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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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

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Keywords

Crossrefs

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.

Programs

  • Magma
    [1] cat [&*[ 9^k-1: k in [1..n] ]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015
    
  • Mathematica
    Abs@QPochhammer[9, 9, Range[0, 10]] (* Vladimir Reshetnikov, Nov 20 2015 *)
  • PARI
    a(n) = prod(i=1, n, 9^i-1); \\ Altug Alkan, Dec 24 2015

Formula

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

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Keywords

Comments

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

Crossrefs

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).

Programs

  • Magma
    [1] cat [&*[11^k-1: k in [1..n]]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015
  • Maple
    seq(mul(11^i-1,i=1..n),n=0..20; # Robert Israel, Nov 24 2015
  • Mathematica
    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 *)
  • PARI
    a(n)=prod(i=1,n,11^i-1) \\ Anders Hellström, Nov 21 2015
    

Formula

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

A027880 a(n) = Product_{i=1..n} (12^i - 1).

Original entry on oeis.org

1, 11, 1573, 2716571, 56328099685, 14016177372718235, 41852067359921313500005, 1499635200191700040518673659035, 644815685260091508353787979063721364325, 3327107302821620489265827570792988872583047378075
Offset: 0

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Comments

In general, Product_{i=1..n} (q^i-1) ~ c * q^(n*(n+1)/2), where c = Product_{k >= 1} (1-1/q^k). - Vaclav Kotesovec, Nov 21 2015

Crossrefs

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), A027879 (q=11).
Cf. A132268.

Programs

  • Magma
    [1] cat [&*[12^k-1: k in [1..n]]: n in [1..11]]; // Vincenzo Librandi, Dec 24 2015
  • Mathematica
    FoldList[Times,1,12^Range[10]-1] (* Harvey P. Dale, Mar 01 2015 *)
    Abs@QPochhammer[12, 12, Range[0, 30]] (* G. C. Greubel, Nov 24 2015 *)
  • PARI
    a(n) = prod(k=1, n, 12^k - 1) \\ Altug Alkan, Nov 25 2015
    

Formula

a(n) ~ c * 12^(n*(n+1)/2), where c = Product_{k>=1} (1-1/12^k) = 0.909726268905994888636362046977080249120791691941... . - Vaclav Kotesovec, Nov 21 2015
(11)^n*(13)^(floor(n/2))|a(n) for n>=0. - G. C. Greubel, Nov 24 2015
Equals 12^(binomial(n+1,2))*(1/12;1/12){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Dec 24 2015
G.f.: Sum_{n>=0} 12^(n*(n+1)/2)*x^n / Product_{k=0..n} (1 + 12^k*x). - Ilya Gutkovskiy, May 22 2017
Sum_{n>=0} (-1)^n/a(n) = A132268. - Amiram Eldar, May 07 2023

A377484 a(n) = Product_{d|n, d>1} (d - 1).

Original entry on oeis.org

1, 1, 2, 3, 4, 10, 6, 21, 16, 36, 10, 330, 12, 78, 112, 315, 16, 1360, 18, 2052, 240, 210, 22, 53130, 96, 300, 416, 6318, 28, 146160, 30, 9765, 640, 528, 816, 1570800, 36, 666, 912, 560196, 40, 639600, 42, 27090, 39424, 990, 46, 37456650, 288, 42336, 1600, 45900, 52, 1874080, 2160
Offset: 1

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Author

Ridouane Oudra, Oct 29 2024

Keywords

Examples

			a(12) = (2-1)*(3-1)*(4-1)*(6-1)*(12-1) = 1*2*3*5*11 = 330.
		

Crossrefs

Programs

  • Maple
    with(numtheory): seq(mul(d-1, d in divisors(n) minus {1}), n=1..80);
  • Mathematica
    a[n_] := Times @@ (Rest@ Divisors[n] - 1); Array[a, 60] (* Amiram Eldar, Nov 01 2024 *)
  • PARI
    a(n) = my(d=divisors(n)); prod(k=2, #d, d[k]-1); \\ Michel Marcus, Oct 30 2024

Formula

a(n) = Product_{k=2..A000005(n)} (A027750(n,k) - 1).
a(p^n) = Product_{k=1..n} (p^k - 1), where p is prime, and n an integer.
a(2^n) = A005329(n).
a(3^n) = A027871(n).
a(5^n) = A027872(n).
a(7^n) = A027875(n).
a(11^n) = A027879(n).
From Amiram Eldar, Nov 02 2024: (Start)
a(n) = n-1 if and only if n is in A175787 (i.e., n = 4 or n is prime).
a(n) == 1 (mod 2) if and only if n is a power of 2 (A000079). (End)

A132324 Decimal expansion of Product_{k>=1} (1+1/3^k).

