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.

Previous Showing 11-17 of 17 results.

A159083 Products of 7 consecutive integers.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 5040, 40320, 181440, 604800, 1663200, 3991680, 8648640, 17297280, 32432400, 57657600, 98017920, 160392960, 253955520, 390700800, 586051200, 859541760, 1235591280, 1744364160, 2422728000, 3315312000, 4475671200, 5967561600, 7866331200
Offset: 0

Views

Author

Zerinvary Lajos, Apr 05 2009

Keywords

Links

Crossrefs

Cf. A052762, A052787, A053625, A173333.
Equals A008279(n,7) (for n>=7).

Programs

  • Magma
    I:=[0,0,0,0,0,0,0,5040]; [n le 8 select I[n] else 8*Self(n-1) - 28*Self(n-2) +56*Self(n-3) -70*Self(n-4) +56*Self(n-5) -28*Self(n-6) +8*Self(n-7) -Self(n-8): n in [1..30]]; // G. C. Greubel, Jun 28 2018
  • Maple
    G(x):=x^7*exp(x): f[0]:=G(x): for n from 1 to 36 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..33);
  • Mathematica
    Table[Times@@(n+Range[0,6]),{n,-6,25}] (* or *) LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{0,0,0,0,0,0,0,5040},30] (* Harvey P. Dale, Apr 07 2018 *)
  • PARI
    my(x='x+O('x^30)); concat([0,0,0,0,0,0,0], Vec(5040*x^7/(1-x)^8)) \\ G. C. Greubel, Jun 28 2018

Formula

E.g.f.: x^7*exp(x).
For n>=8: a(n) = A173333(n,n-7). - Reinhard Zumkeller, Feb 19 2010
G.f.: 5040*x^7/(1-x)^8. - Colin Barker, Mar 27 2012
From Amiram Eldar, Mar 08 2022: (Start)
a(n) = n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*(n-6) = n!/(n-7)!.
Sum_{n>=7} 1/a(n) = 1/4320.
Sum_{n>=7} (-1)^(n+1)/a(n) = 4*log(2)/45 - 1327/21600. (End)

A162995 A scaled version of triangle A162990.

Original entry on oeis.org

1, 3, 1, 12, 4, 1, 60, 20, 5, 1, 360, 120, 30, 6, 1, 2520, 840, 210, 42, 7, 1, 20160, 6720, 1680, 336, 56, 8, 1, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
Offset: 1

Views

Author

Johannes W. Meijer, Jul 27 2009

Keywords

Comments

We get this scaled version of triangle A162990 by dividing the coefficients in the left hand columns by their 'top-values' and then taking the square root.
T(n,k) = A173333(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Feb 19 2010
T(n,k) = A094587(n+1,k+1), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

Examples

			The first few rows of the triangle are:
[1]
[3, 1]
[12, 4, 1]
[60, 20, 5, 1]

Links

Crossrefs

Cf. A094587.
A056542(n) equals the row sums for n>=1.
A001710, A001715, A001720, A001725, A001730, A049388, A049389, A049398, A051431 are related to the left hand columns.
A000012, A009056, A002378, A007531, A052762, A052787, A053625 and A159083 are related to the right hand columns.

Programs

  • Haskell
    a162995 n k = a162995_tabl !! (n-1) !! (k-1)
a162995_row n = a162995_tabl !! (n-1)
a162995_tabl = map fst $ iterate f ([1], 3)
   where f (row, i) = (map (* i) row ++ [1], i + 1)
-- Reinhard Zumkeller, Jul 04 2012
  • Maple
    a := proc(n, m): (n+1)!/(m+1)! end: seq(seq(a(n, m), m=1..n), n=1..9); # Johannes W. Meijer, revised Nov 23 2012
  • Mathematica
    Table[(n+1)!/(m+1)!, {n, 10}, {m, n}] (* Paolo Xausa, Mar 31 2024 *)

Formula

a(n,m) = (n+1)!/(m+1)! for n = 1, 2, 3, ..., and m = 1, 2, ..., n.

A001095 a(n) = n + n*(n-1)*(n-2)*(n-3)*(n-4).

