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.

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A259151 Decimal expansion of phi(exp(-8*Pi)), where phi(q) = Product_{n>=1} (1-q^n) is the Euler modular function.

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Jun 19 2015

Keywords

Examples

			0.999999999987838443290442788140998270959486945673852198543872725583699...
		

Crossrefs

Cf. A048651 phi(1/2), A100220 phi(1/3), A100221 phi(1/4), A100222 phi(1/5), A132034 phi(1/6), A132035 phi(1/7), A132036 phi(1/8), A132037 phi(1/9), A132038 phi(1/10), A292862 phi(exp(-Pi/8)), A292863 phi(exp(-Pi/4)), A259147 phi(exp(-Pi/2)), A259148 phi(exp(-Pi)), A259149 phi(exp(-2*Pi)), A292888 phi(exp(-3*Pi)), A259150 phi(exp(-4*Pi)), A292905 phi(exp(-5*Pi)), A363018 phi(exp(-6*Pi)), A363117 phi(exp(-7*Pi)), A363118 phi(exp(-9*Pi)), A363019 phi(exp(-10*Pi)), A363081 phi(exp(-11*Pi)), A363020 phi(exp(-12*Pi)), A363178 phi(exp(-13*Pi)), A363119 phi(exp(-14*Pi)), A363179 phi(exp(-15*Pi)), A292864 phi(exp(-16*Pi)), A363120 phi(exp(-18*Pi)), A363021 phi(exp(-20*Pi)).

Programs

  • Mathematica
    phi[q_] := QPochhammer[q, q]; RealDigits[phi[Exp[-8*Pi]], 10, 103] // First

Formula

phi(q) = QPochhammer(q,q) = (q;q)_infinity.
phi(q) also equals theta'(1, 0, sqrt(q))^(1/3)/(2^(1/3)*q^(1/24)), where theta' is the derivative of the elliptic theta function theta(a,u,q) w.r.t. u.
phi(exp(-8*Pi)) = (sqrt(2) - 1)^(1/4)*exp(Pi/3)*(Gamma(1/4)/(2^(29/16)*Pi^(3/4))).
A259151 = A259147 * exp(5*Pi/16)/2. - Vaclav Kotesovec, Jul 03 2017

A132037 Decimal expansion of Product_{k>0} (1-1/9^k).

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Aug 14 2007

Keywords

Examples

			0.8765603540359642058360198...
		

Crossrefs

Programs

  • Mathematica
    digits = 103; NProduct[1-1/9^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    RealDigits[QPochhammer[1/9], 10, 100][[1]] (* Amiram Eldar, May 09 2023 *)
  • PARI
    prodinf(k=1, 1 - 1/(9^k)) \\ Amiram Eldar, May 09 2023

Formula

Equals exp(-Sum_{n>0} sigma_1(n)/(n*9^n)) = exp(-Sum_{n>0} A000203(n)/(n*9^n)).
Equals Sum_{n>=0} A010815(n)/9^n. - R. J. Mathar, Mar 04 2009
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(Pi/log(3)) * exp(log(3)/12 - Pi^2/(12*log(3))) * Product_{k>=1} (1 - exp(-2*k*Pi^2/log(3))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027877(n). (End)

A079555 Decimal expansion of Product_{k>=1} (1 + 1/2^k) = 2.384231029031371...

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Jan 25 2003

Keywords

Examples

			2.38423102903137172414989928867839723877161951650843345769...
		

Crossrefs

Programs

  • Mathematica
    digits = 105; NProduct[(1 + 1/2^k), {k, 1, Infinity}, WorkingPrecision -> digits+10, NProductFactors -> 200] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 14 2013 *)
    N[QPochhammer[-1/2,1/2]] (* G. C. Greubel, Dec 05 2015 *)
    1/N[QPochhammer[1/2, 1/4]] (* Gleb Koloskov, Apr 04 2021 *)
  • PARI
    prodinf(n=1,1+2.^-n) \\ Charles R Greathouse IV, May 27 2015
    
