A135356 -id:A135356 - cp's OEIS frontend

A133265 Diagonal of the A135356 triangle.

2, 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, 2, 14, 2, 16, 2, 18, 2, 20, 2, 22, 2, 24, 2, 26, 2, 28, 2, 30, 2, 32, 2, 34, 2, 36, 2, 38, 2, 40, 2, 42, 2, 44, 2, 46, 2, 48, 2, 50, 2, 52, 2, 54, 2, 56, 2, 58, 2, 60, 2, 62, 2, 64, 2, 66, 2, 68, 2, 70, 2, 72, 2, 74, 2, 76, 2, 78, 2, 80

Offset: 0

Paul Curtz, Dec 20 2007

Regular continued fraction expansion of 2*sin(1/2)/( cos(1/2) - sin(1/2) ) = 2.40822 34423 35827 84841 ... = 2 + 1/(2 + 1/(2 + 1/(4 + 1/(2 + 1/(6 + 1/(2 + 1/(8 + 1/(2 + ... )))))))). Cf. A019425. - Peter Bala, Feb 15 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A019425.

[(n+3+(n-1)*(-1)^(n+1))/2: n in [0..80]]; // Vincenzo Librandi, Aug 30 2011

A133265 := n -> (n+2+(n-2)*(-1)^n)/2: # Peter Luschny, Aug 30 2011

Table[(n + 3 + (n - 1) (-1)^(n + 1))/2, {n, 0, 79}] (* or *)
Table[Mod[(2 n + 5), (n (3 + (-1)^n) - (-1)^n + 7)/2], {n, 0, 79}] (* or *)
CoefficientList[Series[2 (1 + x - x^2)/((1 - x)^2*(1 + x)^2), {x, 0, 79}], x] (* Michael De Vlieger, Nov 18 2016 *)

Vec(2*(1 + x - x^2) / ((1 - x)^2 * (1 + x)^2) + O(x^100)) \\ Colin Barker, Nov 17 2016

2*(A057979 without 1, 0, first two terms).
a(n) = (n+3+(n-1)*(-1)^(n+1))/2. - Vincenzo Librandi, Aug 30 2011
a(n) = (2*n + 5) mod (n*(3 + (-1)^n) - (-1)^n + 7)/2. - Lechoslaw Ratajczak, Nov 17 2016
From Colin Barker, Nov 17 2016: (Start)
a(n) = 2*a(n-2) - a(n-4) for n>3.
G.f.: 2*(1 + x - x^2) / ((1 - x)^2 * (1 + x)^2).
(End)

A140343 a(n)=4a(n-1)-7a(n-2)+6a(n-3)-3a(n-4), n>4.

Original entry on oeis.org

0, 0, 0, 0, 1, 4, 9, 14, 14, 0, -41, -122, -243, -364, -364, 0, 1093, 3280, 6561, 9842, 9842, 0, -29525, -88574, -177147, -265720, -265720, 0, 797161, 2391484, 4782969, 7174454, 7174454, 0, -21523361, -64570082, -129140163, -193710244, -193710244, 0, 581130733

Offset: 0

Paul Curtz, May 29 2008

This is the main sequence representing the degenerate case of sequences which equal their sixth differences, where, besides the generic a(n)=6a(n-1)-15a(n-2)+20a(n-3)-15a(n-4)+6a(n-5), cf. A135356, there is also a shorter recurrence. Another sequence of this kind is A134581.

Index entries for linear recurrences with constant coefficients, signature (4,-7,6,-3).

Cf. A135356.

