A323397 a(n) = (4^n + 15*n - 1)/9.

0, 2, 5, 12, 35, 122, 465, 1832, 7295, 29142, 116525, 466052, 1864155, 7456562, 29826185, 119304672, 477218615, 1908874382, 7635497445, 30541989692, 122167958675, 488671834602, 1954687338305, 7818749353112, 31274997412335, 125099989649222, 500399958596765

Offset

0,2

Comments

Conjecture: satisfies a linear recurrence having signature (6, -9, 4). (This is correct, see Formula section.)

References

Roman Andronov, How can I prove that 4^n+15n-1 is divisible by 9?, Quora Digest (Nov. 17, 2018).

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (6,-9,4).

Formula

a(n+1) = 4*a(n) - 5*n + 2, with a(0)=0, a(1)=2. This implies a(n+2) = 5*a(n+1) - 4*a(n) - 5, and also that a(n+3) = 6*a(n+2) - 9*a(n+1) + 4*a(n). - N. J. A. Sloane, Jan 13 2019

G.f.: x*(2 - 7*x) / ((1 - x)^2*(1 - 4*x)). - Colin Barker, Jan 19 2019

Mathematica

Table[(4^n+15n-1)/9, {n, 0, 40}]

Prog

(PARI) concat(0, Vec(x*(2 - 7*x) / ((1 - x)^2*(1 - 4*x)) + O(x^30))) \\ Colin Barker, Jan 19 2019

Crossrefs

Sequence in context: A000105 A055192 A108555 * A292169 A283799 A225798

Adjacent sequences: A323394 A323395 A323396 * A323398 A323399 A323400

Keyword

nonn,easy

Author

Harvey P. Dale, Jan 13 2019

Status

approved