  • PARI
    1/prodinf(n=0, 1-2^(-2*n-1)) \\ Gleb Koloskov, Apr 04 2021

Formula

(1/2)*lim sup Product_{k=0..floor(log_2(n)), (1 + 1/floor(n/2^k))} for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132369(n)/A098844(n) for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132269(n)/n^((1+log_2(n))/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132270(n)/n^((log_2(n)-1)/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
exp(sum{n>0, 2^(-n)*sum{k|n, -(-1)^k/k}})=exp(sum{n>0, A000593(n)/(n*2^n)}). - Hieronymus Fischer, Aug 20 2007
(1/2)*lim sup A132269(n+1)/A132269(n)=2.3842310290313717241498992886... for n-->oo. - Hieronymus Fischer, Aug 20 2007
Equals (-1/2; 1/2){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 05 2015
2 + Sum_{k>1} 1/(Product_{i=2..k} (2^i-1)) = 2 + 1/3 + 1/(3*7) + 1/(3*7*15) + 1/(3*7*15*31) + 1/(3*7*15*31*63) + ... (conjecture). - Werner Schulte, Dec 22 2016
From Peter Bala, Dec 15 2020: (Start)
The above conjecture of Schulte follows by setting x = 1/2 and t = -1 in the identity Product_{k >= 1} (1 - t*x^k) = Sum_{n >= 0} (-1)^n*x^(n*(n+1)/2)*t^n/( Product_{k = 1..n} 1 - x^k ), due to Euler.
Constant C = 1 + Sum_{n >= 0} (1/2)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
C = 2 + Sum_{n >= 0} (1/4)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*C = 7 + Sum_{n >= 0} (1/8)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*7*C = 50 + Sum_{n >= 0} (1/16)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
3*7*15*C = 751 + Sum_{n >= 0} (1/32)^(n+1)*Product_{k = 1..n} (1 + 1/2^k).
(End)
Equals 1/(1-P), where P is the Pell constant from A141848. - Gleb Koloskov, Apr 04 2021
Equals Sum_{k>=0} A000009(k)/2^k. - Vaclav Kotesovec, Sep 15 2021
From Amiram Eldar, Feb 19 2022: (Start)
Equals (sqrt(2)/2) * exp(log(2)/24 + Pi^2/(12*log(2))) * Product_{k>=1} (1 - exp(-2*(2*k-1)*Pi^2/log(2))) (McIntosh, 1995).
Equals (1/2) * A081845.
Equals Sum_{n>=0} 1/A005329(n). (End)

A100222 Decimal expansion of Product_{k>=1} (1-1/5^k).

Original entry on oeis.org

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

Views

Author

Eric W. Weisstein, Nov 09 2004

Keywords

Examples

			0.76033279587123242010148829629266515947434392887320...
		

Crossrefs

Programs

  • Mathematica
    (5^(1/24)*EllipticThetaPrime[1, 0, 1/Sqrt[5]]^(1/3))/2^(1/3)
    N[QPochhammer[1/5,1/5]] (* G. C. Greubel, Dec 01 2015 *)
  • PARI
    prodinf(k=1, 1 - 1/(5^k)) \\ Amiram Eldar, May 09 2023

Formula

Equals exp(-Sum_{k>0} sigma_1(k)/(k*5^k)) = exp(-Sum_{k>0} A000203(k)/(k*5^k)). - Hieronymus Fischer, Aug 07 2007
Equals (1/5; 1/5){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 01 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/log(5)) * exp(log(5)/24 - Pi^2/(6*log(5))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/log(5))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027872(n). (End)

A081845 Decimal expansion of Product_{k>=0} (1 + 1/2^k).

Original entry on oeis.org

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

Views

Author

Benoit Cloitre, Apr 09 2003

Keywords

Comments

Twice the product in A079555.

Examples

			4.76846205806274344829979857....
		

Crossrefs

Programs

  • Mathematica
    digits = 105; NProduct[1 + 1/2^k, {k, 0, Infinity}, WorkingPrecision -> digits+5, NProductFactors -> digits] // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Mar 04 2013 *)
    N[QPochhammer[-1, 1/2], 100] (* Vaclav Kotesovec, Dec 13 2015 *)
    2*N[QPochhammer[-1/2, 1/2], 200] (* G. C. Greubel, Dec 20 2015 *)
  • PARI
    prodinf(k=0,1/2^k,1) \\ Hugo Pfoertner, Feb 21 2020