Join[{0},LinearRecurrence[{4,-7,6,-3},{0,0,0,1},40]] (* Harvey P. Dale, Mar 11 2015 *)

O.g.f.: x^4/((x^2-x+1)(3x^2-3x+1)). - R. J. Mathar, Jul 10 2008

Edited and corrected by R. J. Mathar, Jul 10 2008

A101508 Product of binomial matrix and the Mobius matrix A051731.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 8, 4, 3, 1, 16, 8, 6, 4, 1, 32, 16, 11, 10, 5, 1, 64, 32, 21, 20, 15, 6, 1, 128, 64, 42, 36, 35, 21, 7, 1, 256, 128, 85, 64, 70, 56, 28, 8, 1, 512, 256, 171, 120, 127, 126, 84, 36, 9, 1, 1024, 512, 342, 240, 220, 252, 210, 120, 45, 10, 1, 2048, 1024, 683, 496, 385, 463, 462, 330, 165, 55, 11, 1

Offset: 0

Paul Barry, Dec 05 2004

Row sums are A101509. Diagonal sums are A101510.
The matrix inverse appears to be A128313. - R. J. Mathar, Mar 22 2013
Read as upper triangular matrix, this can be seen as "recurrences in A135356 applied to A023531" [Paul Curtz, Mar 03 2017]. - The columns are: A000079, A131577, A024495, A000749, A139761, ... Column n differs after the (n+1)-th nonzero term on from the binomial coefficients C(k,n). - M. F. Hasler, Mar 05 2017

			Rows begin
  1;
  2,1;
  4,2,1;
  8,4,3,1;
  16,8,6,4,1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

A101508 := proc(n,k)
    a := 0 ;
    for i from 0 to n do
        if modp(i+1,k+1) = 0 then
            a := a+binomial(n,i) ;
        end if;
    end do:
    return a;
end proc: # R. J. Mathar, Mar 22 2013

t[n_, k_] := Sum[If[Mod[i + 1, k + 1] == 0, Binomial[n, i], 0], {i, 0, n}]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 24 2014 *)

T(n,k)=sum(i=0,n, if((i+1)%(k+1)==0, binomial(n, i))) \\ M. F. Hasler, Mar 05 2017

T(n, k) = Sum_{i=0..n} if(mod(i+1, k+1)=0, binomial(n, i), 0).

Rows have g.f. x^k/((1-x)^(k+1)-x^(k+1)).

A138112 a(n)=3a(n-1)-4a(n-2)+2a(n-3)-a(n-4), a(0)=a(1)=a(2)=0, a(3)=1, a(4)=3.

Original entry on oeis.org

0, 0, 0, 1, 3, 5, 5, 0, -13, -34, -55, -55, 0, 144, 377, 610, 610, 0, -1597, -4181, -6765, -6765, 0, 17711, 46368, 75025, 75025, 0, -196418, -514229, -832040, -832040, 0, 2178309, 5702887, 9227465, 9227465, 0, -24157817, -63245986, -102334155, -102334155

Offset: 0

Paul Curtz, May 04 2008

sign

Obeys also the recurrence a(n)=5a(n-1)-10a(n-2)+10a(n-3)-5a(n-4)+2a(n-5), so the sequence is identical to its fifth differences (cf. A135356). a(n) = A138110(0,n): if A138110 is interpreted as an array with five rows, this is the top row.
The first differences are represented by A100334(n-1).
The 2nd differences are represented by A103311(n).
The 3rd differences are essentially represented by -A138003(n-2).
The 4th differences are represented by -A105371(n).
A102312 contains the absolute values of the terms which occur in pairs, for example a(5)=a(6)=5=A102312(1), a(10)=a(11)= -55 = -A102312(2).
Inverse BINOMIAL transform yields two zeros followed by A105384. - R. J. Mathar, Jul 04 2008

Index entries for linear recurrences with constant coefficients, signature (3, -4, 2, -1).

Cf. A138003, A103311, A105371.

