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.

User: Justin Gaetano

Justin Gaetano's wiki page.

Justin Gaetano has authored 3 sequences.

A295867 Numbers of the form A000217(n)*A007494(n) that are divisible by 3.

Original entry on oeis.org

0, 9, 30, 60, 120, 189, 432, 630, 825, 1122, 1404, 2205, 2760, 3264, 3978, 4617, 6300, 7392, 8349, 9660, 10800, 13689, 15498, 17052, 19140, 20925, 25344, 28050, 30345, 33390, 35964, 42237, 46020, 49200, 53382, 56889, 65340, 70380, 74589, 80088, 84672, 95625, 102102, 107484, 114480, 120285, 134064, 142158
Offset: 0

Views

Author

Justin Gaetano, Feb 13 2018

Keywords

Examples

			For n = 1, let m = floor((6*1+4)/5) = 2, a(1) = (3*2/2)*binomial(2+1,2) = 3*3 = 9.
For n = 2, let m = floor((6*2+4)/5) = 3, a(2) = ((3*3+1)/2)*binomial(3+1,2) = 5*6 = 30.

Links

Crossrefs

Cf. A000217, A007494, A047248.

Programs

  • Mathematica
    Select[Array[Floor[(3 # + 1)/2] (# + 1) #/2 &, 58, 0], Divisible[#, 3] &] (* Michael De Vlieger, Feb 17 2018 *)
LinearRecurrence[{1, 0, 0, 0, 3, -3, 0, 0, 0, -3, 3, 0, 0, 0, 1, -1}, {0, 9, 30, 60, 120, 189, 432, 630, 825, 1122, 1404, 2205, 2760, 3264, 3978, 4617, 6300}, 48] (* Robert G. Wilson v, Feb 19 2018 *)
  • PARI
    a(n) = my(k=n%5); 162/125*n^3 + if(k==0, 27/25*n^2, k==1, 459/125*n^2 + 396/125*n + 108/125, k==2, 423/125*n^2 + 339/125*n + 84/125, k==3, 297/125*n^2 + 144/125*n + 21/125, 261/125*n^2 + 111/125*n + 12/125) \\ Charles R Greathouse IV, Feb 20 2018
  • PARI
    concat(0, Vec(3*x*(3 + 7*x + 10*x^2 + 20*x^3 + 23*x^4 + 72*x^5 + 45*x^6 + 35*x^7 + 39*x^8 + 25*x^9 + 33*x^10 + 8*x^11 + 3*x^12 + x^13) / ((1 - x)^4*(1 + x + x^2 + x^3 + x^4)^3) + O(x^60))) \\ Colin Barker, Mar 01 2018

Formula

Let m = floor((6*n+4)/5), a(n) = (3*m/2)*binomial(m+1,2) if m is even, otherwise ((3*m+1)/2)*binomial(m+1,2).
Conjectures from Colin Barker, Feb 15 2018: (Start)
G.f.: 3*x*(3 + 7*x + 10*x^2 + 20*x^3 + 23*x^4 + 72*x^5 + 45*x^6 + 35*x^7 + 39*x^8 + 25*x^9 + 33*x^10 + 8*x^11 + 3*x^12 + x^13) / ((1 - x)^4*(1 + x + x^2 + x^3 + x^4)^3).
a(n) = a(n-1) + 3*a(n-5) - 3*a(n-6) - 3*a(n-10) + 3*a(n-11) + a(n-15) - a(n-16) for n>15.
(End)
Colin Barker's conjecture is true. This is a cubic quasipolynomial of order 5: a(n) = 162/125*n^3 + 27/25*n^2 if n is 0 mod 5, 162/125*n^3 + 261/125*n^2 + 111/125*n + 12/125 if n is 4 mod 5, 162/125*n^3 + 297/125*n^2 + 144/125*n + 21/125 if n is 3 mod 5, 162/125*n^3 + 423/125*n^2 + 339/125*n + 84/125 if n is 2 mod 5, and 162/125*n^3 + 459/125*n^2 + 396/125*n + 108/125 if n is 1 mod 5. Generally a(n) = 162/125*n^3 + O(n^2). - Charles R Greathouse IV, Feb 20 2018

A299412 Pentagonal pyramidal numbers divisible by 3.