Formula

lim sup Product_{k=0..floor(log_2(n))} (1 + 1/floor(n/2^k)) for n-->oo. - Hieronymus Fischer, Aug 20 2007
lim sup A132369(n)/A098844(n) for n-->oo. - Hieronymus Fischer, Aug 20 2007
lim sup A132269(n)/n^((1+log_2(n))/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
lim sup A132270(n)/n^((log_2(n)-1)/2) for n-->oo. - Hieronymus Fischer, Aug 20 2007
2*exp(Sum_{n>0} 2^(-n)*Sum_{k|n} -(-1)^k/k) = 2*exp(Sum_{n>0} A000593(n)/(n*2^n)). - Hieronymus Fischer, Aug 20 2007
lim sup A132269(n+1)/A132269(n) = 4.76846205806274344... for n-->oo. - Hieronymus Fischer, Aug 20 2007
Sum_{k>=1} (-1)^(k+1) * 2^k / (k*(2^k-1)) = log(A081845) = 1.562023833218500307570359922772014353168080202860122... . - Vaclav Kotesovec, Dec 13 2015
Equals 2*(-1/2; 1/2){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 20 2015
Equals 1 + Sum_{n>=1} 2^n/((2-1)*(2^2-1)*...*(2^n-1)). - Robert FERREOL, Feb 21 2020
From Peter Bala, Jan 18 2021: (Start)
Constant C = 3*Sum_{n >= 0} (1/2)^n/Product_{k = 1..n} (2^k - 1).
Faster converging series:
C = (2*3*5)/(2^3)*Sum_{n >= 0} (1/4)^n/Product_{k = 1..n} (2^k - 1),
C = (2*3*5*9)/(2^6)*Sum_{n >= 0} (1/8)^n/Product_{k = 1..n} (2^k - 1),
C = (2*3*5*9*17)/(2^10)*Sum_{n >= 0} (1/16)^n/Product_{k = 1..n} (2^k - 1), and so on. The sequence [2,3,5,9,17,...] is A000051. (End)
From Amiram Eldar, Mar 20 2022: (Start)
Equals sqrt(2) * exp(log(2)/24 + Pi^2/(12*log(2))) * Product_{k>=1} (1 - exp(-2*(2*k-1)*Pi^2/log(2))) (McIntosh, 1995).
Equals 1/A083864. (End)
Equals lim_{n->oo} A020696(2^n)/A006125(n+1) (Sándor, 2021). - Amiram Eldar, Jun 29 2022

A132328 a(n) = Product_{k>0} (1+floor(n/3^k)).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 3, 3, 8, 8, 8, 10, 10, 10, 12, 12, 12, 21, 21, 21, 24, 24, 24, 27, 27, 27, 80, 80, 80, 88, 88, 88, 96, 96, 96, 130, 130, 130, 140, 140, 140, 150, 150, 150, 192, 192, 192, 204, 204, 204, 216, 216, 216, 399, 399, 399, 420, 420, 420, 441, 441, 441, 528
Offset: 0

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-3 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product (1+d(m)d(m-1)d(m-2)...d(2)d(1))*(1+d(m)d(m-1)d(m-2)...d(2))*...*(1+d(m)d(m-1)d(m-2))*(1+d(m)d(m-1))*(1+d(m)).

Examples

			a(12)=(1+floor(12/3^1))*(1+floor(12/3^2))=5*2=10; a(19)=21 since 19=201(base-3) and so a(19)=(1+20)*(1+2)(base-3)=7*3=21.
		

Crossrefs

For formulas regarding a general parameter p (i.e. terms 1+floor(n/p^k)) see A132272.
For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

Programs

  • Maple
    f:= proc(n) option remember; local t;
      t:= floor(n/3);
      (1+t)*procname(t)
    end proc:
    f(0):= 1: f(1):= 1: f(2):= 1:
    map(f, [$0..100]); # Robert Israel, Oct 20 2020
  • Mathematica
    (* Using definition *)
    Table[Product[1 + Floor[n/3^k], {k, IntegerLength[n, 3] - 1}], {n, 0, 100}]
    (* Using recurrence -- faster *)
    a[0] = 1; a[n_] := a[n] = (1 + #)*a[#] & [Floor[n/3]];
    Table[a[n], {n, 0, 100}] (* Paolo Xausa, Sep 23 2024 *)

