A081142 -id:A081142 - cp's OEIS frontend

A027476 Third column of A027467.

1, 45, 1350, 33750, 759375, 15946875, 318937500, 6150937500, 115330078125, 2114384765625, 38058925781250, 674680957031250, 11806916748046875, 204350482177734375, 3503151123046875000, 59553569091796875000

Offset: 3

Olivier Gérard

nonn

Vincenzo Librandi, Table of n, a(n) for n = 3..200
Index entries for linear recurrences with constant coefficients, signature (45,-675,3375).

Sequences similar to the form q^(n-2)*binomial(n, 2): A000217 (q=1), A001788 (q=2), A027472 (q=3), A038845 (q=4), A081135 (q=5), A081136 (q=6), A027474 (q=7), A081138 (q=8), A081139 (q=9), A081140 (q=10), A081141 (q=11), A081142 (q=12), this sequence (q=15).

[(n-1)*(n-2)/2 * 15^(n-3): n in [3..20]]; // Vincenzo Librandi, Dec 29 2012

seq((15)^(n-3)*binomial(n-1, 2), n=3..30) # G. C. Greubel, May 13 2021

Table[(n-1)*(n-2)/2 * 15^(n-3), {n, 3, 30}] (* Vincenzo Librandi, Dec 29 2012 *)

[(15)^(n-3)*binomial(n-1,2) for n in (3..30)] # G. C. Greubel, May 13 2021

Numerators of sequence a[3,n] in (a[i,j])^4 where a[i,j] = binomial(i-1, j-1)/2^(i-1) if j<=i, 0 if j>i.
a(n) = 15^(n-3)*binomial(n-1, 2).
From G. C. Greubel, May 13 2021: (Start)
a(n) = 45*a(n-1) - 675*a(n-2) + 3375*a(n-3).
G.f.: x^3/(1 - 15*x)^3.
E.g.f.: (-2 + (2 - 30*x + 225*x^2)*exp(15*x))/6750. (End)
From Amiram Eldar, Jan 06 2022: (Start)
Sum_{n>=3} 1/a(n) = 30 - 420*log(15/14).
Sum_{n>=3} (-1)^(n+1)/a(n) = 480*log(16/15) - 30. (End)

A081130 Square array of binomial transforms of (0,0,1,0,0,0,...), read by antidiagonals.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 3, 0, 0, 0, 1, 6, 6, 0, 0, 0, 1, 9, 24, 10, 0, 0, 0, 1, 12, 54, 80, 15, 0, 0, 0, 1, 15, 96, 270, 240, 21, 0, 0, 0, 1, 18, 150, 640, 1215, 672, 28, 0, 0, 0, 1, 21, 216, 1250, 3840, 5103, 1792, 36, 0, 0, 0, 1, 24, 294, 2160, 9375, 21504, 20412, 4608, 45, 0

Offset: 0

Paul Barry, Mar 08 2003

Rows, of the square array, are three-fold convolutions of sequences of powers.

			The array begins as:
  0,  0,  0,   0,   0,    0, ...
  0,  0,  0,   0,   0,    0, ...
  0,  1,  1,   1,   1,    1, ... A000012
  0,  3,  6,   9,  12,   15, ... A008585
  0,  6, 24,  54,  96,  150, ... A033581
  0, 10, 80, 270, 640, 1250, ... A244729
The antidiagonal triangle begins as:
  0;
  0, 0;
  0, 0, 0;
  0, 0, 1, 0;
  0, 0, 1, 3,  0;
  0, 0, 1, 6,  6,  0;
  0, 0, 1, 9, 24, 10, 0;

G. C. Greubel, Antidiadoganal rows n = 0..50, flattened

Main diagonal: A081131.
Rows: A000012 (n=2), A008585 (n=3), A033581 (n=4), A244729 (n=5).
Columns: A000217 (k=1), A001788 (k=2), A027472 (k=3), A038845 (k=4), A081135 (k=5), A081136 (k=6), A027474 (k=7), A081138 (k=8), A081139 (k=9), A081140 (k=10), A081141 (k=11), A081142 (k=12), A027476 (k=15).

[k eq n select 0 else (n-k)^(k-2)*Binomial(k,2): k in [0..n], n in [0..12]]; // G. C. Greubel, May 14 2021

Table[If[k==n, 0, (n-k)^(k-2)*Binomial[k, 2]], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 14 2021 *)

T(n, k)=if (k==0, 0, k^(n-2)*binomial(n, 2));
seq(nn) = for (n=0, nn, for (k=0, n, print1(T(k, n-k), ", ")); );
seq(12) \\ Michel Marcus, May 14 2021

flatten([[0 if (k==n) else (n-k)^(k-2)*binomial(k,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 14 2021

T(n, k) = k^(n-2)*binomial(n, 2), with T(n, 0) = 0 (square array).
T(n, n) = A081131(n).
Rows have g.f. x^3/(1-k*x)^n.
From G. C. Greubel, May 14 2021: (Start)
T(k, n-k) = (n-k)^(k-2)*binomial(k,2) with T(n, n) = 0 (antidiagonal triangle).
Sum_{k=0..n} T(n, n-k) = A081197(n). (End)

Term a(5) corrected by G. C. Greubel, May 14 2021

cp's OEIS Frontend

A027476 Third column of A027467.

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