Original entry on oeis.org

0, 1, 2, 3, 4, 125, 726, 2527, 6728, 15129, 30250, 55451, 95052, 154453, 240254, 360375, 524176, 742577, 1028178, 1395379, 1860500, 2441901, 3160102, 4037903, 5100504, 6375625, 7893626, 9687627, 11793628, 14250629, 17100750, 20389351
Offset: 0

Views

Author

N. J. A. Sloane, Ray Wills (rwills(AT)vmprofs.estec.esa.nl)

Keywords

Links

Crossrefs

Equals A052787(n) + n.

Programs

  • GAP
    List([0..35], n-> n + 120*Binomial(n,5)); # G. C. Greubel, Aug 26 2019
  • Magma
    [n + n*(n-1)*(n-2)*(n-3)*(n-4): n in [0..35]]; // Vincenzo Librandi, Apr 30 2011
  • Maple
    seq(n + 5!*binomial(n,5), n=0..35); # G. C. Greubel, Aug 26 2019
  • Mathematica
    Table[n+Times@@(n-Range[0,4]),{n,0,40}] (* or *)  LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,2,3,4,125},40] (* Harvey P. Dale, Oct 08 2017 *)
  • PARI
    vector(35, n, (n-1) + 5!*binomial(n-1,5)) \\ G. C. Greubel, Aug 26 2019
  • Sage
    [n + 120*binomial(n,5) for n in (0..35)] # G. C. Greubel, Aug 26 2019

Formula

G.f.: x*(1 - 4*x + 6*x^2 - 4*x^3 + 121*x^4)/(1-x)^6. - Colin Barker, Jun 25 2012
From G. C. Greubel, Aug 26 2019: (Start)
a(n) = n + 5!*binomial(n,5).
E.g.f.: x*(1 + x^4)*exp(x). (End)

Extensions

More terms from James Sellers, Sep 19 2000

A293617 Array of triangles read by ascending antidiagonals, T(m, n, k) = Pochhammer(m, k) * Stirling2(n + m, k + m) with m >= 0, n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 2, 1, 0, 1, 10, 3, 7, 3, 0, 1, 15, 4, 25, 12, 2, 0, 1, 21, 5, 65, 30, 6, 1, 0, 1, 28, 6, 140, 60, 12, 15, 7, 0, 1, 36, 7, 266, 105, 20, 90, 50, 12, 0, 1, 45, 8, 462, 168, 30, 350, 195, 60, 6, 0, 1, 55, 9, 750, 252, 42, 1050, 560, 180, 24, 1, 0
Offset: 0

Views

Author

Peter Luschny, Oct 20 2017

Keywords

Examples

			Array starts:
m\j| 0   1  2     3       4       5       6       7       8       9      10
---|-----------------------------------------------------------------------
m=0| 1,  0, 0,    0,      0,      0,      0,      0,      0,      0,      0
m=1| 1,  1, 1,    1,      3,      2,      1,      7,     12,      6,      1
m=2| 1,  3, 2,    7,     12,      6,     15,     50,     60,     24,     31
m=3| 1,  6, 3,   25,     30,     12,     90,    195,    180,     60,    301
m=4| 1, 10, 4,   65,     60,     20,    350,    560,    420,    120,   1701
m=5| 1, 15, 5,  140,    105,     30,   1050,   1330,    840,    210,   6951
m=6| 1, 21, 6,  266,    168,     42,   2646,   2772,   1512,    336,  22827
m=7| 1, 28, 7,  462,    252,     56,   5880,   5250,   2520,    504,  63987
m=8| 1, 36, 8,  750,    360,     72,  11880,   9240,   3960,    720, 159027
m=9| 1, 45, 9, 1155,    495,     90,  22275,  15345,   5940,    990, 359502
   A000217, A001296,A027480,A002378,A001297,A293475,A033486,A007531,A001298
.
m\j| ...      11      12      13      14
---|-----------------------------------------
m=0| ...,      0,      0,      0,      0, ... [A000007]
m=1| ...,     15,     50,     60,     24, ... [A028246]
m=2| ...,    180,    390,    360,    120, ... [A053440]
m=3| ...,   1050,   1680,   1260,    360, ... [A294032]
m=4| ...,   4200,   5320,   3360,    840, ...
m=5| ...,  13230,  13860,   7560,   1680, ...
m=6| ...,  35280,  31500,  15120,   3024, ...
m=7| ...,  83160,  64680,  27720,   5040, ...
m=8| ..., 178200, 122760,  47520,   7920, ...
m=9| ..., 353925, 218790,  77220,  11880, ...
         A293476,A293608,A293615,A052762, ...
.
The parameter m runs over the triangles and j indexes the triangles by reading them by rows. Let T(m, n) denote the row [T(m, n, k) for 0 <= k <= n] and T(m) denote the triangle [T(m, n) for n >= 0]. Then for instance T(2) is the triangle A053440, T(3, 2) is row 2 of A294032 (which is [25, 30, 12]) and T(3, 2, 1) = 30.
.
Remark: To adapt the sequences A028246 and A053440 to our enumeration use the exponential generating functions exp(x)/(1 - y*(exp(x) - 1)) and exp(x)*(2*exp(x) - y*exp(2*x) + 2*y*exp(x) - 1 - y)/(1 - y*(exp(x) - 1))^2 instead of those indicated in their respective entries.

