A246695 Row sums of the triangular array A246694.

1, 3, 9, 18, 35, 57, 91, 132, 189, 255, 341, 438, 559, 693, 855, 1032, 1241, 1467, 1729, 2010, 2331, 2673, 3059, 3468, 3925, 4407, 4941, 5502, 6119, 6765, 7471, 8208, 9009, 9843, 10745, 11682, 12691, 13737, 14859, 16020, 17261, 18543, 19909, 21318, 22815

Offset: 0

Author

Clark Kimberling, Sep 01 2014

Keywords

• nonn
• easy

Comments

Also partial sums of A257083. - Reinhard Zumkeller, Apr 17 2015

Examples

			First 5 rows of A246694 preceded by sums
sum = 1: ...... 1
sum = 3: ...... 1 ... 2
sum = 9: ...... 3 ... 2 ... 4
sum = 18: ..... 3 ... 5 ... 4 ... 6
sum = 35: ..... 7 ... 5 ... 8 ... 6 ... 9
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A246694, A246705, A246706.

Cf. A257083, A377802.

Programs

Haskell

a246695 n = a246695_list !! n
a246695_list = scanl1 (+) a257083_list
-- Reinhard Zumkeller, Apr 17 2015

Mathematica

z = 25; t[0, 0] = 1; t[1, 0] = 1; t[1, 1] = 2;
t[n_, 0] := If[OddQ[n], t[n - 1, n - 2] + 1, t[n - 1, n - 1] + 1];
t[n_, 1] := If[OddQ[n], t[n - 1, n - 1] + 1, t[n - 1, n - 2] + 1];
t[n_, k_] := t[n, k - 2] + 1; A246695 = Table[Sum[t[n, k], {k, 0, n}], {n, 0, z}]

Formula

Conjectured linear recurrence: a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6), with a(0) = 1, a(1) = 3, a(2) = 9, a(3) = 18, a(4) = 35, a(5) = 57, a(6) = 91.

Conjectured g.f.: (1 + x + 2*x^2 + x^3 + x^4)/((x - 1)^4*(x + 1)^2).

Conjecture: a(n) = (1/8)*(n + 1)*((-1)^n + 2*n^2 + 4*n + 7). - Eric Simon Jacob, Jul 19 2023 [This conjecture is correct; compare A377802, note offset 1. - Werner Schulte, Nov 22 2024]

Extensions

Corrected and edited by M. F. Hasler, Nov 17 2014