A308124 a(n) = (2 + 7*4^n)/3.

3, 10, 38, 150, 598, 2390, 9558, 38230, 152918, 611670, 2446678, 9786710, 39146838, 156587350, 626349398, 2505397590, 10021590358, 40086361430, 160345445718, 641381782870, 2565527131478, 10262108525910, 41048434103638, 164193736414550, 656774945658198, 2627099782632790

Offset: 0

Author

Paul Curtz, Jul 23 2019

Keywords

• nonn
• easy

Comments

Consider A092808 and its differences:

1, 0, 3, 1, 11, 5, 43, 21, 171, ...

-1, 3, -2, 10, -6, 38, -22, 150, ... = b(n).

a(n) is the second bisection of b(n). The first is A047849.

a(n) mod 9 is the period 9 sequence: repeat [3, 1, 2, 6, 4, 5, 0, 7, 8].

b(n) + b(n+1) = A135520(n).

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,-4).

Crossrefs

Cf. A000302, A001045, A002042, A092808, A047849, A135520, A141495.

Programs

Mathematica

LinearRecurrence[{5,-4},{3,10},30] (* Paolo Xausa, Nov 13 2023 *)
(2+7*4^Range[0,30])/3 (* Harvey P. Dale, Aug 15 2025 *)

PARI

a(n) = (2 + 7*4^n)/3; \\ Stefano Spezia, Jul 23 2019

PARI

Vec((3 - 5*x) / ((1 - x)*(1 - 4*x)) + O(x^40)) \\ Colin Barker, Jul 23 2019

Formula

a(n) = 4*a(n-1) - 2 for n=1,2,... , a(0) = 3.

a(n+1) = a(n) + A002042(n).

Binomial transform of A141495(n+1) = 3, 7, 21, ....

From Colin Barker, Jul 23 2019: (Start)

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

a(n) = 5*a(n-1) - 4*a(n-2) for n>1.

(End)

a(n+2) = a(n) + 35*A000302(n) for n=0,1,2, ... .

Extensions

a(14)-a(25) from Stefano Spezia, Jul 23 2019