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A101120
Records in A101119, which forms the nonzero differences of A006519 and A003484.
4
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
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
KEYWORD
nonn,easy
AUTHOR
Simon Plouffe and Paul D. Hanna, Dec 02 2004
STATUS
approved