Original entry on oeis.org

0, 6, 18, 75, 126, 288, 405, 726, 936, 1470, 1800, 2601, 3078, 4200, 4851, 6348, 7200, 9126, 10206, 12615, 13950, 16896, 18513, 22050, 23976, 28158, 30420, 35301, 37926, 43560, 46575, 53016, 56448, 63750, 67626, 75843, 80190, 89376, 94221, 104430, 109800, 121086, 127008, 139425, 145926, 159528, 166635
Offset: 0

Views

Author

Justin Gaetano, Feb 20 2018

Keywords

Examples

			The first 6 pentagonal pyramidal numbers are 0, 1, 6, 18, 40, 75; of these, 0, 6, 18, 75 are divisible by 3.

Links

Crossrefs

Cf. A001318, A002411, A007494, A117748, A269416.

Programs

  • Magma
    [IsEven(n) select (3*n/2)^2*(3*n/2+1)/2 else ((3*n+1)/2)^2*((3*n+1)/2+1)/2: n in [0..50] ]; // Vincenzo Librandi, Mar 14 2018
  • Maple
    f:= proc(n) if n::even then (3*n/2)^2*(3*n/2+1)/2 else
((3*n+1)/2)^2*((3*n+1)/2+1)/2 fi end proc:
map(f, [$0..100]); # Robert Israel, Feb 28 2018
  • Mathematica
    Array[((3 #1 + #2)/2)^2*((3 #1 + #2)/2 + 1)/2 & @@ {#, Boole@ OddQ@ #} &, 47, 0] (* Michael De Vlieger, Feb 21 2018 *)
LinearRecurrence[{1,3,-3,-3,3,1,-1},{0,6,18,75,126,288,405},50] (* Harvey P. Dale, Jul 16 2021 *)
  • PARI
    lista(nn) = {for (n=0, nn, if (!(n^2*(n+1)/2 % 3), print1(n^2*(n+1)/2, ", ")););} \\ Michel Marcus, Feb 21 2018
  • PARI
    x='x+O('x^99); concat(0, Vec(3*x*(3*x^4+5*x^3+13*x^2+4*x+2)/((x-1)^4*(x+1)^3))) \\ Altug Alkan, Mar 14 2018

Formula

a(n) = A007494(n)*A117748(n).
a(n) = (3*n/2)^2*(3*n/2+1)/2 if n even.
a(n) = ((3*n+1)/2)^2*((3*n+1)/2+1)/2 if n odd.
From Omar E. Pol, Feb 21 2018: (Start)
a(n) = 3*A001318(n)*A007494(n).
a(n) = A001318(n)*abs(A269416(n-1)), n >= 1. (End)
G.f.: 3*x*(3*x^4 + 5*x^3 + 13*x^2 + 4*x + 2)/((x-1)^4*(x+1)^3). - Robert Israel, Feb 28 2018

A293656 a(n) = binomial(n+1,2)*n!/n!!.

Original entry on oeis.org

1, 3, 12, 30, 120, 315, 1344, 3780, 17280, 51975, 253440, 810810, 4193280, 14189175, 77414400, 275675400, 1579253760, 5892561675, 35300966400, 137493105750, 858370867200, 3478575575475, 22562891366400, 94870242967500, 637646929920000, 2774954606799375
Offset: 1

Views

Author

Justin Gaetano, Feb 06 2018

Keywords

Comments

It appears that the difference between a(n)/a(n-1) and a(n-1)/a(n-2) approaches some factor, 5 < x < 7, as n --> oo.
It appears that 3|a(n) for n > 1.

Examples

			For n = 6, a(6) = binomial(6+1,2)/(6!!/6!) = 315.

References

  • L. Euler and J. L. Lagrange, Elements of Algebra, J. Johnson and Co. 1810. See pages 332-335.

Crossrefs

Cf. A000217, A000142, A006882.

Programs

  • Mathematica
    Table[Binomial[n+1,2] n!/n!!,{n,30}] (* Harvey P. Dale, Jul 27 2021 *)

Formula

a(n) = ((n*(n+1))/2)/(Product_{i=0..floor((n-1)/2),n-2*i}/Product_{i=1..n}).
From Chai Wah Wu, Feb 07 2018: (Start)
a(n) = n*(n+1)!!/2.
a(n)/a(n-1) = ((n+1)!!/n!!)*(n/(n-1)) = n/b*(n-1) if n is even and n*Pi/(2*b*(n-1)) if n is odd where b = Integral_{x=0..(Pi/2)} sin^(n+1)*x dx.
Since b -> 0 as n -> oo, a(n)/a(n-1) is unbounded as n -> oo. On the other hand, a(n)/a(n-1) and a(n-1)/a(n-2) differ by a multiplicative factor of approximately Pi/2.
(End)

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

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