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A349118
Row sums of a triangle based on A261327.
0
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).
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
KEYWORD
nonn,easy
AUTHOR
Paul Curtz, Nov 08 2021
STATUS
approved