Links

Crossrefs

A000217(n) = T(n, 1, 0), A001296(n) = T(n, 2, 0), A027480(n) = T(n, 2, 1),
A002378(n) = T(n, 2, 2), A001297(n) = T(n, 3, 0), A293475(n) = T(n, 3, 1),
A033486(n) = T(n, 3, 2), A007531(n) = T(n, 3, 3), A001298(n) = T(n, 4, 0),
A293476(n) = T(n, 4, 1), A293608(n) = T(n, 4, 2), A293615(n) = T(n, 4, 3),
A052762(n) = T(n, 4, 4), A052787(n) = T(n, 5, 5), A000225(n) = T(1, n, 1),
A028243(n) = T(1, n, 2), A028244(n) = T(1, n, 3), A028245(n) = T(1, n, 4),
A032180(n) = T(1, n, 5), A228909(n) = T(1, n, 6), A228910(n) = T(1, n, 7),
A000225(n) = T(2, n, 0), A007820(n) = T(n, n, 0).
A028246(n,k) = T(1, n, k), A053440(n,k) = T(2, n, k), A294032(n,k) = T(3, n, k),
A293926(n,k) = T(n, n, k), A124320(n,k) = T(n, k, k), A156991(n,k) = T(k, n, n).
Cf. A293616.

Programs

  • Maple
    A293617 := proc(m, n, k) option remember:
if m = 0 then 0^n elif k < 0 or k > n then 0 elif n = 0 then 1 else
(k+m)*A293617(m,n-1,k) + k*A293617(m,n-1,k-1) + A293617(m-1,n,k) fi end:
for m in [$0..4] do for n in [$0..6] do print(seq(A293617(m, n, k), k=0..n)) od od;
# Sample uses:
A027480 := n -> A293617(n, 2, 1): A293608 := n -> A293617(n, 4, 2):
# Flatten:
a := proc(n) local w; w := proc(k) local t, s; t := 1; s := 1;
while t <= k do s := s + 1; t := t + s od; [s - 1, s - t + k] end:
seq(A293617(n - k, w(k)[1], w(k)[2]), k=0..n) end: seq(a(n), n = 0..11);
  • Mathematica
    T[m_, n_, k_] := Pochhammer[m, k] StirlingS2[n + m, k + m];
For[m = 0, m < 7, m++, Print[Table[T[m, n, k], {n,0,6}, {k,0,n}]]]
A293617Row[m_, n_] := Table[T[m, n, k], {k,0,n}];
(* Sample use: *)
A293926Row[n_] := A293617Row[n, n];

Formula

T(m,n,k) = (k + m)*T(m, n-1, k) + k*T(m, n-1, k-1) + T(m-1, n, k) with boundary conditions T(0, n, k) = 0^n; T(m, n, k) = 0 if k<0 or k>n; and T(m, 0, k) = 0^k.
T(m,n,k) = Pochhammer(m, k)*binomial(n + m, k + m)*NorlundPolynomial(n - k, -k - m).

A054778 5n*(5n-1)*(5n-2)*(5n-3)*(5n-4).

