A005329 -id:A005329 - cp's OEIS frontend

A027637 a(n) = Product_{i=1..n} (4^i - 1).

1, 3, 45, 2835, 722925, 739552275, 3028466566125, 49615367752825875, 3251543125681443718125, 852369269595510700600441875, 893773106866112632882108339078125, 3748755223447856814435325652920396921875

Offset: 0

N. J. A. Sloane

The q-analog of double factorials (A000165) evaluated at q=2. - Michael Somos, Sep 12 2014
3^n*5^(floor(n/2))|a(n) for n>=1. - G. C. Greubel, Nov 21 2015
Given probability p = 1/4^n that an outcome will occur at the n-th stage of an infinite process, then starting at n=1, 1-a(n)/A053763(n+1) is the probability that the outcome has occurred up to and including the n-th iteration. The limiting ratio is 1-A100221 ~ 0.3114625. - Bob Selcoe, Mar 01 2016

G. C. Greubel, Table of n, a(n) for n = 0..50

Cf. A000165.
Sequences of the form q-Pochhammer(n, q, q): A005329 (q=2), A027871 (q=3), this sequence (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).
Cf. A053763, A100221.

[1] cat [&*[4^k-1: k in [1..n]]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

A027637 := proc(n)
    mul( 4^i-1,i=1..n) ;
end proc:
seq(A027637(n),n=0..8) ;

A027637 = Table[Product[4^i - 1, {i, n}], {n, 0, 9}] (* Alonso del Arte, Nov 14 2011 *)
a[ n_] := If[ n < 0, 0, Product[ (q^(2 k) - 1) / (q - 1), {k, n}] /. q -> 2]; (* Michael Somos, Sep 12 2014 *)
Abs@QPochhammer[4, 4, Range[0, 10]] (* Vladimir Reshetnikov, Nov 20 2015 *)

a(n) = prod(i=1, n, 4^i-1); \\ Michel Marcus, Nov 21 2015

from sage.combinat.q_analogues import q_pochhammer
def A027637(n): return (-1)^n*q_pochhammer(n, 4, 4)
[A027637(n) for n in (0..15)] # G. C. Greubel, Aug 04 2022

a(n) ~ c * 2^(n*(n+1)), where c = Product_{k>=1} (1-1/4^k) = A100221 = 0.688537537120339715456514357293508184675549819378... . - Vaclav Kotesovec, Nov 21 2015
a(n) = 4^(binomial(n+1,2))*(1/4;1/4){n} = (4; 4){n}, where (a;q){n} is the q-Pochhammer symbol. - _G. C. Greubel, Dec 24 2015
G.f.: Sum_{n>=0} 4^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 4^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A100221. - Amiram Eldar, May 07 2023

A027878 a(n) = Product_{i=1..n} (10^i - 1).

Original entry on oeis.org

1, 9, 891, 890109, 8900199891, 890011088900109, 890010198889020099891, 8900101098880002109889900109, 890010100987899112108987901010099891, 890010100097889011121088788901111989989900109

Offset: 0

N. J. A. Sloane

nonn

G. C. Greubel, Table of n, a(n) for n = 0..50

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027879 (q=11), A027880 (q=12).

Cf. A132038, A132326.

[1] cat [&*[10^k-1: k in [1..n]]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

Table[Product[10^i-1,{i,n}],{n,0,10}] (* Harvey P. Dale, Aug 15 2011 *)
Abs@QPochhammer[10, 10, Range[0, 30]] (* G. C. Greubel, Nov 24 2015 *)

a(n) = prod(k=1, n, 10^k - 1) \\ Altug Alkan, Nov 25 2015

a(n) ~ c * 10^(n*(n+1)/2), where c = Product_{k>=1} (1-1/10^k) = A132038 = 0.890010099998999000000100009999999989999900000000... . - Vaclav Kotesovec, Nov 21 2015
3^n*(11)^(floor(n/2)) divides a(n) for n>=0. - G. C. Greubel, Nov 24 2015
Equals 10^(binomial(n+1,2))*(1/10;1/10){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
G.f.: Sum_{n>=0} 10^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 10^k*x). - Ilya Gutkovskiy, May 22 2017
From Amiram Eldar, May 07 2023: (Start)
Sum_{n>=0} 1/a(n) = A132326.
Sum_{n>=0} (-1)^n/a(n) = A132038. (End)

A027872 a(n) = Product_{i=1..n} (5^i - 1).

