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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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

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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

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).

A222170 a(n) = n^2 + 2*floor(n^2/3).

Original entry on oeis.org

0, 1, 6, 15, 26, 41, 60, 81, 106, 135, 166, 201, 240, 281, 326, 375, 426, 481, 540, 601, 666, 735, 806, 881, 960, 1041, 1126, 1215, 1306, 1401, 1500, 1601, 1706, 1815, 1926, 2041, 2160, 2281, 2406, 2535, 2666, 2801, 2940, 3081, 3226, 3375, 3526, 3681, 3840
Offset: 0

Views

Author

Bruno Berselli, Aug 08 2013

Keywords

Comments

Also, a(n) = n^2 + floor(2*n^2/3), since 2*floor(n^2/3) = floor(2*n^2/3).

Crossrefs

Subsequence of A008851.
Cf. A004773 (numbers of the type n+floor(n/3)), A008810 (numbers of the type n^2-2*floor(n^2/3)), A047220 (numbers of the type n+floor(2*n/3)), A184637 (numbers of the type n^2+floor(n^2/3), except the first two).

Programs

  • Magma
    [n^2+2*Floor(n^2/3): n in [0..50]];
    
  • Magma
    I:=[0,1,6,15,26]; [n le 5 select I[n] else 2*Self(n-1)-Self(n-2)+Self(n-3)-2*Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
  • Mathematica
    Table[n^2 + 2 Floor[n^2/3], {n, 0, 50}]
    CoefficientList[Series[x (1 + x) (1 + 3 x + x^2) / ((1 + x + x^2) (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
    LinearRecurrence[{2, -1, 1, -2, 1}, {0, 1, 6, 15, 26}, 50] (* Hugo Pfoertner, Jan 17 2023 *)

Formula

G.f.: x*(1+x)*(1 + 3*x + x^2)/((1 + x + x^2)*(1-x)^3).
a(n) = a(-n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
a(n) = floor(5*n^2/3). - Wesley Ivan Hurt, Mar 16 2015
a(n) = a(n-3) + 5*(2n-3) [Tadeusz Dorozinski]. - Eduard Baumann, Jan 18 2023

A272915 a(n) = n + floor(5*n/6).

Original entry on oeis.org

0, 1, 3, 5, 7, 9, 11, 12, 14, 16, 18, 20, 22, 23, 25, 27, 29, 31, 33, 34, 36, 38, 40, 42, 44, 45, 47, 49, 51, 53, 55, 56, 58, 60, 62, 64, 66, 67, 69, 71, 73, 75, 77, 78, 80, 82, 84, 86, 88, 89, 91, 93, 95, 97, 99, 100, 102, 104, 106, 108, 110, 111, 113, 115, 117, 119
Offset: 0

Views

Author

Bruno Berselli, Jun 15 2016

Keywords

Comments

Equivalently, numbers congruent to {0, 1, 3, 5, 7, 9} mod 11.
In general, n + floor((k-1)*n/k) provides the numbers congruent to {0, 1, 3, 5, ..., 2*k-3} mod (2*k-1) for k>1.

Crossrefs

Cf. similar sequences with formula n+floor((k-1)*n/k): A032766 (k=2), A047220 (k=3), A047392 (k=4), A187318 (k=5).

Programs

  • Magma
    [n+Floor(5*n/6): n in [0..70]];
  • Mathematica
    Table[n + Floor[5 n/6], {n, 0, 70}]
  • Maxima
    makelist(n+floor(5*n/6), n, 0, 70);
    
  • PARI
    vector(70, n, n--; n+floor(5*n/6))
    
  • Python
    [n+int(5*n/6) for n in range(70)]
    
  • Sage
    [n+floor(5*n/6) for n in range(70)];
    

Formula

G.f.: x*(1 + 2 x + 2 x^2 + 2 x^3 + 2 x^4 + 2 x^5)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5)).
a(n) = a(n-1) + a(n-6) - a(n-7).
a(6*k + r) = 11*k + 2*r - (1 - (-1)^a(r))/2, with r = 0..5.
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