m=6), A352125 (m=8), A094888 (m=10).

RealDigits[N[Pi/3,6! ]] (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)

Pi/3 \\ Charles R Greathouse IV, Sep 08 2011

N=150; default(realprecision, 3+N); digits(Pi*10^N\3) \\ M. F. Hasler, Jul 08 2016

A third of A000796, a sixth of A019692, the square root of A100044.
Sum_{k >= 0} (-1)^k/(6k+1) + (-1)^k/(6k+5). - Charles R Greathouse IV, Sep 08 2011
Product_{k >= 1}(1-(6k)^(-2))^(-1). - Fred Daniel Kline, May 30 2013
From Peter Bala, Feb 05 2015: (Start)
Pi/3 = Sum {k >= 0} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k = 2F1(1/2,1/2;3/2;1/4). Similar series expansions hold for Pi^2 (A002388), Pi^3 (A091925) and Pi/(2*sqrt(2)) (A093954.)
The integer sequences A(n) := 4^n*(2*n + 1)! and B(n) := A(n)*( Sum {k = 0..n} binomial(2*k,k)*1/(2*k + 1)*(1/16)^k ) both satisfy the second-order recurrence equation u(n) = (20*n^2 + 4*n + 1)*u(n-1) - 8*(n - 1)*(2*n - 1)^3*u(n-2). From this observation we can obtain the continued fraction expansion Pi/3 = 1 + 1/(24 - 8*3^3/(89 - 8*2*5^3/(193 - 8*3*7^3/(337 - ... - 8*(n - 1)*(2*n - 1)^3/((20*n^2 + 4*n + 1) - ... ))))). Cf. A002388 and A093954. (End)
Equals Sum_{k >= 1} arctan(sqrt(3)*L(2k)/L(4k)) where L=A000032. See also A005248 and A056854. - Michel Marcus, Mar 29 2016
Equals Product_{n >= 1} A016910(n) / A136017(n). - Fred Daniel Kline, Jun 09 2016
Equals Integral_{x=-oo..oo} sech(x)/3 dx. - Ilya Gutkovskiy, Jun 09 2016
From Peter Bala, Nov 16 2016: (Start)
Euler's series transformation applied to the series representation Pi/3 = Sum_{k >= 0} (-1)^k/(6*k + 1) + (-1)^k/(6*k + 5) given above by Greathouse produces the faster converging series Pi/3 = (1/2) * Sum_{n >= 0} 3^n*n!*( 1/(Product_{k = 0..n} (6*k + 1)) + 1/(Product_{k = 0..n} (6*k + 5)) ).
The series given above by Greathouse is the case n = 0 of the more general result Pi/3 = 9^n*(2*n)! * Sum_{k >= 0} (-1)^(k+n)*( 1/(Product_{j = -n..n} (6*k + 1 + 6*j)) + 1/(Product_{j = -n..n} (6*k + 5 + 6*j)) ) for n = 0,1,2,.... Cf. A003881. See the example section for notes on the case n = 1.(End)
Equals Product_{p>=5, p prime} p/sqrt(p^2-1). - Dimitris Valianatos, May 13 2017
Equals A019699/4 or A019693/2. - Omar E. Pol, Aug 17 2019
Equals Integral_{x >= 0} (sin(x)/x)^4 = 1/2 + Sum_{n >= 0} (sin(n)/n)^4, by the Abel-Plana formula. - Peter Bala, Nov 05 2019
Equals Integral_{x=0..oo} 1/(1 + x^6) dx. - Bernard Schott, Mar 12 2022
Pi/3 = -Sum_{n >= 1} i/(n*P(n, 1/sqrt(-3))*P(n-1, 1/sqrt(-3))), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The first twenty terms of the series gives the approximation Pi/3 = 1.04719755(06...) correct to 8 decimal places. - Peter Bala, Mar 16 2024
Equals Integral_{x >= 0} (2*x^2 + 1)/((x^2 + 1)*(4*x^2 + 1)) dx. - Peter Bala, Feb 12 2025

A002544 a(n) = binomial(2n+1,n)(n+1)^2.

