A130706 -id:A130706 - cp's OEIS frontend

A169585 A000004 preceded by 1, 3.

1, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Klaus Brockhaus, Dec 02 2009

Inverse binomial transform of A016777; second inverse binomial transform of A053220; third inverse binomial transform of A027471 without first term; fourth inverse binomial transform of A081039.

Cf. A000004 (zero sequence), A016777 (3*n+1), A053220 ((3*n-1)*2^(n-2)), A027471 ((n-1)*3^(n-2)), A081039 ((3*n+4)*4^(n-1), a(0)=1, a(1)=7), A130706 (1, 2, 0, 0, 0, ...), A166926 (1, 2, 4, 0, 0, 0, ...), A130779 (1, 1, 2, 0, 0, 0, ...).

PARI
```
{concat([1, 3], vector(103))}
```

a(0) = 1, a(1) = 3, a(n) = 0 for n > 1.
G.f.: 1+3*x.
a(n) = 3^n mod 9. - Ridouane Oudra, Apr 09 2025

A350530 Square array read by antidiagonals downwards: T(n,k) is the number of sequences of length n with terms in 0..k such that the (n-1)-st difference is zero, but no earlier iterated difference is zero, n, k >= 1.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 0, 0, 0, 1, 4, 2, 0, 0, 0, 1, 5, 4, 0, 0, 0, 0, 1, 6, 8, 0, 0, 0, 0, 0, 1, 7, 12, 4, 0, 0, 0, 0, 0, 1, 8, 18, 12, 8, 4, 0, 0, 0, 0, 1, 9, 24, 28, 36, 28, 4, 0, 0, 0, 0, 1, 10, 32, 52, 84, 116, 48, 16, 0, 0, 0, 0

Offset: 1

Pontus von Brömssen, Jan 03 2022

For fixed n, T(n,k) is a quasi-polynomial of degree n-1 in k. For example, T(4,k) = (8/27)*k^3 - 2*k^2 + b(k)*k + c(k), where b and c are periodic with period 3.

			Array begins:
  n\k|  0  1  2  3  4   5    6     7     8      9     10
  ---+--------------------------------------------------
   1 |  1  1  1  1  1   1    1     1     1      1      1
   2 |  0  1  2  3  4   5    6     7     8      9     10
   3 |  0  0  0  2  4   8   12    18    24     32     40
   4 |  0  0  0  0  0   4   12    28    52     84    132
   5 |  0  0  0  0  0   8   36    84   176    332    568
   6 |  0  0  0  0  4  28  116   308   704   1396   2548
   7 |  0  0  0  0  4  48  232   728  2104   4940  11008
   8 |  0  0  0  0 16 100  556  1936  7092  19908  49364
   9 |  0  0  0  0 12 176 1348  6588 23356  74228 202504
  10 |  0  0  0  0  8 268 2492 15544 72820 259800 842688
For n = 4 and k = 6, the following T(4,6) = 12 sequences are counted: 1454, 1564, 2125, 2565, 3126, 3236, 4541, 4651, 5212, 5652, 6213, 6323.

Pontus von Brömssen, Antidiagonals n = 1..19, flattened

Cf. A200154, A350365, A350529.
Rows: A000012 (n=1), A001477 (n=2), A007590 (n=3).
Columns: A000007 (k=0), A019590 (k=1), A130706 (k=2).

def A350530_col(k,nmax):
    d = []
    c = [0]*nmax
    while 1:
        if not d or all(d[-1][:-1]):
            if d and d[-1][-1] == 0:
                c[len(d)-1] += 1 + (0 != 2*d[0][0] != k+1)
            elif len(d) < nmax:
                d.append([-1])
                for i in range(len(d)-1):
                    d[-1].append(d[-1][-1]-d[-2][i])
        while d and d[-1][0] == k:
            d.pop()
        if not d or len(d) == 1 and 2*d[0][0] >= k: return c
        for i in range(len(d)):
            d[-1][i] += 1

A136487 Coefficients of polynomial recursion p(n,x) = (x-1)(p(n-1,x) - (x+1)p(n-2,x)), p(0,x) = 1, p(1,x) = x+1.

Original entry on oeis.org

1, 1, 1, 1, 1, -1, -1, -1, 0, 2, 0, -1, 2, 0, -4, 0, 2, -3, 2, 7, -4, -5, 2, 1, 5, -5, -11, 11, 7, -7, -1, 1, -8, 12, 16, -28, -8, 20, 0, -4, 13, -25, -20, 60, -2, -46, 12, 12, -3, -1, -21, 50, 19, -120, 38, 92, -50, -24, 15, 2, -1

Offset: 0

Roger L. Bagula, Mar 21 2008

Only coefficients of x^k for k <= degree of p(n,x) are included. With this then, since p(2,x) = 0, row 2 is empty.

