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

2,2

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

Table of n, a(n) for n=2..50.

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

Formula

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

Mathematica

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

Crossrefs

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

Cf. also A008590, A010731, A261327.

Sequence in context: A133172 A363686 A075453 * A073845 A169697 A092525

Adjacent sequences: A349115 A349116 A349117 * A349119 A349120 A349121

Keyword

nonn,easy

Author

Paul Curtz, Nov 08 2021

Status

approved