Original entry on oeis.org

1, 12, 90, 560, 3150, 16632, 84084, 411840, 1969110, 9237800, 42678636, 194699232, 878850700, 3931426800, 17450721000, 76938289920, 337206098790, 1470171918600, 6379820115900, 27569305764000, 118685861314020, 509191949220240, 2177742427450200, 9287309860732800

Offset: 0

N. J. A. Sloane, Simon Plouffe

Coefficients for numerical differentiation.
Take the first n integers 1,2,3..n and find all combinations with repetitions allowed for the first n of them. Find the sum of each of these combinations to get this sequence. Example for 1 and 2: 1,2,1+1,1+2,2+2 gives sum of 12=a(2). - J. M. Bergot, Mar 08 2016
Let cos(x) = 1 -x^2/2 +x^4/4!-x^6/6!.. = Sum_i (-1)^i x^(2i)/(2i)! be the standard power series of the cosine, and y = 2*(1-cos(x)) = 4*sin^2(x/2) = x^2 -x^4/12 +x^6/360 ...= Sum_i 2*(-1)^(i+1) x^(2i)/(2i)! be a closely related series. Then this sequence represents the reversion x^2 = Sum_i 1/a(i) *y^(i+1). - R. J. Mathar, May 03 2022

C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=0..200
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table, Phil. Mag., 33 (1942), 1-12 (plus tables) [Annotated scanned copy]
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
A. Petojevic and N. Dapic, The vAm(a,b,c;z) function, Preprint 2013.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135. [Annotated scanned copy]
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)
R. Shenton and A. W. Kemp, An S-fraction and ln^2(1+x), Journal of Computational and Applied Mathematics, 26 (1989) 367-370 North-Holland.
T. R. Van Oppolzer, Lehrbuch zur Bahnbestimmung der Kometen und Planeten, Vol. 2, Engelmann, Leipzig, 1880, p. 21.
Mats Vermeeren, A dynamical solution to the Basel problem, arXiv preprint arXiv:1506.05288 [math.CA], 2015.

Cf. A085373, A002457, A002674, A202543.
Equals A002736/2.
A diagonal of A331430.

seq((n+1)^2*(binomial(2*n+2, n+1))/2, n=0..29); # Zerinvary Lajos, May 31 2006

Table[Binomial[2n+1,n](n+1)^2,{n,0,20}] (* Harvey P. Dale, Mar 23 2011 *)

PARI
```
a(n)=binomial(2*n+1,n)*(n+1)^2
```

x='x+O('x^99); Vec((1+2*x)/(1-4*x)^(5/2)) \\ Altug Alkan, Jul 09 2016

from sympy import binomial
def a(n): return binomial(2*n + 1, n)*(n + 1)**2 # Indranil Ghosh, Apr 18 2017

G.f.: (1 + 2x)/(1 - 4x)^(5/2).
a(n-1) = sum(i_1 + i_2 + ... + i_n) where the sum is over 0 <= i_1 <= i_2 <= ... <= i_n <= n; a(n) = (n+1)^2 C(2n+1, n). - David Callan, Nov 20 2003
a(n) = (n+1)^2 * binomial(2*n+2,n+1)/2. - Zerinvary Lajos, May 31 2006
Asymptotics: a(n)-> (1/64) * (128*n^2+176*n+41) * 4^n * n^(-1/2)/(sqrt(Pi)), for n->infinity. - Karol A. Penson, Aug 05 2013
G.f.: 2F1(3/2,2;1;4x). - R. J. Mathar, Aug 09 2015
a(n) = A002457(n)*(n+1). - R. J. Mathar, Aug 09 2015
a(n) = A000217(n)*A000984(n). - J. M. Bergot, Mar 10 2016
a(n-1) = A001791(n)*n*(n+1)/2. - Anton Zakharov, Jul 04 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
E.g.f.: ((1 + 2*x)*(1 + 8*x)*BesselI(0,2*x) + 2*x*(3 + 8*x)*BesselI(1,2*x))*exp(2*x).
Sum_{n>=0} 1/a(n) = Pi^2/9 = A100044. (End)
From Peter Bala, Apr 18 2017: (Start)
With x = y^2/(1 + y) we have log^2(1 + y) = Sum_{n >= 0} (-1)^n*x^(n+1)/a(n). See Shenton and Kemp.
Series reversion ( Sum_{n >= 0} (-1)^n*x^(n+1)/a(n) ) = Sum_{n >= 1} 2*x^n/(2*n)! = Sum_{n >= 1} x^n/A002674(n). (End)
D-finite with recurrence n^2*a(n) -2*(n+1)*(2*n+1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)^2 = A202543^2. - Amiram Eldar, May 14 2022

A214549 Decimal expansion of 4*Pi^2/27.

