A106510 -id:A106510 - cp's OEIS frontend

A365084 G.f. satisfies A(x) = 1 + x*A(x) / (1 + x)^5.

1, 1, -4, 6, 6, -49, 95, 24, -592, 1417, -414, -6809, 20142, -14831, -73353, 274761, -311105, -715647, 3607624, -5463428, -5785294, 45588556, -87189477, -25565196, 552659892, -1305250324, 340413165, 6379267117, -18606431142, 13202513476, 69064770845

Offset: 0

Seiichi Manyama, Aug 21 2023

Index entries for linear recurrences with constant coefficients, signature (-4,-10,-10,-5,-1).

Cf. A106510, A127896, A365083.

Cf. A079675.

LinearRecurrence[{-4, -10, -10, -5, -1}, {1, 1, -4, 6, 6, -49}, 1 + 30] (* Robert P. P. McKone, Aug 21 2023 *)

a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n+4*k-1, n-k));

G.f.: A(x) = 1/( 1 - x/(1+x)^5 ).
a(n) = -4*a(n-1) - 10*a(n-2) - 10*a(n-3) - 5*a(n-4) - a(n-5) for n > 5.
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n+4*k-1,n-k).

A145462 Eigentriangle, row sums = the Padovan sequence, A000931.

Original entry on oeis.org

1, 1, 1, -1, 1, 2, 0, -1, 2, 2, 1, 0, -2, 2, 3, -1, 1, 0, -2, 3, 4, 0, -1, 2, 0, -3, 4, 5, 1, 0, -2, 2, 0, -4, 5, 7, -1, 1, 0, -2, 3, 0, -5, 7, 9, 0, -1, 2, 0, -3, 4, 0, -7, 9, 12, 1, 0, -2, 2, 0, -4, 5, 0, -9, 12, 16, -1, 1, 0, -2, 3, 0, -5, 7, 0, -12, 16, 21

Offset: 6

Gary W. Adamson, Oct 10 2008

Right border = Padovan sequence starting with offset 6.
Row sums = Padovan sequence starting with offset 7.
Sum of n-th row terms = rightmost term of next row.

			First few rows of the triangle =
1;
1, 1;
-1, 1, 2;
0, -1, 2, 2;
1, 0, -2, 2, 3;
-1, 1, 0, -2, 3, 4;
0, -1, 2, 0, -3, 4, 5;
1, 0, -2, 2, 0, -4, 5, 7;
-1, 1, 0, -2, 3, 0, -5, 7, 9;
0, -1, 2, 0, -3, 4, 0, -7, 9, 12;
1, 0, -2, 2, 0, -4, 5, 0, -9, 12, 16;
...
Example: Row 10 = (1, 0, -2, 2, 3) with A000931(10) = 3, rightmost term. This row = the termwise products of (1, 0, -1, 1, 1) and (1, 1, 2, 2, 3); where the Padovan sequence starting with offset 6 = (1, 1, 2, 2, 3, 4, 5, 7, 9,...).

A000931, A106510

Triangle read by rows, T(n,k) = M * (A000931 * 0^(n-k)). M = an infinite lower triangular matrix with A106510 in every column: (1, 1, -1, 0, 1, -1, 0, 1, -1,...); and A000931 is a diagonalized infinite lower triangular matrix with the Padovan sequence starting with offset 6: (1, 1, 2, 2, 3, 4, 5, 7, 9,...) as the main diagonal and the rest zeros.

A274922 a(n) = (-1)^n * n if n>0, a(0) = 1.

Original entry on oeis.org

1, -1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15, 16, -17, 18, -19, 20, -21, 22, -23, 24, -25, 26, -27, 28, -29, 30, -31, 32, -33, 34, -35, 36, -37, 38, -39, 40, -41, 42, -43, 44, -45, 46, -47, 48, -49, 50, -51, 52, -53, 54, -55, 56, -57, 58, -59

Offset: 0

Michael Somos, Dec 28 2016

This is a divisibility sequence.

			G.f. = 1 - x + 2*x^2 - 3*x^3 + 4*x^4 - 5*x^5 + 6*x^6 - 7*x^7 + 8*x^8 + ...

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

Cf. A004277, A005408, A028310, A060576, A106510.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1 + x+x^2)/(1+2*x+x^2))); // G. C. Greubel, Jul 29 2018

a[ n_] := If[ n < 1, Boole[n == 0], (-1)^n n];
a[ n_] := SeriesCoefficient[ (1 + x + x^2) / (1 + 2*x + x^2), {x, 0, n}];
LinearRecurrence[{-2,-1},{1,-1,2},60] (* Harvey P. Dale, Mar 30 2019 *)

PARI
```
{a(n) = if( n<1, n==0, (-1)^n * n)};
```

{a(n) = if( n<0, 0, polcoeff( (1 + x + x^2) / (1 + 2*x + x^2) + x * O(x^n), n))};

