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.

Showing 1-10 of 11 results. Next

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

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Author

Keywords

Examples

			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, ...
    ...
		

Crossrefs

Rows n=3..5 are A069778, A069779, A218503.
Main diagonal gives A347611.

Programs

  • Maple
    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
  • Mathematica
    (* Returns the rectangular array *) Table[Table[QFactorial[n, q], {q, 0, 6}], {n, 0, 6}] (* Geoffrey Critzer, May 21 2017 *)
  • PARI
    T(n,q)=prod(k=1, n, ((q^k - 1) / (q - 1))) \\ Andrew Howroyd, Feb 19 2018

Formula

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

Extensions

Name edited by Michel Marcus, Sep 08 2021

A015001 q-factorial numbers for q=3.

Original entry on oeis.org

1, 1, 4, 52, 2080, 251680, 91611520, 100131391360, 328430963660800, 3232089113385932800, 95424198983606279987200, 8452007576574959037306265600, 2245867453247498115393020895232000, 1790317944898228845164815929864036352000
Offset: 0

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Keywords

Comments

a(n) is the number of maximal chains in the lattice of subspaces of an n-dimensional vector space over GF(3). - Geoffrey Critzer, Sep 07 2022

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (3^n-1)*Self(n-1)/2: n in [1..15]]; // Vincenzo Librandi, Oct 22 2012
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==((3^n - 1) * a[n-1])/2}, a, {n,15}] (* Vincenzo Librandi, Oct 27 2012 *)
    Table[QFactorial[n, 3], {n, 15}] (* Bruno Berselli, Aug 14 2013 *)

Formula

a(n) = Product_{k=1..n} (q^k - 1) / (q - 1).
a(0) = 1, a(n) = (3^n - 1)*a(n-1)/2. - Vincenzo Librandi, Oct 27 2012
a(n) = (Product_{i=0..n-1} (q^n-q^i))/((q-1)^n*q^binomial(n,2)) = A053290(n)/(A000079(n)*A047656(n)). - Geoffrey Critzer, Sep 07 2022
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A003462(k).
a(n) ~ c * 3^(n*(n+1)/2)/2^n, where c = A100220. (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A015002 q-factorial numbers for q=4.

Original entry on oeis.org

1, 1, 5, 105, 8925, 3043425, 4154275125, 22686496457625, 495586515116818125, 43304845277422684580625, 15136126045591163828042953125, 21161832960467051739150680807015625, 118345540457280742481284963098558216328125, 2647344887069536899904944217513732945696167890625
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (4^n-1)*Self(n-1)/3: n in [1..15]]; // Vincenzo Librandi, Oct 22 2012
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==((4^n - 1) * a[n-1])/3}, a, {n, 15}] (* Vincenzo Librandi, Oct 27 2012 *)
    Table[QFactorial[n, 4], {n, 15}] (* Bruno Berselli, Aug 14 2013 *)

Formula

a(n) = Product_{k=1..n} (q^k - 1) / (q - 1) with q=4.
a(0) = 1, a(n) = (4^n-1)*a(n-1)/3. - Vincenzo Librandi, Oct 27 2012
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A002450(k).
a(n) ~ c * 2^(n*(n+1))/3^n, where c = A100221. (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A015004 q-factorial numbers for q=5.

Original entry on oeis.org

1, 1, 6, 186, 29016, 22661496, 88515803376, 1728802155736656, 168827903320618878336, 82435457461295106532780416, 201258420458750640859769304304896, 2456767777551003294245070550498298923776, 149949204558598784020761783280669552547300269056
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (5^n-1)*Self(n-1)/4: n in [1..15]]; // Vincenzo Librandi, Oct 25 2012
    
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==((5^n - 1)* a[n-1])/4}, a, {n, 15}] (* Vincenzo Librandi, Oct 25 2012 *)
    Table[QFactorial[n, 5], {n, 15}] (* Bruno Berselli, Aug 14 2013 *)
  • PARI
    a(n) = { my(q=5); prod(k=1, n, ((q^k - 1) / (q - 1))) } \\ Andrew Howroyd, Feb 18 2018

Formula

a(n) = Product_{k=1..n} ((q^k - 1) / (q - 1)) where q = 5.
a(0) = 1, a(n) = (5^n-1)*a(n-1)/4. - Vincenzo Librandi, Oct 25 2012
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A003463(k).
a(n) ~ c * 5^(n*(n+1)/2)/4^n, where c = A100222. (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A015006 q-factorial numbers for q=7.

Original entry on oeis.org

1, 1, 8, 456, 182400, 510902400, 10017774259200, 1375009641495014400, 1321109263548409835520000, 8885253784030448738183147520000, 418310711031156574478261944188764160000, 137856159231156714984163673320892478249861120000
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (7^n-1)*Self(n-1)/6: n in [1..15]]; // Vincenzo Librandi, Oct 25 2012
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==((7^n - 1) * a[n-1])/6}, a, {n, 15}] (* Vincenzo Librandi, Oct 25 2012 *)
    Table[QFactorial[n, 7], {n, 15}] (* Bruno Berselli, Aug 14 2013 *)

Formula

a(n) = Product_{k=1..n} (7^k-1)/(7-1).
a(0) = 1, a(n) = (7^n - 1)*a(n-1)/6. - Vincenzo Librandi, Oct 25 2012
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A023000(k).
a(n) ~ c * 7^(n*(n+1)/2)/6^n, where c = A132035. (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A015007 q-factorial numbers for q=8.

