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 1-3 of 3 results.

A317980 a(n) = Product_{i=1..n} floor(5*i/2).

Original entry on oeis.org

2, 10, 70, 700, 8400, 126000, 2142000, 42840000, 942480000, 23562000000, 636174000000, 19085220000000, 610727040000000, 21375446400000000, 790891516800000000, 31635660672000000000, 1328697748224000000000, 59791398670080000000000, 2810195737493760000000000
Offset: 1

Views

Author

Vaclav Kotesovec, Oct 02 2018

Keywords

Comments

If p > 2 and p is odd, then Product_{i=1..n} floor(p*i/2) ~ (p/2)^n * n! * 2^(1/(2*p)) * sqrt(Pi) / (Gamma(1/2 - 1/(2*p)) * n^(1/(2*p))).

Links

Crossrefs

Cf. A010786, A180736, A275062, A319948, A319950.

Programs

  • Mathematica
    Table[Product[Floor[i*5/2], {i, 1, n}], {n, 1, 20}]
RecurrenceTable[{4 a[n] - 10 a[n - 1] - 5 (n - 1) (5 n - 6) a[n - 2] == 0, a[1] == 2, a[2] == 10}, a, {n, 1, 20}] (* Bruno Berselli, Oct 03 2018 *)
FoldList[Times,Floor[5*Range[20]/2]] (* Harvey P. Dale, Sep 17 2020 *)

Formula

a(n) ~ (5/2)^n * n! * 2^(1/10) * sqrt(Pi) / (Gamma(2/5) * n^(1/10)).
Recurrence: 4*a(n) - 10*a(n-1) - 5*(n - 1)*(5*n - 6)*a(n-2) = 0, with n >= 3. - Bruno Berselli, Oct 03 2018

A319949 a(n) = Product_{i=1..n} floor(4*i/3).

Original entry on oeis.org

1, 2, 8, 40, 240, 1920, 17280, 172800, 2073600, 26956800, 377395200, 6038323200, 102651494400, 1847726899200, 36954537984000, 776045297664000, 17072996548608000, 409751917166592000, 10243797929164800000, 266338746158284800000
Offset: 1

Views

Author

Vaclav Kotesovec, Oct 02 2018

Keywords

Links

Crossrefs

Cf. A004773, A010786, A180736, A275062, A319948, A319950.

Programs

  • Mathematica
    Table[Product[Floor[i*4/3], {i, 1, n}], {n, 1, 20}]
RecurrenceTable[{27*(3*n - 7)*a[n] == 54*(2*n - 5)*a[n-1] + 12*(12*n^2 - 42*n + 35)*a[n-2] + 8*(n-2)*(2*n - 5)*(3*n - 4)*(4*n - 9)*a[n-3], a[1]==1, a[2]==2, a[3]==8}, a, {n, 1, 20}]
FoldList[Times,Floor[4 Range[20]/3]] (* Harvey P. Dale, Mar 21 2024 *)
  • PARI
    a(n) = prod(i=1, n, (4*i)\3); \\ Michel Marcus, Oct 03 2018

Formula

a(n) ~ (4/3)^n * n! * 2*sqrt(Pi) / (3^(1/4) * Gamma(1/4) * n^(1/4)).
Recurrence: 27*(3*n - 7)*a(n) = 54*(2*n - 5)*a(n-1) + 12*(12*n^2 - 42*n + 35)*a(n-2) + 8*(n-2)*(2*n - 5)*(3*n - 4)*(4*n - 9)*a(n-3).

A319950 a(n) = Product_{i=1..n} floor(5*i/3).

Original entry on oeis.org

1, 3, 15, 90, 720, 7200, 79200, 1029600, 15444000, 247104000, 4447872000, 88957440000, 1868106240000, 42966443520000, 1074161088000000, 27928188288000000, 781989272064000000, 23459678161920000000, 727250023019520000000, 23999250759644160000000
Offset: 1

Views

Author

Vaclav Kotesovec, Oct 02 2018

Keywords

Comments

If p > 3 and gcd(p,3)=1 then Product_{i=1..n} floor(i*p/3) ~ (p/3)^n * n! * 2*Pi * 3^(1/p - 1/2) / (c(p) * n^(1/p)), where
c(p) = Gamma(2/3 - 2/(3*p)) * Gamma(1/3 - 1/(3*p)) if mod(p, 3) = 1,
c(p) = Gamma(1/3 - 2/(3*p)) * Gamma(2/3 - 1/(3*p)) if mod(p, 3) = 2.
In general, if q > 1, p > q and gcd(p,q)=1, then Product_{i=1..n} floor(i*p/q) ~ c(p,q) * (p/q)^n * n! / n^((q-1)/(2*p)), where c(p,q) is a constant.

Crossrefs

Cf. A010786, A047220, A180736, A275062, A319948, A319949, A317980.

Programs

  • Mathematica
    Table[Product[Floor[i*5/3], {i, 1, n}], {n, 1, 20}]
RecurrenceTable[{27*(15*n - 32)*a[n] == 675*(n-2)*a[n-1] + 15*(75*n^2 - 255*n + 194)*a[n-2] + 5*(n-2)*(5*n - 12)*(5*n - 11)*(15*n - 17)*a[n-3], a[1]==1, a[2]==3, a[3]==15}, a, {n, 1, 20}]
  • PARI
    a(n) = prod(i=1, n, (5*i)\3); \\ Michel Marcus, Oct 03 2018

Formula

a(n) ~ (5/3)^n * n! * 2*Pi / (3^(3/10) * Gamma(1/5) * Gamma(3/5) * n^(1/5)).
Recurrence: 27*(15*n - 32)*a(n) = 675*(n-2)*a(n-1) + 15*(75*n^2 - 255*n + 194)*a(n-2) + 5*(n-2)*(5*n - 12)*(5*n - 11)*(15*n - 17)*a(n-3).
Showing 1-3 of 3 results.

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