Original entry on oeis.org

1, 4, 6, 2, 1, 6, 3, 6, 1, 4, 9, 7, 6, 2, 0, 1, 2, 7, 6, 8, 6, 4, 3, 6, 9, 0, 3, 7, 0, 1, 8, 6, 8, 9, 0, 5, 7, 0, 8, 3, 5, 1, 1, 0, 2, 3, 2, 9, 4, 9, 3, 1, 9, 4, 4, 6, 5, 3, 8, 2, 9, 5, 3, 7, 2, 1, 7, 7, 8, 4, 4, 1, 8, 1, 3, 6, 1, 7, 8, 5, 5, 4, 5, 1, 8, 7, 8, 1, 2, 4, 4, 9, 9

Offset: 1

R. J. Mathar, Jul 20 2012

Represents the value of the Dirichlet series for A011655 (principal Dirichlet character mod 3) at s=2.

Equals the asymptotic mean of the abundancy index of the numbers that are not divisible by 3 (A001651). - Amiram Eldar, May 12 2023

			1.4621636149762012768643690370186...

G. C. Greubel, Table of n, a(n) for n = 1..10000
R. J. Mathar, Table of Dirichlet L-Series, arXiv:1008.2547 [math.NT], 2010-2015, Table 22.
Index entries for transcendental numbers.

Cf. A001651, A011655, A100044.

using Nemo
R = RealField(310)
t = const_pi(RR) + const_pi(RR); s = t * t
s / RR(27) |> println # Peter Luschny, Mar 13 2018

R:= RealField(); 4*Pi(R)^2/27; // G. C. Greubel, Mar 08 2018

R:=RealField(106); SetDefaultRealField(R); n:=4*Pi(R)^2/27; Reverse(Intseq(Floor(10^105*n))); // Bruno Berselli, Mar 13 2018

Maple
```
evalf(4*Pi^2/27) ;
```

RealDigits[(4Pi^2)/27,10,120][[1]] (* Harvey P. Dale, Dec 20 2012 *)

4*Pi^2/27 \\ G. C. Greubel, Mar 08 2018

Equals (4/3)*A100044.
Equals Sum_{n>=0} (1/(3*n+1)^2 + 1/(3*n+2)^2).
From Peter Luschny, May 13 2020: (Start)
Equals (8/9) * Sum_(k>=1) 1/k^2 =8/9 *A013661.
Equals -(16/9) * Sum_(k>=1) (-1)^k/k^2 = -16/9 * A072691.
Equals (64/27) * ( Integral_{x=0..1} sqrt(1 - x^2) )^2 = 64/27 * A091476. (End)
Equals Integral_{x=0..oo} log(x)/(x^3 - 1) dx. - Amiram Eldar, Aug 12 2020

A120325 Period 6: repeat [0, 0, 1, 0, 1, 0].

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0

Offset: 0

Paolo P. Lava and Giorgio Balzarotti, Jun 21 2006

Dirichlet series for the principal character mod 6: L(s,chi) = Sum_{n>=1} a(n+3)/n^s = (1 + 1/6^s - 1/2^s - 1/3^s) Riemann-zeta(s), e.g., L(2,chi) = A100044, L(4,chi) = 5*Pi^4/486, L(6,chi) = 91*Pi^6/87480. See Jolley eq (313) and arXiv:1008.2547 L(m=6,r=1,s). - R. J. Mathar, Jul 31 2010

			a(0) = (1/3)*(sin(0) + sin(0))^2 = 0.
a(1) = (1/3)*(sin(Pi/6) + sin(7*Pi/6))^2 = (1/3)*(1/2 - 1/2)^2 = 0.
a(2) = (1/3)*(sin(Pi/3) + sin(7*Pi/3))^2 = (1/3)*((sqrt(3))/2 + (sqrt(3))/2)^2 = 1.
a(3) = (1/3)*(sin(Pi/2) + sin(7*Pi/2))^2 = (1/3)*(1 - 1)^2 = 0.
a(4) = (1/3)*(sin(2*Pi/3) + sin(14*Pi/3))^2 = (1/3)*((sqrt(3))/2 + (sqrt(3))/2)^2 = 1.
a(5) = (1/3)*(sin(5*Pi/6) + sin(35*Pi/6))^2 = (1/3)*(1/2 - 1/2)^2 = 0.

