A185342 Triangle of successive recurrences in columns of A117317(n).

2, 4, -4, 6, -12, 8, 8, -24, 32, -16, 10, -40, 80, -80, 32, 12, -60, 160, -240, 192, -64, 14, -84, 280, -560, 672, -448, 128, 16, -112, 448, -1120, 1792, -1792, 1024, -256, 18, -144, 672, -2016, 4032, -5376, 4608, -2304, 512, 20, -180, 960, -3360, 8064

Offset: 0

Author

Paul Curtz, Jan 26 2012

Keywords

• sign
• tabl

Comments

A117317 (A):

1

2 1

4 5 1

8 16 9 1

16 44 41 14 1

32 112 146 85 20 1

64 272 456 377 155 27 1

have for their columns successive signatures

(2) (4,-4) (6,-12,8) (8,-24, 32, -16) (10,-40,80,-80,32) i.e. a(n).

Take based on abs(A133156) (B):

1

2 0

4 1 0

8 4 0 0

16 12 1 0 0

32 32 6 0 0 0

64 80 24 1 0 0 0.

The recurrences of successive columns are also a(n). a(n) columns: A005843(n+1), A046092(n+1), A130809, A130810, A130811, A130812, A130813.

A053220 + A001787 = A014480.

Examples

			Triangle T(n,k),for 1<=k<=n, begins :
2                                         (1)
4    -4                                   (2)
6   -12   8                               (3)
8   -24  32   -16                         (4)
10  -40  80   -80   32                    (5)
12  -60 160  -240  192   -64              (6)
14  -84 280  -560  672  -448  128         (7)
16 -112 448 -1120 1792 -1792 1024 -256    (8)
Successive rows can be divided by A171977.
		

Links

• G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Crossrefs

Cf. For (A): A053220, A056243. For (B): A000079, A001787, A001788, A001789. For A193862: A115068 (a Coxeter group). For (2): A014480 (also (3),(4),(5),..); also A053220 and A001787.

Cf. A007318.

Programs

Mathematica

Table[(-1)*Binomial[n, k]*(-2)^k, {n, 1, 20}, {k, 1, n}] // Flatten (* G. C. Greubel, Jun 27 2017 *)

PARI

for(n=1,20, for(k=1,n, print1((-2)^(k+1)*binomial(n,k)/2, ", "))) \\ G. C. Greubel, Jun 27 2017

Formula

Take A133156(n) without 1's or -1's double triangle (C)=

2

4

8 -4

16 -12

32 -32 6

64 -80 24

128 -192 80 -8

256 -448 240 -40

512 -1024 672 -160 10;

a(n) is increasing odd diagonals and increasing (sign changed) even diagonals. Rows sum of (C) = A201629 (?) Another link between Chebyshev polynomials and cos( ).

Absolute values: A013609(n) without 1's. Also 2*A193862 = (2*A002260)*A135278.

T(n,k) = T(n-1,k) - 2*T(n-1,k-1) for k>1, T(n,1) = 2*n = 2*T(n-1,1) - T(n-2,1). - Philippe Deléham, Feb 11 2012

T(n,k) = (-1)* Binomial(n,k)*(-2)^k, 1<=k<=n. - Philippe Deléham, Feb 11 2012