A099571 -id:A099571 - cp's OEIS frontend

A054450 Triangle of partial row sums of unsigned triangle A049310(n,m), n >= m >= 0 (Chebyshev S-polynomials).

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 4, 4, 1, 1, 8, 8, 5, 5, 1, 1, 13, 12, 12, 6, 6, 1, 1, 21, 21, 17, 17, 7, 7, 1, 1, 34, 33, 33, 23, 23, 8, 8, 1, 1, 55, 55, 50, 50, 30, 30, 9, 9, 1, 1, 89, 88, 88, 73, 73, 38, 38, 10, 10, 1, 1, 144, 144, 138, 138, 103, 103, 47, 47, 11, 11, 1, 1

Offset: 0

Wolfdieter Lang, Apr 27 2000 and May 08 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is Fib(z)/(1-x*z/(1-z^2)) where Fib(x)=1/(1-x-x^2) = g.f. for A000045(n+1) (Fibonacci numbers without 0).

This is the first member of the family of Riordan-type matrices obtained from the unsigned convolution matrix A049310 by repeated application of the partial row sums procedure.

			Triangle begins as:
   1;
   1,  1;
   2,  1,  1;
   3,  3,  1,  1;
   5,  4,  4,  1,  1;
   8,  8,  5,  5,  1,  1;
  13, 12, 12,  6,  6,  1,  1;
  21, 21, 17, 17,  7,  7,  1,  1;
  34, 33, 33, 23, 23,  8,  8,  1,  1;
  55, 55, 50, 50, 30, 30,  9,  9,  1, 1;
  89, 88, 88, 73, 73, 38, 38, 10, 10, 1, 1;
  ...
Fourth row polynomial (n=3): p(3,x) = 3 + 3*x + x^2 + x^3.

Cf. A000027, A000045, A029907 (row sums), A038793, A038794, A038795, A038796.

Cf. A049310, A052952, A053121, A054451, A054453, A099571, A099572, A152948.

A049310:= func< n,k | ((n+k) mod 2) eq 0 select (-1)^(Floor((n+k)/2)+k)*Binomial(Floor((n+k)/2), k) else 0 >;
A054450:= func< n,k | (&+[Abs(A049310(n,j)): j in [k..n]]) >;
[A054450(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

A049310[n_, k_]:= A049310[n, k]= If[n<0, 0, If[k==n, 1, A049310[n-1, k-1] - A049310[n-2, k] ]];
A054450[n_, k_]:= A054450[n, k]= Sum[Abs[A049310[n,j]], {j,k,n}];
Table[A054450[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

@CachedFunction
def A049310(n, k):
    if (n<0): return 0
    elif (k==n): return 1
    else: return A049310(n-1, k-1) - A049310(n-2, k)
def A054450(n,k): return sum( abs(A049310(n,j)) for j in (k..n) )
flatten([[A054450(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

T(n, m) = Sum_{k=m..n} |A049310(n, k)| (sequence of partial row sums in column m).

Column m recursion: T(n, m) = Sum_{j=m..n} T(j-1, m)*|A049310(n-j, 0)| + |A049310(n, m)|, n >= m >= 0, a(n, m) := 0 if n

T(n, 0) = A000045(n+1).

T(n, 1) = A052952(n-1).

T(n, 2) = A054451(n-2).

Sum_{k=0..n} T(n, k) = A029907(n) = A054453(n, 0).

G.f. for column m: Fib(x)*(x/(1-x^2))^m, m >= 0, with Fib(x) = g.f. A000045(n+1).

The corresponding square array has T(n, k) = Sum_{j=0..floor(k/2)} binomial(n+k-j, j). - Paul Barry, Oct 23 2004

From G. C. Greubel, Jul 25 2022: (Start)

T(n, 3) = A099571(n-3).

T(n, 4) = A099572(n-4).

T(n, n) = T(n, n-1) = A000012(n).

T(n, n-2) = A000027(n), n >= 2.

T(n, n-3) = A000027(n), n >= 3.

T(n, n-4) = A152948(n), n >= 4.

T(n, n-5) = A152948(n), n >= 5.

T(n, n-6) = A038793(n), n >= 6.

T(n, n-8) = A038794(n), n >= 8.

T(n, n-10) = A038795(n), n >= 10.

T(n, n-12) = A038796(n), n >= 12. (End)

A099572 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+4, k).