Original entry on oeis.org

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

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Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

Half the constant A132323.

Examples

			1.56493401856701153793884910...
		

Crossrefs

Programs

  • Mathematica
    digits = 105; NProduct[1+1/3^k, {k, 1, Infinity}, NProductFactors -> 100, WorkingPrecision -> digits+5] // N[#, digits+5]& // RealDigits[#, 10, digits]& // First (* Jean-François Alcover, Feb 18 2014 *)
    N[QPochhammer[-1/3,1/3]] (* G. C. Greubel, Dec 01 2015 *)

Formula

(1/2)*lim sup Product{k=0..floor(log_3(n))} (1+1/floor(n/3^k)) for n-->oo.
(1/2)*lim sup A132327(n)/A132027(n) for n-->oo.
(1/2)*lim sup A132327(n)/n^((1+log_3(n))/2) for n-->oo.
(1/2)*lim sup A132328(n)/n^((log_3(n)-1)/2) for n-->oo.
exp(Sum_{n>0} 3^(-n)*Sum_{k|n} -(-1)^k/k) = exp(Sum_{n>0} A000593(n)/(n*3^n)).
(1/2)*lim sup A132327(n+1)/A132327(n) = 1.56493401856701153793884910... for n-->oo.
Equals (-1/3; 1/3){infinity}, where (a;q){infinity} is the q-Pochhammer symbol. - G. C. Greubel, Dec 01 2015
From Amiram Eldar, Feb 19 2022: (Start)
Equals (sqrt(2)/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).
Equals Sum_{n>=0} 1/A027871(n). (End)

A290000 a(n) = Product_{k=1..n-1} (3^k + 1).

Original entry on oeis.org

1, 1, 4, 40, 1120, 91840, 22408960, 16358540800, 35792487270400, 234870301468364800, 4623187014103292723200, 272999193182799435304960000, 48361261073946554365403054080000, 25701205307660304745058529866383360000, 40976048450930207702360695570691784048640000
Offset: 0

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Author

Ilya Gutkovskiy, Jun 06 2020

Keywords

Crossrefs

Sequences of the form Product_{j=1..n-1} (m^j + 1): A000012 (m=0), A011782 (m=1), A028362 (m=2), this sequence (m=3), A309327 (m=4).

Programs

  • Magma
    [n lt 3 select 1 else (&*[3^j +1: j in [1..n-1]]): n in [1..20]]; // G. C. Greubel, Feb 21 2021
  • Mathematica
    Table[Product[3^k + 1, {k, 1, n - 1}], {n, 0, 14}]
  • PARI
    a(n) = prod(k=1, n-1, 3^k + 1); \\ Michel Marcus, Jun 06 2020
    
  • Sage
    from sage.combinat.q_analogues import q_pochhammer
    [1]+[3^(binomial(n,2))*q_pochhammer(n-1, -1/3, 1/3) for n in (1..20)] # G. C. Greubel, Feb 21 2021
    

Formula

G.f. A(x) satisfies: A(x) = 1 + x * A(3*x) / (1 - x).
G.f.: Sum_{k>=0} 3^(k*(k - 1)/2) * x^k / Product_{j=0..k-1} (1 - 3^j*x).
a(0) = 1; a(n) = Sum_{k=0..n-1} 3^k * a(k).
a(n) ~ c * 3^(n*(n - 1)/2), where c = Product_{k>=1} (1 + 1/3^k) = 1.564934018567011537938849... = A132324.
a(n) = 3^(binomial(n+1,2))*(-1/3;1/3){n}, where (a;q){n} is the q-Pochhammer symbol. - G. C. Greubel, Feb 21 2021

A028692 23-factorial numbers.