Original entry on oeis.org

0, 120, 30240, 360360, 1860480, 6375600, 17100720, 38955840, 78960960, 146611080, 254251200, 417451320, 655381440, 991186560, 1452361680, 2071126800, 2884801920, 3936182040, 5273912160, 6952862280, 9034502400, 11587277520, 14686982640, 18417137760
Offset: 0

Views

Author

Henry Bottomley, May 19 2000

Keywords

Links

Programs

  • Mathematica
    With[{f=Times@@(5n-Range[0,4])},Table[f,{n,0,30}]] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{0,120,30240,360360,1860480,6375600},30] (* Harvey P. Dale, Aug 29 2012 *)
  • PARI
    concat(0, Vec(120*x*(126*x^4+1246*x^3+1506*x^2+246*x+1)/(x-1)^6 + O(x^100))) \\ Colin Barker, Sep 13 2014

Formula

a(n) = A052787(5n) = 120*binomial(5*n, 5).
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6), with a(0)=0, a(1)=120, a(2)=30240, a(3)=360360, a(4)=1860480, a(5)=6375600. - Harvey P. Dale, Aug 29 2012
G.f.: 120*x*(126*x^4+1246*x^3+1506*x^2+246*x+1) / (x-1)^6. - Colin Barker, Sep 13 2014

Extensions

More terms from Colin Barker, Sep 13 2014

A300847 a(n) = 12*binomial(n, 5).

Original entry on oeis.org

0, 0, 0, 0, 0, 12, 72, 252, 672, 1512, 3024, 5544, 9504, 15444, 24024, 36036, 52416, 74256, 102816, 139536, 186048, 244188, 316008, 403788, 510048, 637560, 789360, 968760, 1179360, 1425060, 1710072, 2038932, 2416512, 2848032, 3339072, 3895584, 4523904, 5230764
Offset: 0

Views

Author

Eric W. Weisstein, Mar 13 2018

Keywords

Comments

Also the number of 5-cycles in the complete graph K_n for n >= 1.

Links

Crossrefs

Cf. A000389, A052787.

Programs

  • Mathematica
    Table[12 Binomial[n, 5], {n, 0, 20}]
12 Binomial[Range[0, 20], 5]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 0, 0, 0, 12, 72}, {0, 20}]
CoefficientList[Series[12 x^5/(x - 1)^6, {x, 0, 20}], x]
  • PARI
    a(n) = 12*binomial(n, 5); \\ Altug Alkan, Mar 13 2018

Formula

G.f.: 12*x^5/(x - 1)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) + 6*a(n-4) - a(n-5).
a(n) = A052787(n)/10 = 12*A000389(n).
a(n) = (n - 4)*(n - 3)*(n - 2)*(n - 1)*n/10.

A385631 Products of five consecutive integers whose prime divisors are consecutive primes starting at 2.

Original entry on oeis.org

120, 720, 2520, 6720, 15120, 30240, 55440, 240240, 360360
Offset: 1

Views

Author

Ken Clements, Jul 05 2025

Keywords

Examples

			a(1) = 120 = 1*2*3*4*5 = 2^3 * 3^1 * 5^1.
a(2) = 720 = 2*3*4*5*6 = 2^4 * 3^2 * 5^1.
a(3) = 2520 = 3*4*5*6*7 = 2^3 * 3^2 * 5^1 * 7^1.
a(4) = 6720 = 4*5*6*7*8 = 2^6 * 3^1 * 5^1 * 7^1.
a(5) = 15120 = 5*6*7*8*9 = 2^4 * 3^3 * 5^1 * 7^1.
a(6) = 30240 = 6*7*8*9*10 = 2^5 * 3^3 * 5^1 * 7^1.
a(7) = 55440 = 7*8*9*10*11 = 2^4 * 3^2 * 5^1 * 7^1 * 11^1.
a(8) = 240240 = 10*11*12*13*14 = 2^4 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1.
a(9) = 360360 = 11*12*13*14*15 = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1.

Crossrefs

Intersection of A052787 and A055932.
Cf. A217056, A385189, A385415.

Programs

  • Mathematica
    Select[(#*(# + 1)*(# + 2)*(# + 3)*(# + 4)) & /@ Range[12], PrimePi[(f = FactorInteger[#1])[[-1, 1]]] == Length[f] &] (* Amiram Eldar, Jul 05 2025 *)
  • Python
    from sympy import prime, primefactors
def is_pi_complete(n): # Check for complete set of
    factors = primefactors(n) # prime factors
    return factors[-1] == prime(len(factors))
def aupto(limit):
    result = []
    for i in range(1, limit+1):
        n = i * (i+1) * (i+2) * (i+3) * (i+4)
        if is_pi_complete(n):
            result.append(n)
    return result
print(aupto(1000))
Previous Showing 11-17 of 17 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.