A101120 Records in A101119, which forms the nonzero differences of A006519 and A003484.

7, 22, 52, 112, 239, 494, 1004, 2024, 4071, 8166, 16356, 32736, 65503, 131038, 262108, 524248, 1048535, 2097110, 4194260, 8388560, 16777167, 33554382, 67108812, 134217672, 268435399, 536870854, 1073741764, 2147483584, 4294967231, 8589934526, 17179869116, 34359738296

Offset

1,1

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (3,-2,0,1,-3,2).

Formula

a(n) = A101119(2^(n-1)) for n>=1.

a(n) = 2^(n+3) - 2^((n-1)(mod 4)) - 8*floor((n+3)/4).

a(n) = 2^(n+3) - A003485(n+3). - Johannes W. Meijer, Oct 31 2012

From Chai Wah Wu, Apr 15 2017: (Start)

a(n) = 3*a(n-1) - 2*a(n-2) + a(n-4) - 3*a(n-5) + 2*a(n-6) for n > 6.

G.f.: x*(-x - 7)/((x - 1)^2*(x + 1)*(2*x - 1)*(x^2 + 1)). (End)

E.g.f.: (exp(x)*(32*exp(x) - 8*x - 27) - 4*cos(x) - cosh(x) - 2*sin(x) + sinh(x))/4. - Stefano Spezia, Jun 06 2023

Mathematica

LinearRecurrence[{3, -2, 0, 1, -3, 2}, {7, 22, 52, 112, 239, 494}, 30] (* Harvey P. Dale, Jan 23 2023 *)

Prog

(PARI) a(n)=2^(n+3)-2^((n-1)%4)-8*((n+3)\4)
(Python)
def A101120(n): return (1<<(n+3))-(1<<((n-1)&3))-(((n+3)&-4)<<1) # Chai Wah Wu, Jul 10 2022

Crossrefs

Cf. A003484, A006519, A101119, A101121, A101122.

Sequence in context: A041215 A060822 A011926 * A151717 A188377 A213585

Adjacent sequences: A101117 A101118 A101119 * A101121 A101122 A101123

Keyword

nonn,easy

Author

Simon Plouffe and Paul D. Hanna, Dec 02 2004

Status

approved