CoefficientList[Series[x^3/(1-3x+4x^2-2x^3+x^4),{x,0,45}],x] (* or *) LinearRecurrence[{3,-4,2,-1},{0,0,0,1},45] (* Harvey P. Dale, Jun 22 2011 *)

O.g.f.: x^3/(1-3x+4x^2-2x^3+x^4). - R. J. Mathar, Jul 04 2008

Edited and extended by R. J. Mathar, Jul 04 2008

A140344 Catalan triangle A009766 prepended by n zeros in its n-th row.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 1, 2, 2, 0, 0, 0, 1, 3, 5, 5, 0, 0, 0, 0, 1, 4, 9, 14, 14, 0, 0, 0, 0, 0, 1, 5, 14, 28, 42, 42, 0, 0, 0, 0, 0, 0, 1, 6, 20, 48, 90, 132, 132, 0, 0, 0, 0, 0, 0, 0, 1, 7, 27, 75, 165, 297, 429, 429, 0, 0, 0, 0, 0, 0, 0, 0, 1, 8, 35, 110, 275, 572, 1001, 1430, 1430

Offset: 0

Paul Curtz, May 29 2008

The triangle's n-th row is also related to recurrences for sequences f(n) which p-th differences, p=n+2: The denominator of the generating function contains a factor 1-2x in these cases.
This factor may be "lifted" either by looking at auxiliary sequences f(n+1)-2f(n) or by considering the corresponding "degenerate" shorter recurrences right away. In the case p=4, the recurrence is f(n)=4f(n-1)-6f(n-2)+4f(n-3) from the 4th row in A135356, the denominator in the g.f. is 1-4x+6x^2-4x^3=(1-2x)(1-2x+2x^2), which yields the degenerate recurrence f(n)=2f(n-1)-2f(n-2) from the 2nd factor and leaves the first three coefficients of 1/(1-2x+2x^2)=1+2x+2x^2+.. in row 2.
A000749 is an example which follows the recurrence but not the degenerate recurrence, but still A000749(n+1)-2A000749(n) = 0, 0, 1, 2, 2,.. starts with the 3 coefficients. A009545 follows both recurrences and starts with the three nonzero terms because there is only a power of x in the numerator of the g.f.
In the case p=5, the recurrence is f(n)=5f(n-1)-10f(n-2)+10f(n-3)-5f(n-4)+2f(n-5), the denominator in the g.f. is 1-5x+10x^2-10x^3+5x^4-2x^5= (1-2x)(1-3x+4x^2-2x^3+x^4), where 1/(1-3x+4x^2-2x^3+x^4) = 1+3x+5x^2+5x^3+... and the 4 coefficients populate row 3.
A049016 obeys the main recurrence but not the degenerate recurrence f(n)=3f(n-1)-4f(n-2)+2f(n-3)-f(n-4), yet A049016(n+1)-2A049016(n)=1, 3, 5, 5,.. starts with the 4 coefficients. A138112 obeys both recurrences and is constructed to start with the 4 coefficients themselves.
In the nomenclature of Foata and Han, this is the doubloon polynomial triangle d_{n,m}(0), up to index shifts. - R. J. Mathar, Jan 27 2011

			Triangle starts
1;
0,1,1;
0,0,1,2,2;
0,0,0,1,3,5,5;
0,0,0,0,1,4,9,14,14;

D. Foata, G.-N. Han, The doubloon polynomial triangle, Ramanujan J. 23 (2010), 107-126

Cf. A135356, A130020, A139687, A140343 (p=6), A140342 (p=7).

Table[Join[Array[0&, n], Table[Binomial[n+k, n]*(n-k+1)/(n+1), {k, 0, n}]], {n, 0, 8}] // Flatten (* Jean-François Alcover, Dec 16 2014 *)

Edited by R. J. Mathar, Jul 10 2008

A099855 a(n) = n2^n - 2^(n/2)sin(Pi*n/4).

Original entry on oeis.org

0, 1, 6, 22, 64, 164, 392, 904, 2048, 4592, 10208, 22496, 49152, 106560, 229504, 491648, 1048576, 2227968, 4718080, 9960960, 20971520, 44041216, 92276736, 192940032, 402653184, 838856704, 1744822272, 3623870464, 7516192768

Offset: 0

Paul Barry, Oct 28 2004

Related to binomial transform of A002265. Sequence is identical to its fourth differences (cf. A139756, A137221). See also A097064, A135035, A038504, A135356. - Paul Curtz, Jun 18 2008

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-14,16,-8).

