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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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)

A000707 Number of permutations of [1,2,...,n] with n-1 inversions.

Original entry on oeis.org

1, 1, 2, 6, 20, 71, 259, 961, 3606, 13640, 51909, 198497, 762007, 2934764, 11333950, 43874857, 170193528, 661386105, 2574320659, 10034398370, 39163212165, 153027659730, 598577118991, 2343628878849, 9184197395425, 36020235035016, 141376666307608
Offset: 1

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Author

Keywords

Comments

Same as number of submultisets of size n-1 of the multiset with multiplicities [1,2,...,n-1]. - Joerg Arndt, Jan 10 2011. Stated another way, a(n-1) is the number of size n "multisubsets" (see example) of M = {a^1,b^2,c^3,d^4,...,#^n!}. - Geoffrey Critzer, Apr 01 2010, corrected by Jacob Post, Jan 03 2011
For a more general result (taking multisubset of any size) see A008302. - Jacob Post, Jan 03 2011
The number of ordered submultisets is found in A129481; credit for this observation should go to Marko Riedel at Mathematics Stack Exchange (see link). - J. M. Bergot, Aug 12 2016
The number of ordered submultisets is found in A129481. - J. M. Bergot, Aug 12 2016
For n>0: a(n) is the number of compositions of n-1 into n-1 nonnegative parts such that the i-th part is not larger than i. a(4) = 6: [0,0,3], [0,1,2], [0,2,1], [1,0,2], [1,1,1], [1,2,0]. - Alois P. Heinz, Jun 26 2023

Examples

			a(4) = 6 because there are 6 multisubsets of {a,b,b,c,c,c} with cardinality =3: {a,b,b}, {a,b,c}, {a,c,c}, {b,b,c}, {b,c,c}, {c,c,c}. - _Geoffrey Critzer_, Apr 01 2010, corrected by _Jacob Post_, Jan 03 2011
G.f. = x + x^2 + 2*x^3 + 6*x^4 + 20*x^5 + 71*x^6 + 259*x^7 + 961*x^8 + ...
		

References

  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 241.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 5.14., p.356
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, p. 15.
  • E. Netto, Lehrbuch der Combinatorik. 2nd ed., Teubner, Leipzig, 1927, p. 96.
  • 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).

Crossrefs

One of the diagonals of triangle in A008302.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n>i*(i+1)/2, 0,
          `if`(n=0, 1, add(b(n-j, i-1), j=0..min(n, i))))
        end:
    a:= n-> b(n-1$2):
    seq(a(n), n=1..27);  # Alois P. Heinz, Jun 26 2023
  • Mathematica
    Table[SeriesCoefficient[ Series[Product[Sum[x^i, {i, 0, k}], {k, 0, n}], {x, 0, 20}], n], {n, 1, 20}] (* Geoffrey Critzer, Apr 01 2010 *)
    a[ n_] := SeriesCoefficient[ Product[ Sum[ x^i, {i, 0, k}], {k, 0, n}], {x, 0, n}]; (* Michael Somos, Aug 15 2016 *)
  • PARI
    {a(n) = my(v); if( n<1, 0, sum(k=0, n!-1, v = numtoperm(n, k); n-1 == sum(i=1, n-1, sum(j=i+1, n, v[i]>v[j]))))}; /* Michael Somos, Aug 15 2016 */

Formula

See A008302 for g.f.
a(n) = 2^(2*n-2)/sqrt(Pi*n)*Q*(1+O(n^(-1))), where Q is a digital search tree constant, Q = Product_{n>=1} (1 - 1/(2^n)) = QPochhammer[1/2, 1/2] = 0.288788095... (see A048651), corrected and extended by Vaclav Kotesovec, Mar 16 2014

Extensions

More terms from James Sellers, Dec 16 1999
Asymptotic formula from Barbara Haas Margolius (margolius(AT)math.csuohio.edu), May 31 2001
Better definition from Joerg Arndt, Jan 10 2011

A028365 Order of general affine group over GF(2), AGL(n,2).

Original entry on oeis.org

1, 2, 24, 1344, 322560, 319979520, 1290157424640, 20972799094947840, 1369104324918194995200, 358201502736997192984166400, 375234700595146883504949480652800, 1573079924978208093254925489963584716800
Offset: 0

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Author

Keywords

Comments

For n > 0, a(n) = v(n+1)/v(n), where v = A203305 is the Vandermonde determinant of the first n of the numbers -2^j - 1; see the Mathematica section. - Clark Kimberling, Jan 01 2012

References

  • J. M. Borwein, D. H. Bailey and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 54 (1.64).