Original entry on oeis.org

1, 4, 96, 11904, 7428096, 23205371904, 362560730628096, 28324694519589371904, 11064305472020078810628096, 21609960560733744406929189371904, 211034749490954911990173458030810628096

Offset: 0

N. J. A. Sloane

nonn

Given probability p = 1/5^n that an outcome will occur at the n-th stage of an infinite process, then starting at n=1, 1 - a(n)/A109345(n+1) is the probability that the outcome has occurred at or before the n-th iteration. The limiting ratio is 1-A100222 ~ 0.2396672. - Bob Selcoe, Mar 01 2016

G. C. Greubel, Table of n, a(n) for n = 0..50

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).

Cf. A109345, A100222.

[1] cat [&*[ 5^k-1: k in [1..n] ]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

A027872 := proc(n)
        mul( 5^i-1, i=1..n) ;
end proc: # R. J. Mathar, Mar 12 2013

Table[Product[(5^k-1),{k,1,n}],{n,0,20}] (* Vaclav Kotesovec, Jul 17 2015 *)
Abs@QPochhammer[5, 5, Range[0, 10]] (* Vladimir Reshetnikov, Nov 20 2015 *)
Join[{1},FoldList[Times,5^Range[10]-1]] (* Harvey P. Dale, Dec 28 2021 *)

a(n) = prod(i=1, n, 5^i-1); \\ Michel Marcus, Nov 21 2015

4^n|a(n) for n >= 1. - G. C. Greubel, Nov 21 2015
a(n) ~ c * 5^(n*(n+1)/2), where c = Product_{k>=1} (1-1/5^k) = A100222 . - Vaclav Kotesovec, Nov 21 2015
a(n) = 5^(binomial(n+1,2))*(1/5; 1/5){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 23 2015
a(n) = Product_{i=1..n} A024049(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 5^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 5^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A100222. - Amiram Eldar, May 07 2023

A027873 a(n) = Product_{i=1..n} (6^i - 1).

Original entry on oeis.org

1, 5, 175, 37625, 48724375, 378832015625, 17674407688984375, 4947685316415841015625, 8310206472731792807458984375, 83747726219216824716765369541015625

Offset: 0

N. J. A. Sloane

nonn

G. C. Greubel, Table of n, a(n) for n = 0..50

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).

Cf. A132034.

[1] cat [&*[ 6^k-1: k in [1..n] ]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

Table[Product[(6^k-1),{k,1,n}],{n,0,20}] (* Vaclav Kotesovec, Jul 17 2015 *)
Abs@QPochhammer[6, 6, Range[0, 10]] (* Vladimir Reshetnikov, Nov 20 2015 *)
FoldList[Times,Join[{1},6^Range[10]-1]] (* Harvey P. Dale, Oct 13 2017 *)

a(n) = prod(i=1, n, 6^i-1); \\ Michel Marcus, Nov 21 2015

5^n|a(n) for n>=0. - G. C. Greubel, Nov 20 2015
a(n) ~ c * 6^(n*(n+1)/2), where c = Product_{k>=1} (1-1/6^k) = A132034 = 0.805687728162164940923750215496298968917997628693... . - Vaclav Kotesovec, Nov 21 2015
a(n) = 6^(binomial(n+1,2))*(1/6;1/6){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
a(n) = Product_{i=1..n} A024062(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 6^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 6^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132034. - Amiram Eldar, May 07 2023

A027875 a(n) = Product_{i=1..n} (7^i - 1).

Original entry on oeis.org

1, 6, 288, 98496, 236390400, 3972777062400, 467389275837235200, 384914699001548351078400, 2218956256804125934296760320000, 89542886518308517126993353029713920000

Offset: 0

N. J. A. Sloane

nonn

G. C. Greubel, Table of n, a(n) for n = 0..50

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).

Cf. A132035.

[1] cat [&*[ 7^k-1: k in [1..n] ]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

Abs@QPochhammer[7, 7, Range[0, 10]] (* Vladimir Reshetnikov, Nov 20 2015 *)
Table[Product[7^k-1,{k,n}],{n,0,10}] (* Harvey P. Dale, Jul 28 2022 *)

a(n) = prod(i=1, n, 7^i-1); \\ Michel Marcus, Nov 21 2015

2*(10)^(2m)|a(n) where 4*m <= n <= 4*m+3, for m >= 1. - G. C. Greubel, Nov 20 2015
a(n) ~ c * 7^(n*(n+1)/2), where c = Product_{k>=1} (1-1/7^k) = A132035 = 0.836795407089037871026729798146136241352436435876... . - Vaclav Kotesovec, Nov 21 2015
a(n) = 7^(binomial(n+1,2))*(1/7;1/7){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
a(n) = Product_{i=1..n} A024075(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 7^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 7^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132035. - Amiram Eldar, May 07 2023

A069777 Array of q-factorial numbers n!_q, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 6, 3, 1, 1, 1, 24, 21, 4, 1, 1, 1, 120, 315, 52, 5, 1, 1, 1, 720, 9765, 2080, 105, 6, 1, 1, 1, 5040, 615195, 251680, 8925, 186, 7, 1, 1, 1, 40320, 78129765, 91611520, 3043425, 29016, 301, 8, 1, 1

Offset: 0

Franklin T. Adams-Watters, Apr 07 2002

			Square array begins:
    1,   1,    1,      1,       1,        1,         1, ...
    1,   1,    1,      1,       1,        1,         1, ...
    1,   2,    3,      4,       5,        6,         7, ...
    1,   6,   21,     52,     105,      186,       301, ...
    1,  24,  315,   2080,    8925,    29016,     77959, ...
    1, 120, 9765, 251680, 3043425, 22661496, 121226245, ...
    ...