The same polynomial coefficients may be obtained, without signs, with the use of the recurrence p(x, n) = (x+1)*p(x, n-1) - (x^2-1)*p(x, n-2), and p(x, 0) = 1, p(x, 1) = x-1.

			First few rows are:
    1;
    1,   1;
   {};
    1,   1,  -1,   -1;
   -1,   0,   2,    0, -1;
    2,   0,  -4,    0,  2;
   -3,   2,   7,   -4, -5,   2,   1;
    5,  -5, -11,   11,  7,  -7,  -1,   1;
   -8,  12,  16,  -28, -8,  20,   0,  -4;
   13, -25, -20,   60, -2, -46,  12,  12, -3, -1;
  -21,  50,  19, -120, 38,  92, -50, -24, 15,  2, -1;

Robert Israel, Table of n, a(n) for n = 0..10103(rows 0 to 141, flattened)

Cf. A000045, A130706 (row sums).

m:=12;
function p(n,x)
  if n le 1 then return (x+1)^n;
  else return (x-1)*(p(n-1,x) - (x+1)*p(n-2,x));  end if;
end function;
R:=PowerSeriesRing(Integers(), m+2);
T:= func< n,k | Coefficient(R!( p(n,x) ), k) >;
[1,1,1] cat [T(n,k): k in [0..n], n in [3..m]]; // G. C. Greubel, Jul 31 2023

F:= proc(n) option remember; expand((1-x)*procname(n-1)+(1-x^2)*procname(n-2)) end proc:
F(0):= 1: F(1):= 1+x:
R:=proc(n) local V,j;
 V:= F(n);
 seq(coeff(V,x,j),j=0..degree(V))
end proc:
for i from 0 to 20 do R(i) od; # Robert Israel, Dec 03 2018

P[x,0]= 1; P[x,1]= x+1; P[x_,n_]:= P[x,n]= (x-1)*(P[x,n-1] - (x+1)*P[x,n-2]);
Table[CoefficientList[P[x,n],x],{n,0,10}]//Flatten

def p(n,x):
    if n<2: return (x+1)^n
    else: return (x-1)*(p(n-1,x) - (x+1)*p(n-2,x))
def T(n):
    P. = PowerSeriesRing(QQ)
    return P( p(n,x) ).list()
flatten([T(n) for n in range(13)]) # G. C. Greubel, Jul 31 2023

T(n, k) = coefficient [x^k] ( p(x, n) ), where p(x,n) = (x-1)*p(x,n-1) - (x^2-1)*p(x,n-2), p(x,0) = 1, p(x,1) = x+1.
Sum_{k >= 0} T(n, k) = A130706(n).
From Robert Israel, Dec 03 2018: (Start)
T(n,k) = T(n-1,k-1) - T(n-1,k) - T(n-2,k-2) + T(n-2,k).
G.f. as array: (1-2*x)/(1 + x*(y-1)+x^2*(1-y^2)).
T(n,0) = (-1)^(n+1)*A000045(n-2) for n >= 3. (End)

Edited by Robert Israel, Dec 03 2018

A167891 A000004 preceded by 1, 4, 2.

Original entry on oeis.org

1, 4, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Klaus Brockhaus, Nov 14 2009

Inverse binomial transform of A028387.

Cf. A000004 (zero sequence), A028387 (n+(n+1)^2), A166926 (1, 2, 4, 0, 0, 0, 0, ...), A130706 (1, 2, 0, 0, 0, 0, ...), A130779 (1, 1, 2, 0, 0, 0, 0, ...), A167858 (3, 14, 36, 36, 12, 0, 0, 0, ...), A167876 (1, 3, 4, 2, 0, 0, 0, ...).

PARI
```
{concat([1, 4, 2], vector(100))}
```

a(0) = 1, a(1) = 4, a(2) = 2, a(n) = 0 for n > 2.

G.f.: 1+4*x+2*x^2.

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A169585 A000004 preceded by 1, 3.

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A350530 Square array read by antidiagonals downwards: T(n,k) is the number of sequences of length n with terms in 0..k such that the (n-1)-st difference is zero, but no earlier iterated difference is zero, n, k >= 1.

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A136487 Coefficients of polynomial recursion p(n,x) = (x-1)*(p(n-1,x) - (x+1)*p(n-2,x)), p(0,x) = 1, p(1,x) = x+1.

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A167891 A000004 preceded by 1, 4, 2.

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A136487 Coefficients of polynomial recursion p(n,x) = (x-1)(p(n-1,x) - (x+1)p(n-2,x)), p(0,x) = 1, p(1,x) = x+1.