L. B. W. Jolley, Summation of Series, Dover (1961).

Antti Karttunen, Table of n, a(n) for n = 0..100000
R. J. Mathar, Table of Dirichlet L-Series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015.
Index entries for characteristic functions
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Characteristic function of A047235. One's complement of A093719.

Cf. A100044, A059841, A276086, A358841, A358842, A358754, A358755.

[(n+3)^2 mod (2+((n+1) mod 2)) : n in [0..100]]; // Wesley Ivan Hurt, Oct 31 2014

P:=proc(n)local i,j; for i from 0 by 1 to n do j:=1/3*(sin(i*Pi/6)+sin(7*i*Pi/6))^2; print(j); od; end: P(20);
seq(abs(numtheory[jacobi](n,6)),n=3..150) ; # R. J. Mathar, Jul 31 2010

Table[Mod[(n + 3)^2, (5 + (-1)^n)/2], {n, 0, 100}] (* Wesley Ivan Hurt, Oct 31 2014 *)
PadRight[{},120,{0,0,1,0,1,0}] (* Harvey P. Dale, Oct 05 2016 *)

A120325(n) = ((n%3)&&!(n%2)); \\ Antti Karttunen, Dec 03 2022

def A120325(n): return int(not (n+3) % 6 & 3 ^ 1) # Chai Wah Wu, May 25 2022

a(n) = (1/3)*(sin(n*Pi/6) + sin(7*n*Pi/6))^2.
From R. J. Mathar, Nov 22 2008: (Start)
G.f.: x^2*(1+x^2)/((1+x)*(1-x)*(1+x+x^2)*(1-x+x^2)).
a(n+6) = a(n). (End)
a(n) = ((n+3)*Fibonacci(n+3)) mod 2. - Gary Detlefs, Dec 13 2010
a(n) = 0 if n mod 6 = 0, otherwise a(n) = n mod 2 + (-1)^n. - Gary Detlefs, Dec 13 2010
a(n) = (n+3)^2 mod (5+(-1)^n)/2. - Wesley Ivan Hurt, Oct 31 2014
a(n) = sin(n*Pi/3)^2*(2-4*cos(n*Pi/3))/3. - Wesley Ivan Hurt, Jun 19 2016
E.g.f.: 2*(cosh(x) - cos(sqrt(3)*x/2)*cosh(x/2))/3. - Ilya Gutkovskiy, Jun 20 2016
a(n) = sign((n-3) mod 2) * sign((n-3) mod 3). - Wesley Ivan Hurt, Feb 04 2022
From Antti Karttunen, Dec 03 2022: (Start)
a(n) = 1 - A093719(n).
a(n) = [A276086(n) == 3 (mod 6)], where [ ] is the Iverson bracket.
a(n) = A059841(n) - A358841(n) - A358842(n).
For n >= 1, a(n) = A059841(n) - A358754(n) - A358755(n).
(End)

Data section extended up to a(120) by Antti Karttunen, Dec 03 2022

A353908 Decimal expansion of Pi^2/36.

Original entry on oeis.org

2, 7, 4, 1, 5, 5, 6, 7, 7, 8, 0, 8, 0, 3, 7, 7, 3, 9, 4, 1, 2, 0, 6, 9, 1, 9, 4, 4, 4, 1, 0, 0, 4, 1, 9, 8, 2, 0, 3, 1, 5, 8, 3, 1, 6, 8, 6, 7, 7, 9, 9, 7, 3, 9, 6, 2, 2, 5, 9, 3, 0, 3, 8, 2, 2, 8, 3, 3, 4, 5, 7, 8, 4, 0, 0, 5, 3, 3, 4, 7, 8, 9, 7, 2, 2, 7, 1, 4, 8, 3, 4, 3, 6, 6, 2, 6, 4, 5, 0, 8, 8, 4, 0, 0, 0, 7

Offset: 0

Omar E. Pol, May 10 2022

Ratio between the volume of the stepped pyramid with an infinite number of levels described in A245092 and that of the circumscribed cube (see the first formula).
See also Vaclav Kotesovec's formula (2016) in A175254.
Volume shared by a sphere inscribed in a cube of volume Pi and one of the six pyramids inscribed in the cube. - Omar E. Pol, Sep 01 2024

			0.2741556778080377394120691944410041982031583168677997396225930382283345784...

