A349118 Row sums of a triangle based on A261327.

1, 5, 3, 18, 8, 47, 18, 100, 35, 185, 61, 310, 98, 483, 148, 712, 213, 1005, 295, 1370, 396, 1815, 518, 2348, 663, 2977, 833, 3710, 1030, 4555, 1256, 5520, 1513, 6613, 1803, 7842, 2128, 9215, 2490, 10740, 2891, 12425, 3333, 14278, 3818, 16307, 4348, 18520, 4925

Offset: 2

Author

Paul Curtz, Nov 08 2021

Keywords

• nonn
• easy

Comments

The following triangle has A261327 as its diagonals:

1

5

1 2

5 13

1 2 5

5 13 29

1 2 5 10

5 13 29 53

1 2 5 10 17

5 13 29 53 85

...

a(0) = a(1) = 0.

a(n)'s final digit: neither 4 nor 9.

First full bisection difference table:

0, 1, 3, 8, 18, 35, 61, 98, ... = 0, A081489 = b(n)

1, 2, 5, 10, 17, 26, 37, 50, ... = A002522

1, 3, 5, 7, 9, 11, 13, 15, ... = A005408

2, 2, 2, 2, 2, 2, 2, 2, ... = A007395

0, 0, 0, 0, 0, 0, 0, 0, ... = A000004

Second full bisection difference table:

0, 5, 18, 47, 100, 185, 310, 483, ... = c(n)

5, 13, 29, 53, 85, 125, 173, 229, ... = A078370

8, 16, 24, 32, 40, 48, 56, 64, ... = A008590(n+1)

8, 8, 8, 8, 8, 8, 8, 8, ... = A010731

0, 0, 0, 0, 0, 0, 0, 0, ... = A000004

Both bisections are cubic polynomials.

c(-n) = -c(n).

Links

• Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

Crossrefs

Cf. A002522, A005408, A007395, A078370, A081489 (first bisection).

Cf. also A008590, A010731, A261327.

Programs

Mathematica

LinearRecurrence[{0, 4, 0, -6, 0, 4, 0, -1}, {1, 5, 3, 18, 8, 47, 18, 100}, 50] (* Amiram Eldar, Nov 08 2021 *)

Formula

G.f.: (5*x^5+2*x^4-2*x^3-x^2+5*x+1)/((x-1)^4*(x+1)^4).