Alois P. Heinz, Antidiagonals n = 0..55, flattened
Kent E. Morrison, Integer Sequences and Matrices Over Finite Fields, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.1.
Index entries for sequences related to factorial numbers

Columns q=0..11 are A000012, A000142, A005329, A015001, A015002, A015004, A015005, A015006, A015007, A015008, A015009, A015011.
Rows n=3..5 are A069778, A069779, A218503.
Main diagonal gives A347611.
Cf. A008302, A156173.

A069777 := proc(n,k) local n1: mul(A104878(n1,k), n1=k..n-1) end: A104878 := proc(n,k): if k = 0 then 1 elif k=1 then n elif k>=2 then (k^(n-k+1)-1)/(k-1) fi: end: seq(seq(A069777(n,k), k=0..n), n=0..9); # Johannes W. Meijer, Aug 21 2011
nmax:=9: T(0,0):=1: for n from 1 to nmax do T(n,0):=1:  T(n,1):= (n-1)! od: for q from 2 to nmax do for n from 0 to nmax do T(n+q,q) := product((q^k - 1)/(q - 1), k= 1..n) od: od: for n from 0 to nmax do seq(T(n,k), k=0..n) od; seq(seq(T(n,k), k=0..n), n=0..nmax); # Johannes W. Meijer, Aug 21 2011
# alternative Maple program:
T:= proc(n, k) option remember; `if`(n<2, 1,
      T(n-1, k)*`if`(k=1, n, (k^n-1)/(k-1)))
    end:
seq(seq(T(d-k, k), k=0..d), d=0..10);  # Alois P. Heinz, Sep 08 2021

(* Returns the rectangular array *) Table[Table[QFactorial[n, q], {q, 0, 6}], {n, 0, 6}] (* Geoffrey Critzer, May 21 2017 *)

T(n,q)=prod(k=1, n, ((q^k - 1) / (q - 1))) \\ Andrew Howroyd, Feb 19 2018

T(n,q) = Product_{k=1..n} (q^k - 1) / (q - 1).
T(n,k) = Product_{n1=k..n-1} A104878(n1,k). - Johannes W. Meijer, Aug 21 2011
T(n,k) = Sum_{i>=0} A008302(n,i)*k^i. - Geoffrey Critzer, Feb 26 2025

Name edited by Michel Marcus, Sep 08 2021

A027876 a(n) = Product_{i=1..n} (8^i - 1).

Original entry on oeis.org

1, 7, 441, 225351, 922812345, 30237792108615, 7926625536728661945, 16623330670976050126618695, 278893192683059452825059069034425, 37432410397693271164043156886536608251975

Offset: 0

N. J. A. Sloane

nonn

G. C. Greubel, Table of n, a(n) for n = 0..50

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027877 (q=9), A027878 (q=10), A027879 (q=11), A027880 (q=12).

Cf. A132036.

[1] cat [&*[ 8^k-1: k in [1..n] ]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

seq(mul(8^i-1,i=1..n), n=0..20); # Robert Israel, Dec 24 2015

FoldList[Times,1,8^Range[10]-1] (* Harvey P. Dale, Dec 23 2011 *)

a(n)=prod(i=1,n,8^i-1) \\ Charles R Greathouse IV, Nov 22 2015

a(n) ~ c * 8^(n*(n+1)/2), where c = Product_{k>=1} (1-1/8^k) = A132036 = 0.859405994400702866200758580064418894909484979588... . - Vaclav Kotesovec, Nov 21 2015
7^n | a(n). - G. C. Greubel, Nov 21 2015
It appears that 7^m | a(n) iff 7^m | (7n)!. - Robert Israel, Dec 24 2015
a(n) = 8^(binomial(n+1,2))*(1/8;1/8){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
G.f. g(x) satisfies (1+x) g(x) = 1 + 8 x g(8x). - Robert Israel, Dec 24 2015
a(n) = Product_{i=1..n} A024088(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 8^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 8^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132036. - Amiram Eldar, May 07 2023

A027877 a(n) = Product_{i=1..n} (9^i - 1).