Index entries for transcendental numbers

Cf. A000203, A000796, A002388, A013661, A019673, A021040, A024916, A072691, A086463, A086729, A091476, A100044, A102753, A175254, A195055, A237271, A237593, A245092.

evalf(Pi^2/36, 121);  # Alois P. Heinz, May 11 2022

RealDigits[Pi^2/36, 10, 100][[1]] (* Amiram Eldar, May 11 2022 *)

PARI
```
Pi^2/36
```
PARI
```
zeta(2)/6
```

Equals lim_{n->oo} A175254(n)/n^3.
Equals A002388/36.
Equals A102753/18.
Equals A195055/12.
Equals A091476/9.
Equals A013661/6.
Equals A100044/4.
Equals A072691/3.
Equals A086463/2.
Equals A086729*2.
Equals A019673^2.
Equals A002388*A021040.
Equals Re(dilog((1+sqrt(3)*i)/2)). - Mohammed Yaseen, Jul 03 2024

A104777 Integer squares congruent to 1 mod 6.

Original entry on oeis.org

1, 25, 49, 121, 169, 289, 361, 529, 625, 841, 961, 1225, 1369, 1681, 1849, 2209, 2401, 2809, 3025, 3481, 3721, 4225, 4489, 5041, 5329, 5929, 6241, 6889, 7225, 7921, 8281, 9025, 9409, 10201, 10609, 11449, 11881, 12769, 13225, 14161, 14641, 15625, 16129

Offset: 1

Michael Somos, Mar 24 2005

Exponents of powers of q in expansion of eta(q^24).
Odd squares not divisible by 3. - Reinhard Zumkeller, Nov 14 2015
From Peter Bala, Jan 03 2025: (Start)
Exponents of q in the expansion of q*Product_{n >= 1} (1 - q^(24*n))^5/(1 - q^(48*n))^2 = q - 5*q^(5^2) + 7*q^(7^2) - 11*q^(11^2) + 13*q^(13^2) - 17*q^(17^2) + 19*q^(19)^2 - + ... (a consequence of the quintuple product identity).
Also, exponents in the expansion of q*Product_{n >= 1} (1 - q^(48*n))^13 / ( (1 - q^(24*n))*(1 - q^(96*n)) )^5 = q + 5*q^(5^2) + 7*q^(7^2) + 11*q^(11^2) - 13*q^(13^2) - 17*q^(17^2) - 19*q^(19^2) - 23*q^(23^2) + + + + - - - - ... (see Oliver, Theorem 1.1). (End)

			eta(q^24) = q - q^25 - q^49 + q^121 + q^169 - q^289 - q^361 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Robert J. Lemke Oliver, Eta quotients and theta functions, Advances in Mathematics, Vol. 241, Jul. 2013, pp. 1-17.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1)

Disjoint union of A016922 and A016970.

Cf. A007310, A001318, A033683, A100044.

a104777 = (^ 2) . a007310  -- Reinhard Zumkeller, Nov 14 2015

seq(9*(n-1/2)^2 + 1/4 + (-1)^n * (3*n - 3/2), n = 1 .. 100); # Robert Israel, Dec 12 2014

Select[Range[130]^2,Mod[#,6]==1&] (* or *) LinearRecurrence[{1,2,-2,-1,1},{1,25,49,121,169},50] (* Harvey P. Dale, Mar 09 2017 *)

PARI
```
{a(n) = (3*n - 1 - n%2)^2};
```

A033683(a(n)) = 1.
G.f.: ( -1-24*x-22*x^2-24*x^3-x^4 ) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Feb 20 2011
a(n) = A007310(n)^2 = 1 + 24*A001318(n-1).
a(n) = 9*n^2 - 9*n + 5/2 + (-1)^n * (3*n - 3/2).  a(n+4) = 2*a(n+2) - a(n) + 72. - Robert Israel, Dec 12 2014
a(n) == 1 (mod 24). - Joerg Arndt, Jan 03 2017
Sum_{n>=1} 1/a(n) = Pi^2/9 (A100044). - Amiram Eldar, Dec 19 2020

A352475 Numbers m such that gcd(d(m),6) = 1.