Original entry on oeis.org

1, 22, 11616, 141320256, 39547060439040, 254538406080331591680, 37680818974206486508802211840, 128296611269497862923425473853914480640, 10047034036599529256387830050150921763777884979200, 18096242094820543236399273859296273669601076798103392511590400
Offset: 0

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Programs

  • Mathematica
    FoldList[ #1 (23^#2-1)&, 1, Range[ 20 ] ]
    a[n_] := Abs[QPochhammer[23, 23, n]]; Array[a, 10, 0] (* Amiram Eldar, Jul 14 2025 *)
  • PARI
    a(n) = prod(k = 1, n, 23^k - 1); \\ Amiram Eldar, Jul 14 2025

Formula

From Amiram Eldar, Jul 14 2025: (Start)
a(n) = Product_{k=1..n} (23^k-1).
a(n) ~ c * 23^(n*(n+1)/2), where c = Product_{k>=1} (1 - 1/23^k) = 0.954631535623... . (End)

A028693 24-factorial numbers.

Original entry on oeis.org

1, 23, 13225, 182809175, 60651514035625, 482945140644890444375, 92292253139031982469134515625, 423295781586452233477722435457009484375, 46594416147080909523690749946376478698532878515625, 123093479909646650570543074660375014342475500150254964721484375
Offset: 1

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Crossrefs

Programs

  • Mathematica
    FoldList[ #1 (24^#2-1)&, 1, Range[ 20 ] ]
    a[n_] := Abs[QPochhammer[24, 24, n]]; Array[a, 10, 0] (* Amiram Eldar, Jul 14 2025 *)
  • PARI
    a(n) = prod(k = 1, n, 24^k - 1); \\ Amiram Eldar, Jul 14 2025

Formula

From Amiram Eldar, Jul 14 2025: (Start)
a(n) = Product_{k=1..n} (24^k-1).
a(n) ~ c * 24^(n*(n+1)/2), where c = Product_{k>=1} (1 - 1/24^k) = 0.956597348026... . (End)

A028694 25-factorial numbers.

Original entry on oeis.org

1, 24, 14976, 233985024, 91400166014976, 892579654839833985024, 217914953902301689160166014976, 1330047325845938129350664710839833985024, 202949115880923695556030391039325175289160166014976, 774189437411767935420978172981557217629743778824710839833985024
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Mathematica
    FoldList[ #1 (25^#2-1)&, 1, Range[ 20 ] ]
    a[n_] := Abs[QPochhammer[25, 25, n]]; Array[a, 10, 0] (* Amiram Eldar, Jul 14 2025 *)
  • PARI
    a(n) = prod(k = 1, n, 25^k - 1); \\ Amiram Eldar, Jul 14 2025

Formula

From Amiram Eldar, Jul 14 2025: (Start)
a(n) = Product_{k=1..n} (25^k-1).
a(n) ~ c * 25^(n*(n+1)/2), where c = Product_{k>=1} (1 - 1/25^k) = 0.958400102563... . (End)

A263394 a(n) = Product_{i=1..n} (3^i - 2^i).

Original entry on oeis.org

1, 5, 95, 6175, 1302925, 866445125, 1784010512375, 11248186280524375, 215638979183932793125, 12512451767147700321078125, 2190917791975795178520458609375, 1155369543009475708416871245360859375, 1832567448623162714866960405275465241328125
Offset: 1

Views

Author

Bob Selcoe, Mar 02 2016

Keywords

Comments

Generally, for sequences of the form a(n) = Product_{i=1..n} j^i-k^i, where j>k>=1 and n>=1: given probability p=(k/j)^n that an outcome will occur at the n-th stage of an infinite process, then r = 1 - a(n)/j^((n^2+n)/2) is the probability that the outcome has occurred up to and including the n-th iteration. Here, j=3 and k=2, so p=(2/3)^n and r = 1-a(n)/A047656(n+1). The limiting ratio of r ~ 0.9307279.

Crossrefs

Cf. sequences of the form Product_{i=1..n}(j^i - 1): A005329 (j=2), A027871 (j=3), A027637 (j=4), A027872 (j=5), A027873 (j=6), A027875 (j=7),A027876 (j=8), A027877 (j=9), A027878 (j=10), A027879 (j=11), A027880 (j=12).
Cf. sequences of the form Product_{i=1..n}(j^i - k^1), k>1: A269576 (j=4, k=3), A269661 (j=5, k=4).

Programs

Formula

a(n) = Product_{i=1..n} A001047(i).
a(n) ~ c * 3^(n*(n+1)/2), where c = QPochhammer(2/3) = 0.0692720728018644... . - Vaclav Kotesovec, Oct 10 2016
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