A051667 -id:A051667 - cp's OEIS frontend

A050488 a(n) = 3(2^n-1) - 2n.

0, 1, 5, 15, 37, 83, 177, 367, 749, 1515, 3049, 6119, 12261, 24547, 49121, 98271, 196573, 393179, 786393, 1572823, 3145685, 6291411, 12582865, 25165775, 50331597, 100663243, 201326537, 402653127, 805306309, 1610612675, 3221225409, 6442450879, 12884901821, 25769803707

Offset: 0

James Sellers, Dec 26 1999

Number of words of length n+1 where first element is from {0,1,2}, other elements are from {0,1} and sequence does not decrease (for n=2 there are 3*2^2 sequences, but 000, 100, 110, 111, 200, 210, 211 decrease, so a(2) = 12-7 = 5).
Number of subgroups of C_(2^n) X C_(2^n) (see A060724).
Starting with 1 = row sums of triangle A054582. - Gary W. Adamson, Jun 23 2008
Starting with "1" equals the eigensequence of a triangle with integer squares (1, 4, 9, 16, ...) as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010
(1 + 2x + 2x^2 + 2x^3 + ...)*(1 + 3x + 7x^2 + 15x^3 + ...) = (1 + 5x + 15x^2 + 37x^3 + ...). - Gary W. Adamson, Mar 14 2012
The partial sums of A033484. - J. M. Bergot, Oct 03 2012
Binomial transform is 0, 1, 7, 33, ... (shifted A066810); inverse binomial transform is 0, 1, 3, 3, ... (3 repeated). - R. J. Mathar, Oct 05 2012
Define a triangle by T(n,0) = n*(n+1) + 1, T(n,n) = n + 1, and T(r,c) = T(r-1,c-1) + T(r-1,c) otherwise; then a(n+1) is the sum of the terms of row n. - J. M. Bergot, Mar 30 2013
Starting with "1" are also the antidiagonal sums of the array formed by partial sums of integer squares (1, 4, 9, 16, ...). - Luciano Ancora, Apr 24 2015
Sums of 2 adjacent terms in diagonal k=2 of Eulerian triangle A008292. I.e., T(n,2)+T(n-1,2) for n > 0. Also, 4th NW-SE diagonal of A126277. In other words, a(n) = A000295(n) + A000295(n+1). - Gregory Gerard Wojnar, Sep 30 2018

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Adam Buck, Jennifer Elder, Azia A. Figueroa, Pamela E. Harris, Kimberly Harry, and Anthony Simpson, Flattened Stirling Permutations, arXiv:2306.13034 [math.CO], 2023. See p. 14.
Tamas Lengyel, On p-adic properties of the Stirling numbers of the first kind, Journal of Number Theory, 148 (2015) 73-94.
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

A050487(2^m-1).
Equals (1/2) A051667.
Cf. A000079, A000225, A000295, A005408, A008292, A033484, A054852, A060724, A066810, A125165, A126277, A142963, A156919, A156920, A156925.

List([0..30],n->3*(2^n-1)-2*n); # Muniru A Asiru, Oct 26 2018

a050488 n = sum $ zipWith (*) a000079_list (reverse $ take n a005408_list)
-- Reinhard Zumkeller, Jul 24 2015

[3*(2^n-1) - 2*n: n in [0..30]]; // G. C. Greubel, Oct 23 2018

seq(coeff(series(x*(x+1)/((1-x)^2*(1-2*x)),x,n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Oct 26 2018

Table[3(2^n-1)-2n,{n,0,30}] (* or *) LinearRecurrence[{4,-5,2}, {0,1,5}, 40] (* Harvey P. Dale, Apr 09 2018 *)

a(n)=3*(2^n-1)-2*n \\ Charles R Greathouse IV, Sep 24 2015

for n in range(0, 30): print(3*(2**n-1) - 2*n, end=', ') # Stefano Spezia, Oct 27 2018

Row sums of A125165: (1, 5, 15, 37, ...). Binomial transform of [1, 4, 6, 6, 6, ...] = [1, 5, 15, 37, ...]. 4th diagonal from the right of A126777 = (1, 5, 15, ...). - Gary W. Adamson, Dec 23 2006
a(n) = 2*a(n-1) + (2n-1). - Gary W. Adamson, Sep 30 2007
From Johannes W. Meijer, Feb 20 2009: (Start)
a(n+1) = A156920(n+1,1).
a(n+1) = A156919(n+1,1)/2^n.
a(n+1) = A142963(n+2,1)/2.
a(n) = 4a(n-1) - 5a(n-2) + 2a(n-3) for n>2 with a(0) = 0, a(1) = 1, a(2) = 5.
G.f.: z*(1+z)/((1-z)^2*(1-2*z)).
(End)
a(n) = 2*n + 2*a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n+1) = Sum_{k=0..n} A000079(k) * A005408(n-k), convolution of the powers of 2 with the odd numbers. - Reinhard Zumkeller, Mar 08 2012
E.g.f.: exp(x)*(3*exp(x) - 2*x - 3). - Stefano Spezia, May 15 2023

A051666 Rows of triangle formed using Pascal's rule except begin and end n-th row with n^2.

Original entry on oeis.org

0, 1, 1, 4, 2, 4, 9, 6, 6, 9, 16, 15, 12, 15, 16, 25, 31, 27, 27, 31, 25, 36, 56, 58, 54, 58, 56, 36, 49, 92, 114, 112, 112, 114, 92, 49, 64, 141, 206, 226, 224, 226, 206, 141, 64, 81, 205, 347, 432, 450, 450, 432, 347, 205, 81, 100, 286, 552, 779, 882, 900, 882, 779

Offset: 0

Asher Auel

Row sums give 6*2^n - 4*n - 6 (A051667).
Central terms: T(2*n,n) = 2 * A220101(n). - Reinhard Zumkeller, Aug 05 2013
For a closed-form formula for arbitrary left and right borders of Pascal like triangle see A228196. - Boris Putievskiy, Aug 19 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 09 2013

			Triangle begins:
   0;
   1,  1;
   4,  2,  4;
   9,  6,  6,  9;
  16, 15, 12, 15, 16;
  ...

Reinhard Zumkeller, Rows n = 0..100 of table, flattened

Cf. A007318, A000290, A005408, A228196, A228576.

a051666 n k = a051666_tabl !! n !! k
a051666_row n = a051666_tabl !! n
a051666_tabl = map fst $ iterate
   (\(vs, w:ws) -> (zipWith (+) ([w] ++ vs) (vs ++ [w]), ws))
   ([0], [1, 3 ..])
-- Reinhard Zumkeller, Aug 05 2013

T[n_, 0] := n^2; T[n_, n_] := n^2;
T[n_, k_] := T[n, k] = T[n-1, k-1] + T[n-1, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 13 2018 *)

More terms from James Sellers

cp's OEIS Frontend

A050488 a(n) = 3(2^n-1) - 2n.

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A051666 Rows of triangle formed using Pascal's rule except begin and end n-th row with n^2.

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cp's OEIS Frontend

A050488 a(n) = 3*(2^n-1) - 2*n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

Python

Formula

A051666 Rows of triangle formed using Pascal's rule except begin and end n-th row with n^2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Mathematica

Extensions

A050488 a(n) = 3(2^n-1) - 2n.