Original entry on oeis.org

1, 16, 64, 81, 625, 729, 1024, 1296, 2401, 4096, 5184, 10000, 11664, 14641, 15625, 28561, 38416, 40000, 46656, 50625, 59049, 65536, 82944, 83521, 117649, 130321, 153664, 194481, 234256, 250000, 262144, 279841, 331776, 455625, 456976, 531441, 640000, 707281, 746496

Offset: 1

Michael De Vlieger, Mar 26 2022

All terms are square since numbers coprime to 6 are odd.
The square roots of terms are in A001694.
Intersection of A000290 and A336590, i.e., numbers whose prime factorization has only exponents that are congruent to {0, 4} mod 6 (A047233). - Amiram Eldar, Mar 31 2022

Amiram Eldar, Table of n, a(n) for n = 1..10000
Titus Hilberdink, How often is d(n) a power of a given integer?, Journal of Number Theory, Vol. 236 (2022), pp. 261-279.

Cf. A000005, A000290, A001694, A007310, A047233, A076400, A100044, A336590, A350014.

Select[Range[864]^2, GCD[DivisorSigma[0, #], 6] == 1 &] (* or, more efficiently, *)
With[{nn = 864}, Select[Union[Flatten@ Table[a^2*b^3, {b, nn^(1/3)}, {a, Sqrt[nn/b^3]}]]^2, Mod[DivisorSigma[0, #], 3] > 0 &]]

isok(m) = gcd(numdiv(m), 6) == 1; \\ Michel Marcus, Mar 29 2022

m = 100000; seq = direuler(p=2, m, (1 - X^8)/(1 - X^4)/(1 - X^6)); for(n=1, m, if(seq[n] != 0, print1(n, ", "))) \\ Vaclav Kotesovec, May 19 2022

a(n) = A350014(n)^2.
Sum_{n>=1} 1/a(n) = Pi^2/9 (A100044). - Amiram Eldar, Mar 31 2022
The number of terms <= x is (zeta(3/2)/zeta(2))*x^(1/4) + (zeta(2/3)/zeta(4/3))*x^(1/6) + O(x^(1/8 + eps)), for all eps > 0 (Hilberdink, 2022). - Amiram Eldar, May 18 2022

A185717 Expansion of q^(-1) * c(q^2) * (c(q) - c(q^4)) / 9 in powers of q^2 where c() is a cubic AGM theta function.

Original entry on oeis.org

1, 3, 6, 8, 9, 12, 14, 18, 18, 20, 24, 24, 31, 27, 30, 32, 36, 48, 38, 42, 42, 44, 54, 48, 57, 54, 54, 72, 60, 60, 62, 72, 84, 68, 72, 72, 74, 93, 96, 80, 81, 84, 108, 90, 90, 112, 96, 120, 98, 108, 102, 104, 144, 108, 110, 114, 114, 144, 126, 144, 133, 126, 156, 128

Offset: 0

Michael Somos, Feb 10 2011

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

			1 + 3*x + 6*x^2 + 8*x^3 + 9*x^4 + 12*x^5 + 14*x^6 + 18*x^7 + 18*x^8 + ...
q + 3*q^3 + 6*q^5 + 8*q^7 + 9*q^9 + 12*q^11 + 14*q^13 + 18*q^15 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Somos, Introduction to Ramanujan theta functions, 2019.
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions.

Cf. A078708, A100044, A118271, A121443, A124449.

Cf. A000122, A000700, A010054, A121373.

A185717[n_] := SeriesCoefficient[(QPochhammer[q^3, q^3]/QPochhammer[-q^3, q^3])^4*(1/(QPochhammer[q, q^2]*QPochhammer[q^3, q^6])^3), {q, 0, n}];
Table[A185717[n], {n, 0, 50}] (* G. C. Greubel, Jul 10 2017 *)

{a(n) = if( n<0, 0, n = 2*n + 1; sumdiv( n, d, if( (n/d) % 3, 1, 0) * d))}

{a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^3 * eta(x^3 + A)^5 / (eta(x + A)^3 * eta(x^6 + A)), n))}

Expansion of phi(-x^3)^4 / (chi(-x) * chi(-x^3))^3 in powers of x where phi(), chi() are Ramanujan theta functions.
Euler transform of period 6 sequence [ 3, 0, -2, 0, 3, -4, ...].
a(n) = b(2*n + 1) where b(n) is multiplicative with b(2^e) = 0^e, b(3^e) = 3^e, b(p^e) = (p^(e+1) - 1) / (p - 1) if p>3.
G.f. is a period 1 Fourier series which satisfies f(-1 / (12 t)) = 2 (t/i)^2 g(t) where q = exp(2 Pi i t) and g() is g.f. for A118271.
a(3*n + 1) = 3 * a(n). A078708(2*n + 1) = A121443(2*n + 1) = A124449(2*n + 1).
Sum_{k=1..n} a(k) ~ c * n^2, where c = Pi^2/9 = 1.0966227... (A100044). - Amiram Eldar, Dec 28 2023

A195059 Decimal expansion of Pi^2/13.