Euler transform of length 3 sequence [-1, 2, -1].
a(n) = -b(n) where b() is multiplicative with b(2^e) = -(2^e) if e>0, b(p^e) = p^e otherwise.
E.g.f.: 1 - x * exp(-x).
G.f.: (1 + x + x^2) / (1 + 2*x + x^2).
G.f.: (1 - x) * (1 - x^3) / (1 - x^2)^2.
G.f.: 1 / (1 + x / (1 + x / (1 - x / (1 + x)))).
G.f.: 1 - x / (1 + x)^2 = 1 - x /(1 - x)^2 + 4*x^2 / (1 - x^2)^2.
a(n) = (-1)^n * A028310(n). a(2*n) = A004277(n). a(2*n + 1) = - A005408(n).
Convolution inverse of A106510.
A060576(n) = Sum_{k=0..n} binomial(n, k) * a(k).
A028310(n) = Sum_{k=0..n} binomial(n+1, k+1) * a(k).
a(n) = A038608(n), n>0. - R. J. Mathar, May 25 2020

A317023 Square array A(n,k), n >= 0, k >= 0, read by ascending antidiagonals, where the sequence of row n is the expansion of (1-x^(n+1))/((1-x)^(n+1)).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 3, 2, 0, 1, 4, 6, 2, 0, 1, 5, 10, 9, 2, 0, 1, 6, 15, 20, 12, 2, 0, 1, 7, 21, 35, 34, 15, 2, 0, 1, 8, 28, 56, 70, 52, 18, 2, 0, 1, 9, 36, 84, 126, 125, 74, 21, 2, 0, 1, 10, 45, 120, 210, 252, 205, 100, 24, 2, 0, 1, 11, 55, 165, 330, 462, 461, 315, 130, 27, 2, 0, 1, 12, 66

Offset: 0

Werner Schulte, Jul 19 2018

Conjecture: alternating row sums of the triangle give A106510 for n >= 0.

			The square array A(n,k) begins:
  n\k |  0  1  2   3    4    5    6     7     8     9     10
  ====+=====================================================
   0  |  1  0  0   0    0    0    0     0     0     0      0
   1  |  1  2  2   2    2    2    2     2     2     2      2
   2  |  1  3  6   9   12   15   18    21    24    27     30
   3  |  1  4 10  20   34   52   74   100   130   164    202
   4  |  1  5 15  35   70  125  205   315   460   645    875
   5  |  1  6 21  56  126  252  461   786  1266  1946   2877
   6  |  1  7 28  84  210  462  924  1715  2996  4977   7924
   7  |  1  8 36 120  330  792 1716  3432  6434 11432  19412
   8  |  1  9 45 165  495 1287 3003  6435 12870 24309  43749
   9  |  1 10 55 220  715 2002 5005 11440 24310 48620  92377
  10  |  1 11 66 286 1001 3003 8008 19448 43758 92378 184756
  etc.
The triangle T(n,k) begins:
  n\k |  0  1  2   3   4   5   6   7   8   9 10 11 12
  ====+==============================================
   0  |  1
   1  |  1  0
   2  |  1  2  0
   3  |  1  3  2   0
   4  |  1  4  6   2   0
   5  |  1  5 10   9   2   0
   6  |  1  6 15  20  12   2   0
   7  |  1  7 21  35  34  15   2   0
   8  |  1  8 28  56  70  52  18   2   0
   9  |  1  9 36  84 126 125  74  21   2   0
  10  |  1 10 45 120 210 252 205 100  24   2  0
  11  |  1 11 55 165 330 462 461 315 130  27  2  0
  12  |  1 12 66 220 495 792 924 786 460 164 30  2  0
  etc.

Row sums of the triangle give A099036 for n >= 0.
Cf. A000984 (main diagonal), A000012 (column 0), A087156 (column 1).
Cf. A099036, A106510.
In the square array; row 0..12 are: A000007, A040000, A008486, A005893, A008487, A008488, A008489, A008490, A008491, A008492, A008493, A008494, A008495.
A173265 is based on the same square array, but is read by descending antidiagonals with special treatment of column 0.

nmax:=15;; A:=List([0..nmax],n->List([0..nmax],k->Binomial(n+k,k)-Binomial(k-1,k-1-n)));;   b:=List([2..nmax],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->A[b[i][j][2]][b[i][j][1]]))); # Muniru A Asiru, Jul 20 2018

Table[SeriesCoefficient[(1 - x^(# + 1))/((1 - x)^(# + 1)), {x, 0, k}] &[n - k], {n, 0, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Jul 20 2018 *)

T(n,k) = binomial(n+k,k) - binomial(k-1,k-1-n); \\ Michel Marcus, Aug 07 2018

A(n,k) = binomial(n+k,k) - binomial(k-1,k-1-n) for n >= 0 and k >= 0 with binomial(i,j) = 0 if i < j or j < 0.
G.f.: Sum_{k>=0,n>=0} A(n,k)*x^k*y^n = ((1-x)^2)/((1-x-y)*(1-x-x*y)).
Seen as a triangle T(n,k) = A(n-k,k) = binomial(n,k)-binomial(k-1,2*k-1-n) for 0 <= k <= n with binomial(i,j) = 0 if i < j or j < 0.
Mirror image of the triangle equals A173265 except column 0.

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A365084 G.f. satisfies A(x) = 1 + x*A(x) / (1 + x)^5.

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A145462 Eigentriangle, row sums = the Padovan sequence, A000931.

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A274922 a(n) = (-1)^n * n if n>0, a(0) = 1.

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A317023 Square array A(n,k), n >= 0, k >= 0, read by ascending antidiagonals, where the sequence of row n is the expansion of (1-x^(n+1))/((1-x)^(n+1)).

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