Crossrefs

Programs

  • Magma
    [1] cat [(&*[2^(n+1) - 2^(j+1): j in [0..n-1]]): n in [1..20]]; // G. C. Greubel, Aug 31 2023
    
  • Maple
    A028365 := n->2^n*product(2^n-2^'i','i'=0..n-1); # version 1
    A028365 := n->product(2^'j'-1,'j'=1..n)*2^binomial(n+1,2); # version 2
  • Mathematica
    RecurrenceTable[{a[1]==1, a[2]==2, a[3]==24, a[n]==(6a[n-1]^2 a[n-3] - 8a[n-1] a[n-2]^2)/(a[n-2] a[n-3])}, a[n], {n,20}] (* Harvey P. Dale, Aug 03 2011 *)
    (* Next, the connection with Vandermonde determinants *)
    f[j_]:= 2^j - 1; z = 15;
    v[n_]:= Product[Product[f[k] - f[j], {j,k-1}], {k,2,n}]
    Table[v[n], {n,z}]   (* A203303 *)
    Table[v[n+1]/v[n], {n,z}]  (* A028365 *)
    Table[v[n]*v[n+2]/(2*v[n+1])^2, {n,z}]  (* A171499 *) (* Clark Kimberling, Jan 01 2011 *)
    Table[(-1)^n*2^Binomial[n+1,2]*QPochhammer[2,2,n], {n,0,20}] (* G. C. Greubel, Aug 31 2023 *)
  • PARI
    a(n)=if(n<0,0,prod(k=1,n,2^k-1)*2^((n^2+n)/2)) /* Michael Somos, May 09 2005 */
    
  • SageMath
    [product(2^(n+1) - 2^(k+1) for k in range(n)) for n in range(21)] # G. C. Greubel, Aug 31 2023

Formula

a(n) is asymptotic to C*2^(n*(n+1)) where C = 0.288788095086602421278899721... = prod(k>=1, 1-1/2^k) (cf. A048651). - Benoit Cloitre, Apr 11 2003
a(n) = (6*a(n-1)^2*a(n-3) - 8*a(n-1)*a(n-2)^2) / (a(n-2)*a(n-3)). [From Putman Exam]. - Max Alekseyev, May 18 2007
a(n) = 2*A203305(n), n > 0. - Clark Kimberling, Jan 01 2012
From Max Alekseyev, Jun 09 2015: (Start)
a(n) = 2^A000217(n) * A005329(n).
a(n) = 2^n * A002884(n).
a(n) = 2^n * n! * A053601(n). (End)
From G. C. Greubel, Aug 31 2023: (Start)
a(n) = Product_{j=0..n-1} (2^(n+1) - 2^(j+1)).
a(n) = (-1)^n * 2^binomial(n+1,2) * QPochhammer(2,2,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

A132033 Product{0<=k<=floor(log_9(n)), floor(n/9^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 36, 38, 40, 42, 44, 46, 48, 50, 52, 81, 84, 87, 90, 93, 96, 99, 102, 105, 144, 148, 152, 156, 160, 164, 168, 172, 176, 225, 230, 235, 240, 245, 250, 255, 260, 265, 324, 330, 336, 342, 348, 354, 360, 366, 372
Offset: 1

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-9 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 d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(85)=floor(85/9^0)*floor(85/9^1)*floor(85/9^2)=85*9*1=765; a(88)=792 since 88=107(base-9) and so a(88)=107*10*1(base-9)=88*9*1=792.
		

Crossrefs

For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132032(p=8), A067080(p=10), A132263(p=11), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Programs

  • Mathematica
    Table[Product[Floor[n/9^k],{k,0,Floor[Log[9,n]]}],{n,62}] (* James C. McMahon, Mar 03 2025 *)

Formula

Recurrence: a(n)=n*a(floor(n/9)); a(n*9^m)=n^m*9^(m(m+1)/2)*a(n).
a(k*9^m)=k^(m+1)*9^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_9(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_9(n)))/9^((1+floor(log_9(n)))*floor(log_9(n))/2); equality holds for n=k*9^m, 0=0. b(n) can also be written n^(1+floor(log_9(n)))/9^A000217(floor(log_9(n))).
Also: a(n)<=3^(1/4)*n^((1+log_9(n))/2)=1.316074013...*9^A000217(log_9(n)), equality holds for n=3*9^m, m>=0.
a(n)>c*b(n), where c=0.4689451783670236932832800... (see constant A132024).
Also: a(n)>c*2^((1-log_9(2))/2)*n^((1+log_9(n))/2)=0.4689451783670...*1.267747616...*9^A000217(log_9(n)).
lim inf a(n)/b(n)=0.4689451783670236932832800..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_9(n))/2)=0.4689451783670236932832800...*sqrt(2)/2^log_9(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_9(n))/2)=3^(1/4)=1.316074013..., for n-->oo.
lim inf a(n)/a(n+1)=0.4689451783670236932832800... for n-->oo (see constant A132025).

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)

A132263 Product{0<=k<=floor(log_11(n)), floor(n/11^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 275, 280, 285, 290, 295, 300, 305, 310
Offset: 1

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-11 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 d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(50)=floor(50/11^0)*floor(50/11^1)=50*4=200; a(63)=315 since 63=58(base-11) and so a(63)=58*5(base-11)=63*5=315.
		