Original entry on oeis.org

7, 5, 9, 2, 0, 0, 3, 3, 8, 5, 4, 5, 3, 3, 5, 2, 7, 8, 3, 7, 1, 8, 8, 3, 9, 2, 3, 0, 6, 7, 3, 9, 6, 2, 4, 1, 1, 7, 7, 9, 7, 6, 8, 7, 7, 4, 8, 0, 0, 6, 0, 8, 1, 7, 4, 1, 6, 4, 1, 1, 4, 9, 0, 4, 7, 8, 4, 6, 4, 9, 8, 6, 3, 3, 9, 9, 3, 8, 8, 6, 4, 8, 4, 6, 2, 9, 0, 2, 6, 1, 8, 2, 4, 5, 0, 4, 0, 1, 7, 8, 3, 2, 6, 1, 7

Offset: 0

Omar E. Pol, Oct 04 2011

			0.7592003385453352783718839230673962411780...

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for transcendental numbers

Cf. A002388, A013661, A019680, A072691, A100044, A195057.

Pi(RealField(130))^2/13; // G. C. Greubel, Jun 03 2021

RealDigits[Pi^2/13, 10, 105][[1]] (* T. D. Noe, Oct 05 2011 *)

Pi^2/13 \\ Charles R Greathouse IV, Oct 01 2022

numerical_approx(pi^2/13, digits=130) # G. C. Greubel, Jun 03 2021

Extended by T. D. Noe, Oct 05 2011

A214552 Decimal expansion of the Dirichlet L-series of the non-principal character mod 6 evaluated at s=2.

Original entry on oeis.org

9, 7, 6, 6, 2, 8, 0, 1, 6, 1, 2, 0, 6, 0, 7, 8, 7, 1, 0, 8, 3, 9, 8, 4, 2, 8, 7, 0, 3, 0, 1, 1, 5, 4, 4, 5, 4, 5, 6, 4, 1, 7, 9, 2, 0, 6, 8, 1, 6, 0, 6, 7, 7, 5, 2, 7, 7, 6, 2, 5, 0, 7, 8, 7, 0, 8, 6, 0, 8, 7, 3, 0, 8, 1, 4, 5, 2, 2, 7, 7, 2, 6, 1, 6, 0, 8, 6, 9, 6, 3, 5, 4, 0, 2, 6, 2, 3, 2, 6, 2, 7, 6, 3, 0, 2

Offset: 0

R. J. Mathar, Jul 20 2012

The non-principal character is A134667. The constant is sum_{n>=1} A134667(n)/n^s with s=2.

			0.97662801612060787108398...= 1/1^2 -1/5^2 +1/7^2 -1/11^2 + 1/13^2 -1/17^2 +-...

R. J. Mathar, Table of Dirichlet L-series and prime zeta modulo functions for small moduli, arXiv:10008.2547 [math.NT], 2010-2015, Table in section 2.2, value at m=6, r=2, s=2.

Cf. A086724, A086722, A100044.

Maple
```
evalf( (Psi(1,1/6)-Psi(1,5/6))/36) ;
```

RealDigits[ (PolyGamma[1, 1/6] - PolyGamma[1, 5/6])/36, 10, 105] // First  (* Jean-François Alcover, Feb 11 2013, after R. J. Mathar *)

Equals 2/3*4F3(1/2,1,1,2; 5/4,3/2,7/4; 3/4), where 4F3 is the generalized hypergeometric function. - Jean-François Alcover, Dec 16 2014, after R. J. Mathar.

Equals A173973 / 3.6 . - R. J. Mathar, Jun 02 2016

More terms from Jean-François Alcover, Feb 11 2013

cp's OEIS Frontend

A019670 Decimal expansion of Pi/3.

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A002544 a(n) = binomial(2*n+1,n)*(n+1)^2.

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A214549 Decimal expansion of 4*Pi^2/27.

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A120325 Period 6: repeat [0, 0, 1, 0, 1, 0].

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A353908 Decimal expansion of Pi^2/36.

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A104777 Integer squares congruent to 1 mod 6.

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A002544 a(n) = binomial(2n+1,n)(n+1)^2.