A128133 -id:A128133 - cp's OEIS frontend

A128132 A natural number transform, companion to A127701.

1, -1, 2, 0, -1, 3, 0, 0, -1, 4, 0, 0, 0, -1, 5, 0, 0, 0, 0, -1, 6, 0, 0, 0, 0, 0, -1, 7, 0, 0, 0, 0, 0, 0, -1, 8, 0, 0, 0, 0, 0, 0, 0, -1, 9, 0, 0, 0, 0, 0, 0, 0, 0, -1, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 12

Offset: 1

Gary W. Adamson, Feb 15 2007

Binomial transform is A128133.

			Triangle T(n,k) (with rows n >= 1 and columns k = 1..n) begins:
   1;
  -1,  2;
   0, -1,  3;
   0,  0, -1,  4;
   0,  0,  0, -1,  5;
   0,  0,  0,  0, -1, 6;
   ...

Cf. A127701, A128133, A128134.

A128132 := proc(n,k)
    if n = k then
        n;
    elif k = n-1 then
        -1 ;
    else
        0 ;
    end if;
end proc: # R. J. Mathar, Apr 26 2016

{1}~Join~Table[PadLeft[{-1, n}, n], {n, 2, 12}] // Flatten (* Michael De Vlieger, Apr 26 2016 *)

T(n,n) = n.
T(n,n-1) = -1.
T(n,k) = 0 for k <> n, n-1.

A128135 Row sums of A128134.

Original entry on oeis.org

1, 3, 10, 28, 72, 176, 416, 960, 2176, 4864, 10752, 23552, 51200, 110592, 237568, 507904, 1081344, 2293760, 4849664, 10223616, 21495808, 45088768, 94371840, 197132288, 411041792, 855638016, 1778384896, 3690987520, 7650410496, 15837691904, 32749125632, 67645734912, 139586437120, 287762808832

Offset: 1

Gary W. Adamson, Feb 16 2007

Conjecture: a(n)/a(n-1) tends to sqrt(5). (E.g., a(10)/a(9) = 2.235294....)
The conjecture is false. The fraction a(n)/a(n-1) tends to 2 as n grows. - Philipp Zumstein (zuphilip(AT)inf.ethz.ch), Oct 05 2009
This sequence is a subsequence of a greedily and recursively defined sequence (see links). - Sela Fried, Aug 30 2024
For n>=2, a(n) is the total number of ones in runs of ones of length >=3 over all binary strings of length n+1. - Félix Balado, Aug 06 2025

			a(4) = 28 = sum of row 4 of A128134 = 3 + 10 + 11 + 4.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Sela Fried, On integer sequence A128135, 2024.
Sela Fried, Proofs of some Conjectures from the OEIS, arXiv:2410.07237 [math.NT], 2024. See p. 11.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Cf. A128132, A128133, A128134, A134315.

I:=[1, 3, 10]; [n le 3 select I[n] else 4*Self(n-1)-4*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jun 28 2012

CoefficientList[Series[(1-x+2*x^2)/(1-2*x)^2,{x,0,40}],x] (* Vincenzo Librandi, Jun 28 2012 *)
LinearRecurrence[{4,-4},{1,3,10},40] (* Harvey P. Dale, May 26 2023 *)

a(n)=if(n<=2,[1,3][n],2*a(n-1)+2^(n-1)); /* Joerg Arndt, Sep 29 2012 */

Row sums of A128134.
Equals A134315 * [1, 2, 3, ...]. - Gary W. Adamson, Oct 19 2007
a(n) = 2*a(n-1) + 2^(n-1) for n >= 2. - Philipp Zumstein (zuphilip(AT)inf.ethz.ch), Oct 05 2009
From Colin Barker, May 29 2012: (Start)
a(n) = 2^(n - 2)*(2*n - 1) for n > 1.
a(n) = 4*a(n-1) - 4*a(n-2) for n > 3.
G.f.: x*(1 - x + 2*x^2)/(1 - 2*x)^2. (End)
G.f.: (1 - G(0))/2 where G(k) = 1 - (2*k + 2)/(1 - x/(x - (k + 1)/G(k+1))) (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 06 2012
From Amiram Eldar, Aug 05 2020: (Start)
Sum_{n>=1} 1/a(n) = 2*sqrt(2)*arcsinh(1) - 1.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*sqrt(2)*arccot(sqrt(2)) - 1. (End)

More terms from Philipp Zumstein (zuphilip(AT)inf.ethz.ch), Oct 05 2009

Incorrect formula deleted by Colin Barker, May 29 2012

A128134 A128132 * A007318.

Original entry on oeis.org

1, 1, 2, 2, 5, 3, 3, 10, 11, 4, 4, 17, 27, 19, 5, 5, 26, 54, 56, 29, 6, 6, 37, 95, 130, 100, 41, 7, 7, 50, 153, 260, 265, 162, 55, 8, 8, 65, 231, 469, 595, 483, 245, 71, 9, 9, 82, 332, 784, 1190, 1204, 812, 352, 89, 10

Offset: 1

Gary W. Adamson, Feb 15 2007

A007318 * A128132 = A128133. Row sums = A128135: (1, 3, 10, 28, 72, 176, ...).

			Triangle T(n,k) (with rows n >= 1 and columns k = 1..n) begins:
  1;
  1,  2;
  2,  5,  3;
  3, 10, 11,   4;
  4, 17, 27,  19,   5;
  5, 26, 54,  56,  29,  6;
  6, 37, 95, 130, 100, 41, 7;
  ...

Cf. A002522, A007318, A028387, A128132, A128133, A128135, A164845, A332697.

A128132 * A007318 as infinite lower triangular matrices (assuming the top of the Pascal triangle A007318 is shifted from (0,0) to (1,1)).
From Petros Hadjicostas, Jul 26 2020: (Start)
T(n,k) = n*binomial(n-1, k-1) - binomial(n-2, k-1)*[n <> k] for 1 <= k <= n, where [ ] is the Iverson bracket.
Bivariate o.g.f.:  x*y*(1 - x + x^2*(1 + y))/(1 - x*(1 + y))^2.
T(n,k) = T(n-1,k) + T(n-1,k-1) + binomial(n-1,k-1) for 2 <= k <= n with (n,k) <> (2,2).
T(n,k) = 2*T(n-1,k) - T(n-2,k) - T(n-2,k-2) + 2*T(n-1,k-1) - 2*T(n-2,k-1) for 3 <= k <= n with (n,k) <> (3,3).
T(n,1) = n - 1 for n >= 2.
T(n,2) = A002522(n-1) for n >= 2.
T(n,3) = A164845(n-3) for n >= 3.
T(n,4) = A332697(n-3) for n >= 4.
T(n,n) = n for n >= 1.
T(n,n-1) = A028387(n-2) for n >= 2. (End)

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A128132 A natural number transform, companion to A127701.

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A128135 Row sums of A128134.

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A128134 A128132 * A007318.

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