Crossrefs

For formulas regarding a general parameter p (i.e. terms floor(n/p^k)) see A132264.
For the product of terms floor(n/p^k) for p=2 to p=12 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132264(p=12).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Formula

Recurrence: a(n)=n*a(floor(n/11)); a(n*11^m)=n^m*11^(m(m+1)/2)*a(n).
a(k*11^m)=k^(m+1)*11^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_11(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_11(n)))/p^((1+floor(log_11(n)))*floor(log_11(n))/2); equality holds for n=k*11^m, 0=0. b(n) can also be written n^(1+floor(log_11(n)))/11^A000217(floor(log_11(n))).
Also: a(n)<=3^((1-log_11(3))/2)*n^((1+log_11(n))/2)=1.346673852...^((1-log_11(3))/2)*11^A000217(log_11(n)), equality holds for n=3*11^m, m>=0.
a(n)>c*b(n), where c=0.4751041275076031053975644472... (see constant A132265).
Also: a(n)>c*(sqrt(2)/2^log_11(sqrt(2)))*n^((1+log_11(n))/2)=0.607848303...*11^00217(log_11(n)).
lim inf a(n)/b(n)=0.4751041275076031053975644472..., for n-->oo.
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_p(n))/2)=0.4751041275076031...*sqrt(2)/2^log_11(sqrt(2)), for n-->oo.
lim sup a(n)/n^((1+log_p(n))/2)=sqrt(3)/3^log_11(sqrt(3))=1.346673852..., for n-->oo.
lim inf a(n)/a(n+1)=0.4751041275076031053975644472... for n-->oo (see constant A132265).

A132264 Product{0<=k<=floor(log_12(n)), floor(n/12^k)}, n>=1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232, 236, 300, 305, 310, 315
Offset: 1

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-12 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 d(m)d(m-1)d(m-2)...d(2)d(1)d(0)*d(m)d(m-1)d(m-2)...d(2)d(1)*d(m)d(m-1)d(m-2)...d(2)*...*d(m)d(m-1)d(m-2)*d(m)d(m-1)*d(m).

Examples

			a(50)=floor(50/12^0)*floor(50/12^1)=50*4=200.
a(65)=325 since 65=55(base-12) and so a(65)=55*5(base-12)=65*5=325.
		

Crossrefs

For the product of terms floor(n/p^k) for p=2 to p=11 see A098844(p=2), A132027(p=3)-A132033(p=9), A067080(p=10), A132263(p=11).
For the products of terms 1+floor(n/p^k) see A132269-A132272, A132327, A132328.

Formula

The following formulas are given for a general parameter p considering the product of terms floor(n/p^k) for 0<=k<=floor(log_p(n)), where p=12 for this sequence.
Recurrence: a(n)=n*a(floor(n/p)); a(n*p^m)=n^m*p^(m(m+1)/2)*a(n).
a(k*p^m)=k^(m+1)*p^(m(m+1)/2), for 0
Asymptotic behavior: a(n)=O(n^((1+log_p(n))/2)); this follows from the inequalities below.
a(n)<=b(n), where b(n)=n^(1+floor(log_p(n)))/p^((1+floor(log_p(n)))*floor(log_p(n))/2); equality holds for n=k*p^m, 0=0. b(n) can also be written n^(1+floor(log_p(n)))/p^A000217(floor(log_p(n))).
Also: a(n)<=q^((1-log_p(q))/2)*n^((1+log_p(n))/2)=q^((1-log_p(q))/2)*p^A000217(log_p(n)), equality holds for n=q*p^m, m>=0, where q=floor(sqrt(p)+1/2). Also, equality holds for n=(q+1)*p^m, provided p is a A002378-number (in this case we have p=q*(q+1) and so q^((1-log_p(q))/2)=(q+1)^((1-log_p(q+1))/2)).
a(n)>c*b(n), where c=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... (for p=12 see constant A132265).
Also: a(n)>c*(sqrt(2)/2^log_p(sqrt(2)))*n^((1+log_p(n))/2)=0.612870619...*p^A000217(log_p(n)), (p=12).
lim inf a(n)/b(n)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380198334286365820..., for n-->oo (for p=12 see constant A132265).
lim sup a(n)/b(n)=1, for n-->oo.
lim inf a(n)/n^((1+log_p(n))/2)=(sqrt(2)/2^log_p(sqrt(2)))*product{k>0, 1-1/(2*p^k)}=0.612870619..., for n-->oo, (p=12).
lim sup a(n)/n^((1+log_p(n))/2)=sqrt(q)/q^log_p(sqrt(q))=1.358593737..., for n-->oo, (p=12, q=round(sqrt(p))=3).
lim inf a(n)/a(n+1)=product{k>0, 1-1/(2*p^k)}=0.47735217025489380... for n-->oo (for p=12 see constant A132265).
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