Cf. A001787, A002265, A009545, A038504, A097064, A135035, A135356.
Cf. A137221, A139756.
Binomial transform of A047538.

I:=[0,1,6,22]; [n le 4 select I[n] else 6*Self(n-1) -14*Self(n-2) +16*Self(n-3) -8*Self(n-4): n in [1..51]]; // G. C. Greubel, Apr 20 2023

LinearRecurrence[{6,-14,16,-8},{0,1,6,22},30] (* Harvey P. Dale, Mar 22 2018 *)

@CachedFunction
def a(n): # a = A099855
    if (n<5): return (0,1,6,22,64)[n]
    else: return 6*a(n-1) - 14*a(n-2) + 16*a(n-3) - 8*a(n-4)
[a(n) for n in range(51)] # G. C. Greubel, Apr 20 2023

G.f.: x/((1-2*x+2*x^2)*(1-4*x+4*x^2)).
a(n) = Sum_{k=0..n} 2^(k/2)*sin(Pi*k/4)*2^(n-k)*(n-k+1).
a(n) = 6*a(n-1) - 14*a(n-2) + 16*a(n-3) - 8*a(n-4).
a(n) = 2*A001787(n) - A009545(n).

A134658 Triangle read by rows, giving coefficients of extended Jacobsthal recurrence.

Original entry on oeis.org

3, 1, 2, 2, -1, 2, 3, -3, 1, 2, 4, -6, 4, -1, 2, 5, -10, 10, -5, 1, 2, 6, -15, 20, -15, 6, -1, 2, 7, -21, 35, -35, 21, -7, 1, 2, 8, -28, 56, -70, 56, -28, 8, -1, 2, 9, -36, 84, -126, 126, -84, 36, -9, 1, 2, 10, -45, 120, -210, 252, -210, 120, -45, 10, -1, 2

Offset: 0

Paul Curtz, Feb 01 2008

Sequence identical to half its p-th differences from the second term.
This sequence is the second of a family after A135356.
This triangle looks like a Pascal's triangle without first column, and with signs and with additional right diagonal consisting of 2's. - Michel Marcus, Apr 07 2019

			Triangle begins
3;                : A000244 = 1, 3, 9, 27, ... is the main sequence
1, 2;             : A001045 = 0, 1, 1, 3, ... is the main sequence
2, -1, 2;         : 0, 0, (A007910 = 1, 2, 3, ... ) is the main sequence
3, -3, 1, 2;      : 0, 0, 0, 1, 3, 6, 10, 17, ... is the main sequence
4, -6, 4, -1, 2;  : A134987 = 0, 0, 0, 0, 1, ... is the main sequence
...
See signatures of linear recurrence of corresponding sequences.

Cf. A000244, A001045, A007910, A134977 (sum of antidiagonals), A134987, A135356.

T[n_, k_] := T[n, k] = Which[n == 0, 3, k == n, 2, k == 0, n, k == n-1, (-1)^k, True, T[n-1, k] - T[n-1, k-1]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 06 2019 *)

Every row sums to 3. - Jean-François Alcover, Apr 04 2019 (further to a remark e-mailed by Paul Curtz).

In agreement with author, T(0, 0) = 3 and offset 0 by Michel Marcus, Apr 06 2019

A139687 Basis of degenerate cases of sequences identical to its p-th differences. Complement to A140344 which is based on natural Catalan's triangle. Triangle without first term (probable 1) on line.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 3, 5, 5, 1, 3, 6, 9, 9, 1, 4, 10, 19, 28, 28, 1, 4, 10, 20, 34, 48, 48, 1, 5, 15, 35, 69, 117, 165, 165, 1, 5, 15, 35, 70, 125, 200, 275, 275, 1, 6, 21, 56, 126, 251, 451, 726, 1001, 1001, 1, 6, 21, 56, 126, 252, 461, 780, 1209, 1638, 1638

Offset: 0

Paul Curtz, Jun 13 2008

Triangle from A140344:
(1;)
0, 1, 1;
0, 0, 1, 2, 2;
0, 0, 0, 1, 3, 5, 5; see A138112,
0, 0, 0, 0, 1, 4, 9, 14, 14; see A140343,
begins (without 0's) like a(n).