Formula

Recurrence: a(n)=(1+floor(n/3))*a(floor(n/3)); a(3n)=(1+n)*a(n); a(n*3^m)=product{0<=k
a(k*3^m-j)=k^m*3^(m(m-1)/2), for 0=1. a(3^m)=p^(m(m-1)/2)*product{0<=k
a(n)=A132327(floor(n/3))=A132327(n)/(1+n).
Asymptotic behavior: a(n)=O(n^((log_3(n)-1)/p)); this follows from the inequalities below.
a(n)<=A132027(n)/(n+1)*product{0<=k<=floor(log_3(n)), 1+1/3^k}.
a(n)>=A132027(n)/((n+1)*product{0
a(n)A000217(log_3(n))/(n+1), where c=product{k>0, 1+1/3^k}=3.12986803713402307587769821345767... (see constant A132323).
a(n)>n^((1+log_3(n))/2)/(n+1)=3^A000217(log_3(n))/(n+1).
lim sup n*a(n)/A132027(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
lim inf n*a(n)/A132027(n)=1/product{k>0, 1-1/3^k}=1/0.560126077927948944969792243314140014..., for n-->oo (see constant A100220).
lim inf a(n)/n^((1+log_3(n))/2)=1, for n-->oo.
lim sup a(n)/n^((1+log_3(n))/2)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
lim inf a(n+1)/a(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767... for n-->oo (see constant A132323).

A003787 Order of universal Chevalley group A_n (3).

Original entry on oeis.org

1, 24, 5616, 12130560, 237783237120, 42064805779476480, 67034222101339041669120, 961721214905722855895197286400, 124190524600592082795473760093457612800, 144339416867688029764487130056208182629053235200
Offset: 0

Keywords

References

  • J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker and R. A. Wilson, ATLAS of Finite Groups. Oxford Univ. Press, 1985 [for best online version see https://oeis.org/wiki/Welcome#Links_to_Other_Sites], p. xvi.
  • H. S. M. Coxeter and W. O. J. Moser, Generators and Relations for Discrete Groups, 4th ed., Springer-Verlag, NY, reprinted 1984, p. 131.

Programs

  • Magma
    [&*[(3^n - 3^k): k in [0..n-1]]/2: n in [1..10]]; // Vincenzo Librandi, Sep 19 2015
  • Mathematica
    f[m_, n_] := m^(n (n + 1)/2) Product[m^k - 1, {k, 2, n + 1}];
    f[3, #] & /@ Range[0, 9] (* Michael De Vlieger, Sep 18 2015 *)

Formula

Numbers so far appear to equal A053290(n)/2. - Ralf Stephan, Mar 30 2004
a(n) = A(3,n) where A(q,n) = q^(n*(n+1)/2) * Product_{k=2..n+1}(q^k-1). - Sean A. Irvine, Sep 18 2015
a(n) ~ c * 3^(n*(n+2)), where c = (3/2) * A100220 = 0.840189116891... . - Amiram Eldar, Jul 07 2025

Extensions

One more term from Sean A. Irvine, Sep 18 2015

A132036 Decimal expansion of Product_{k>0} (1 - 1/8^k).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 14 2007

Keywords

Examples

			0.8594059944007028662007585800...
		

Programs

  • Mathematica
    digits = 103; NProduct[1-1/8^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    N[QPochhammer[1/8,1/8]] (* G. C. Greubel, Nov 26 2015 *)
  • PARI
    prodinf(x=1, 1-(1/8)^x) \\ Altug Alkan, Dec 01 2015

Formula

Equals exp(-Sum_{n>0} sigma_1(n)/n*(1/8)^n) where sigma_1() is A000203().
Equals (1/8; 1/8){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Nov 26 2015
From Amiram Eldar, May 09 2023: (Start)
Equals sqrt(2*Pi/(3*log(2))) * exp(log(2)/8 - Pi^2/(18*log(2))) * Product_{k>=1} (1 - exp(-4*k*Pi^2/(3*log(2)))) (McIntosh, 1995).
Equals Sum_{n>=0} (-1)^n/A027876(n). (End)

A132323 Decimal expansion of Product_{k>=0} (1+1/3^k).

Original entry on oeis.org

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

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Twice the constant A132324.

Examples

			3.12986803713402307587769821345767...
		