Original entry on oeis.org

1, 1, 6, 7, 23, 30, 73, 103, 211, 314, 581, 895, 1560, 2455, 4135, 6590, 10890, 17480, 28590, 46070, 74946, 121016, 196326, 317342, 514123, 831465, 1346148, 2177613, 3524441, 5702054, 9227311, 14929365, 24157645, 39087010, 63245795, 102332805

Offset: 0

Paul Barry, Oct 23 2004

Fifth column of triangle A054450. In general Sum_{k=0..floor(n/2)} binomial(n-k+r, k), r>=0, will have g.f. 1/((1-x^2)^r*(1-x-x^2)) and for r>0, a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+r-1, r-1)*(1+(-1)^k)/2.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,5,-4,-10,6,10,-4,-5,1,1).

Cf. A000045, A054450, A054451, A052952, A099571.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1-x^2)^4*(1-x-x^2)) )); // G. C. Greubel, Jul 25 2022

Table[Fibonacci(n+5) +(-1)^n*(n^3+9*n^2+35*n+33)/96 -(n^3+21*n^2+155*n+417)/96, {n,0,40}] (* G. C. Greubel, Jul 25 2022 *)

[fibonacci(n+5) + (-1)^n*(n^3+9*n^2+35*n+33)/96 - (n^3+21*n^2+155*n + 417)/96 for n in (0..40)] # G. C. Greubel, Jul 25 2022

G.f.: 1/((1-x^2)^4*(1-x-x^2)). - corrected by R. J. Mathar, Feb 20 2011
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+3, 3)*((1+(-1)^k)/2).
a(n) = Fibonacci(n+5) + (-1)^n*(n^3 + 9*n^2 + 35*n + 33)/96 - (n^3 + 21*n^2 + 155*n + 417)/96. - G. C. Greubel, Jul 25 2022

A099573 Reverse of number triangle A054450.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 5, 1, 1, 5, 5, 8, 8, 1, 1, 6, 6, 12, 12, 13, 1, 1, 7, 7, 17, 17, 21, 21, 1, 1, 8, 8, 23, 23, 33, 33, 34, 1, 1, 9, 9, 30, 30, 50, 50, 55, 55, 1, 1, 10, 10, 38, 38, 73, 73, 88, 88, 89, 1, 1, 11, 11, 47, 47, 103, 103, 138, 138, 144, 144, 1, 1, 12, 12, 57, 57, 141, 141, 211, 211, 232, 232, 233

Offset: 0

Paul Barry, Oct 23 2004

			First few rows of the array:
  1, 1, 2, 3,  5,  8, ... (A000045)
  1, 1, 3, 4,  8, 12, ... (A052952)
  1, 1, 4, 5, 12, 17, ... (A054451)
  1, 1, 5, 6, 17, 23, ... (A099571)
  1, 1, 6, 7, 23, 30, ... (A099572)
  ...
Triangle begins as:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 3, 3;
  1, 1, 4, 4,  5;
  1, 1, 5, 5,  8,  8;
  1, 1, 6, 6, 12, 12, 13;
  1, 1, 7, 7, 17, 17, 21, 21;
  1, 1, 8, 8, 23, 23, 33, 33, 34;
  1, 1, 9, 9, 30, 30, 50, 50, 55, 55;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000045, A029907 (row sums), A052951, A052952, A054450, A054451, A052952.

Cf. A099571, A099572, A099574 (diagonal sums), A099575.

[(&+[Binomial(n-j,j): j in [0..Floor(k/2)]]): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

T[n_, k_]:= Sum[Binomial[n-j,j], {j,0,Floor[k/2]}];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

def A099573(n,k): return sum(binomial(n-j, j) for j in (0..(k//2)))
flatten([[A099573(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

Number triangle T(n, k) = Sum_{j=0..floor(k/2)} binomial(n-j, j) if k <= n, 0 otherwise.
T(n, n) = A000045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A099574(n).
Sum_{k=0..n} T(n, k) = A029907(n+1).
Antidiagonals of the following array: the first row equals the Fibonacci numbers, (1, 1, 2, 3, 5, ...), and the (n+1)-st row is obtained by the matrix-vector product A128174 * n-th row. - Gary W. Adamson, Jan 19 2011
From G. C. Greubel, Jul 25 2022: (Start)
T(n, n-1) = A052952(n-1), n >= 1.
T(n, n-2) = A054451(n-2), n >= 2.
T(n, n-3) = A099571(n-3), n >= 3.
T(n, n-4) = A099572(n-4), n >= 4. (End)

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A054450 Triangle of partial row sums of unsigned triangle A049310(n,m), n >= m >= 0 (Chebyshev S-polynomials).

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A099572 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+4, k).

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A099573 Reverse of number triangle A054450.

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