Cf. A135356, A140344.

First four rows of triangle from second row: 1, 1; 1, 2, 2; see A099087, 1, 3, 5, 5; 1, 3, 6, 9, 9; see A057083 which can be preceded with 3 leading 0's, are, as said, from natural Catalan's triangle A009766. Origin of a(n) explained later.

A140342 a(n)=5a(n-1)-11a(n-2)+13a(n-3)-9a(n-4)+3a(n-5)-a(n-6).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 5, 14, 28, 42, 42, 0, -131, -417, -924, -1652, -2380, -2380, 0, 7753, 25213, 56714, 102256, 147798, 147798, 0, -479779, -1557649, -3499720, -6305992, -9112264, -9112264, 0, 29587889, 96072133, 215873462, 388991876, 562110290, 562110290, 0

Offset: 0

Paul Curtz, May 29 2008

This is the main sequence representing the degenerate case of sequences which equal their seventh difference, where besides the generic a(n)=7a(n-1)-21(n-2)+35a(n-3)-35a(n-4)+21a(n-5)-7a(n-6)+2a(n-7), cf. A135356, there is also a shorter recurrence.

After the first four terms, every seventh term is zero. - Harvey P. Dale, Sep 20 2012

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, -11, 13, -9, 3, -1).

Cf. A135356, A140343.

LinearRecurrence[{5,-11,13,-9,3,-1},{0,0,0,0,0,1},40] (* Harvey P. Dale, Sep 20 2012 *)

O.g.f.: x^5/(1-5x+11x^2-13x^3+9x^4-3x^5+x^6). - R. J. Mathar, Jul 10 2008

Edited and extended by R. J. Mathar, Jul 10 2008

A171373 Binomial transform of A171372.

Original entry on oeis.org

1, 6, 16, 36, 76, 152, 292, 552, 1052, 2052, 4104, 8344, 17044, 34664, 69904, 139808, 278108, 552268, 1098148, 2189908, 4379816, 8776356, 17596496, 35263836, 70598516, 141197032, 282208592, 563931612, 1127077552, 2253369432, 4506738864, 9015534644

Offset: 0

Paul Curtz, Dec 07 2009

nonn

The recurrence shows that the sequence and its successive differences are identical to their fifth differences (see A135356).

Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5, 2).

LinearRecurrence[{5,-10,10,-5,2},{1,6,16,36,76},40] (* Harvey P. Dale, Dec 09 2013 *)

a(n+1)-2*a(n) = 4*A105371(n-1) = 4*A138110(4,n).
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+2*a(n-5).
G.f.: (1+x-4*x^2+6*x^3+x^4)/((1-2*x)*(x^4-2*x^3+4*x^2-3*x+1)).

Edited and extended by R. J. Mathar, Dec 15 2009, Mar 02 2010

cp's OEIS Frontend

A133265 Diagonal of the A135356 triangle.

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A140343 a(n)=4a(n-1)-7a(n-2)+6a(n-3)-3a(n-4), n>4.

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A101508 Product of binomial matrix and the Mobius matrix A051731.

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A138112 a(n)=3a(n-1)-4a(n-2)+2a(n-3)-a(n-4), a(0)=a(1)=a(2)=0, a(3)=1, a(4)=3.

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A140344 Catalan triangle A009766 prepended by n zeros in its n-th row.

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A099855 a(n) = n*2^n - 2^(n/2)*sin(Pi*n/4).

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A134658 Triangle read by rows, giving coefficients of extended Jacobsthal recurrence.

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A099855 a(n) = n2^n - 2^(n/2)sin(Pi*n/4).