Original entry on oeis.org

1, 1, 9, 657, 384345, 1799118945, 67375205371305, 20185139902805378865, 48378633136349277767794425, 927610024989668734297857360967425, 142287668466497494704440569679875994730825, 174605966461872393482359052970987514818406771638225
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (8^n-1)*Self(n-1)/7: n in [1..15]]; // Vincenzo Librandi, Oct 26 2012
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==((8^n - 1) * a[n-1])/7}, a, {n, 15}] (* Vincenzo Librandi, Oct 26 2012 *)
    Table[QFactorial[n, 8], {n, 15}] (* Bruno Berselli, Aug 14 2013 *)

Formula

a(n) = Product_{k=1..n} ((q^k - 1) / (q - 1)), with q=8.
a(0) = 1, a(n) = (8^n-1)*a(n-1)/7. - Vincenzo Librandi, Oct 26 2012
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A023001(k).
a(n) ~ c * 8^(n*(n+1)/2)/7^n, where c = A132036. (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A015008 q-factorial numbers for q=9.

Original entry on oeis.org

1, 1, 10, 910, 746200, 5507702200, 365876657146000, 218747042884536166000, 1177042838234827583459440000, 57001313848230245122464621625840000, 24843911488189148287648216529610193612000000, 97453533413342456299179976631323547842824103012000000
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (9^n - 1)*Self(n-1)/8: n in [1..15]]; // Vincenzo Librandi, Oct 26 2012
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==((9^n - 1) * a[n-1])/8}, a, {n, 15}] (* Vincenzo Librandi, Oct 26 2012 *)
    Table[QFactorial[n, 9], {n, 15}] (* Bruno Berselli, Aug 14 2013 *)

Formula

a(n) = Product_{k=1..n} (9^k - 1) / (9 - 1).
a(0) = 1, a(n) = (9^n - 1)*a(n-1)/8. - Vincenzo Librandi, Oct 26 2012
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A002452(k).
a(n) ~ c * 3^(n*(n+1))/8^n, where c = A132037. (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A015011 q-factorial numbers for q=11.

Original entry on oeis.org

1, 1, 12, 1596, 2336544, 37630041120, 6666387564654720, 12990902775831251994240, 278471536921607824648305285120, 65662131721505488121539650946349537280, 170310659060181679663863033233125976844488908800, 4859161865915056755501262525796512204608930674134393036800
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (11^n-1)*Self(n-1)/10: n in [1..15]]; // Vincenzo Librandi, Oct 26 2012
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==((11^n - 1) * a[n-1])/10}, a, {n, 15}] (* Vincenzo Librandi, Oct 26 2012 *)
    Table[QFactorial[n, 11], {n, 11}] (* Bruno Berselli, Aug 14 2013 *)

Formula

a(n) = Product_{k=1..n} (11^k - 1) / (11 - 1).
a(0) = 1, a(n) = (11^n - 1)*a(n-1)/10. - Vincenzo Librandi, Oct 26 2012
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A016123(k-1).
a(n) ~ c * 11^(n*(n+1)/2)/10^n, where c = A132267. (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A015005 q-factorial numbers for q=6.

Original entry on oeis.org

1, 1, 7, 301, 77959, 121226245, 1131162092095, 63330372050122765, 21274128570193389587095, 42878835824239014254983869205, 518543838148941095553869851505328175, 37625235473766496167083515195884075739704925, 16380389585902052954270520869620904155598347770499975
Offset: 0

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Author

Keywords

Crossrefs

Programs

  • Magma
    [n le 1 select 1 else (6^n-1)*Self(n-1)/5: n in [1..15]]; // Vincenzo Librandi, Oct 25 2012
  • Mathematica
    RecurrenceTable[{a[1]==1, a[n]==((6^n - 1) * a[n-1])/5}, a, {n, 15}] (* Vincenzo Librandi, Oct 25 2012 *)
    Table[QFactorial[n, 6], {n, 15}] (* Bruno Berselli, Aug 14 2013 *)

Formula

a(n) = Product_{k=1..n} (6^k-1)/(6-1).
a(0) = 1, a(n) = (6^n-1)*a(n-1)/5. - Vincenzo Librandi, Oct 25 2012
From Amiram Eldar, Jul 05 2025: (Start)
a(n) = Product_{k=1..n} A003464(k).
a(n) ~ c * 6^(n*(n+1)/2)/5^n, where c = A132034. (End)

Extensions

a(0)=1 prepended by Alois P. Heinz, Sep 08 2021

A334131 Numbers that can be written as a product of distinct repunits.

Original entry on oeis.org

0, 1, 11, 111, 1111, 1221, 11111, 12221, 111111, 122221, 123321, 1111111, 1222221, 1233321, 1356531, 11111111, 12222221, 12333321, 12344321, 13566531, 111111111, 122222221, 123333321, 123444321, 135666531, 135787531, 1111111111, 1222222221
Offset: 1

Views

Author

Ilya Gutkovskiy, Apr 14 2020

Keywords

Examples

			13566531 = 11*111*11111. - _David A. Corneth_, Mar 26 2021
		

Crossrefs

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