Programs

  • Mathematica
    digits = 105; NProduct[1+1/3^k, {k, 0, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+3] // N[#, digits+3]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    2*N[QPochhammer[-1/3,1/3]] (* G. C. Greubel, Dec 01 2015 *)
  • PARI
    prodinf(x=0, 1+(1/3)^x) \\ Altug Alkan, Dec 03 2015

Formula

Equals lim sup_{n->oo} Product_{0<=k<=floor(log_3(n))} (1+1/floor(n/3^k)).
Equals lim sup_{n->oo} A132327(n)/A132027(n).
Equals lim sup_{n->oo} A132327(n)/n^((1+log_3(n))/2).
Equals lim sup_{n->oo} A132328(n)/n^((log_3(n)-1)/2).
Equals 2*exp(Sum_{n>0} 3^(-n) * Sum{k|n} -(-1)^k/k) = 2*exp(Sum_{n>0} A000593(n)/(n*3^n)).
Equals lim sup_{n->oo} A132327(n+1)/A132327(n).
Equals 2*(-1/3; 1/3){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 01 2015
Equals sqrt(2) * exp(log(3)/24 + Pi^2/(12*log(3))) * Product_{k>=1} (1 - exp(-2*(2*k-1)*Pi^2/log(3))) (McIntosh, 1995). - Amiram Eldar, May 25 2023

A132327 a(n) = Product{k>=0} (1 + floor(n/3^k)).

Original entry on oeis.org

1, 2, 3, 8, 10, 12, 21, 24, 27, 80, 88, 96, 130, 140, 150, 192, 204, 216, 399, 420, 441, 528, 552, 576, 675, 702, 729, 2240, 2320, 2400, 2728, 2816, 2904, 3264, 3360, 3456, 4810, 4940, 5070, 5600, 5740, 5880, 6450, 6600, 6750, 8832, 9024, 9216, 9996, 10200
Offset: 0

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-3 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product (1+d(m)d(m-1)d(m-2)...d(2)d(1)d(0))*(1+d(m)d(m-1)d(m-2)...d(2)d(1))*(1+d(m)d(m-1)d(m-2)...d(2))*...*(1+d(m)d(m-1)d(m-2))*(1+d(m)d(m-1))*(1+d(m)).

Examples

			a(12)=(1+floor(12/3^0))*(1+floor(12/3^1))*(1+floor(12/3^2))=13*5*2=130; a(20)=441 since 20=202(base-3) and so
a(20)=(1+202)*(1+20)*(1+2)(base-3)=21*7*3=441.
		

Crossrefs

Cf. A100220, A132027, A132038, A132264, A132269(for p=2), A132271(for p=10).
For formulas regarding a general parameter p (i.e. terms 1+floor(n/p^k)) see A132271.
For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

Programs

  • Mathematica
    Table[Product[1+Floor[n/3^k],{k,0,n}],{n,0,49}] (* James C. McMahon, Mar 07 2025 *)

Formula

Recurrence: a(n)=(1+n)*a(floor(n/3)); a(3n)=(1+3n)*a(n); a(n*3^m)=product{1<=k<=m, 1+n*3^k}*a(n).
a(k*3^m-j)=(k*3^m-j+1)*3^m*p^(m(m-1)/2), for 0=1, a(3^m)=3^(m(m+1)/2)*product{0<=k<=m, 1+1/3^k}, m>=1.
a(n)=A132328(3*n)=(1+n)*A132328(n).
Asymptotic behavior: a(n)=O(n^((1+log_3(n))/2)); this follows from the inequalities below.
a(n)<=A132027(n)*product{0<=k<=floor(log_3(n)), 1+1/3^k}.
a(n)>=A132027(n)/product{1<=k<=floor(log_3(n)), 1-1/3^k}.
a(n)A000217(log_3(n)), where c=product{k>=0, 1+1/p^k}=3.12986803713402307587769821345767... (see constant A132323).
a(n)>n^((1+log_3(n))/2)=3^A000217(log_3(n)).
lim sup a(n)/A132027(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
lim inf a(n)/A132027(n)=1/product{k>0, 1-1/3^k}=1/0.560126077927948944969792243314140014..., for n-->oo (see constant A100220).
lim inf a(n)/n^((1+log_3(n))/2)=1, for n-->oo.
lim sup a(n)/n^((1+log_3(n))/2)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767..., for n-->oo (see constant A132323).
lim inf a(n+1)/a(n)=2*product{k>0, 1+1/3^k}=3.12986803713402307587769821345767... for n-->oo (see constant A132323).
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