Original entry on oeis.org

1, 8, 640, 465920, 3056435200, 180476385689600, 95912370410881024000, 458745798479390789599232000, 19747501938318761090457052119040000, 7650586837724400321220283274999910891520000

Offset: 0

N. J. A. Sloane

nonn

G. C. Greubel, Table of n, a(n) for n = 0..40

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027878 (q=10), A027879 (q=11), A027880 (q=12).

Cf. A132037.

[1] cat [&*[ 9^k-1: k in [1..n] ]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

Abs@QPochhammer[9, 9, Range[0, 10]] (* Vladimir Reshetnikov, Nov 20 2015 *)

a(n) = prod(i=1, n, 9^i-1); \\ Altug Alkan, Dec 24 2015

a(n) ~ c * 3^(n*(n+1)), where c = Product_{k>=1} (1-1/9^k) = A132037 = 0.876560354035964205836019838417862010106635101174... . - Vaclav Kotesovec, Nov 21 2015
From  - G. C. Greubel, Dec 24 2015: (Start)
8^n * 10^(floor(n/2))|a(n), for n>=0.
a(n) = 9^(binomial(n+1,2))*(1/9;1/9){n}, where (a;q){n} is the q-Pochhammer symbol. (End)
a(n) = Product_{i=1..n} A024101(i). - Michel Marcus, Dec 27 2015
G.f.: Sum_{n>=0} 9^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 9^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132037. - Amiram Eldar, May 07 2023

A027879 a(n) = Product_{i=1..n} (11^i - 1).

Original entry on oeis.org

1, 10, 1200, 1596000, 23365440000, 3763004112000000, 6666387564654720000000, 129909027758312519942400000000, 27847153692160782464830528512000000000, 65662131721505488121539650946349537280000000000

Offset: 0

N. J. A. Sloane

nonn

It appears that the number of trailing zeros in a(n) is A191610(n). - Robert Israel, Nov 24 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..43

Cf. A005329 (q=2), A027871 (q=3), A027637 (q=4), A027872 (q=5), A027873 (q=6), A027875 (q=7), A027876 (q=8), A027877 (q=9), A027878 (q=10), A027880 (q=12).

Cf. A132267, A191610.

[1] cat [&*[11^k-1: k in [1..n]]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015

seq(mul(11^i-1,i=1..n),n=0..20; # Robert Israel, Nov 24 2015

FoldList[Times,1,11^Range[10]-1] (* Harvey P. Dale, Aug 13 2013 *)
Abs@QPochhammer[11, 11, Range[0, 40]] (* G. C. Greubel, Nov 24 2015 *)

a(n)=prod(i=1,n,11^i-1) \\ Anders Hellström, Nov 21 2015

10^n|a(n) for n>=0; 12*(10)^(n)|a(n) n>=2. - G. C. Greubel, Nov 21 2015
a(n) ~ c * 11^(n*(n+1)/2), where c = Product_{k>=1} (1-1/11^k) = 0.900832706809715279949862694760647744762491192216... . - Vaclav Kotesovec, Nov 21 2015
E.g.f. E(x) satisfies E'(x) = 11 E(11 x) - E(x). - Robert Israel, Nov 24 2015
Equals 11^(binomial(n+1,2))*(1/11;1/11){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
G.f.: Sum_{n>=0} 11^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 11^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132267. - Amiram Eldar, May 07 2023

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

Benoit Cloitre, Jan 25 2003

			2.38423102903137172414989928867839723877161951650843345769...

G. C. Greubel, Table of n, a(n) for n = 1..1200
Richard J. McIntosh, Some Asymptotic Formulae for q-Hypergeometric Series, Journal of the London Mathematical Society, Vol. 51, No. 1 (1995), pp. 120-136; alternative link.

Cf. A028362, A048651, A081845, A098844, A100220, A132019-A132026, A132034-A132038, A132265-A132268, A132323-A132326, A132269, A132270, A000593.

Cf. A141848. - Gleb Koloskov, Apr 04 2021

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 *)

prodinf(n=1,1+2.^-n) \\ Charles R Greathouse IV, May 27 2015

1/prodinf(n=0, 1-2^(-2*n-1)) \\ Gleb Koloskov, Apr 04 2021

(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)

cp's OEIS Frontend

A027637 a(n) = Product_{i=1..n} (4^i - 1).

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A027878 a(n) = Product_{i=1..n} (10^i - 1).

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A027872 a(n) = Product_{i=1..n} (5^i - 1).

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A027873 a(n) = Product_{i=1..n} (6^i - 1).

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A027875 a(n) = Product_{i=1..n} (7^i - 1).

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A069777 Array of q-factorial numbers n!_q, read by ascending antidiagonals.

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A027876 a(n) = Product_{i=1..n} (8^i - 1).

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