A103897 a(n) = 3*2^(n-1)*(2^n-1).

3, 18, 84, 360, 1488, 6048, 24384, 97920, 392448, 1571328, 6288384, 25159680, 100651008, 402628608, 1610563584, 6442352640, 25769607168, 103078821888, 412316073984, 1649265868800, 6597066620928, 26388272775168, 105553103683584, 422212439900160

Offset

1,1

Comments

Divide the sequence of natural numbers: s0=1,2,3,4,5,6,7,8,9,10,11,12,13,14,... into sections s(n) of length 2*s1-1, where s1=initial digits of s(n): s={1,2},{3,4,5,6},{7,8,9,10,11,12,13,14},... then a(n)=sum of terms of s(n): 3,18,84,...

Sum of the numbers between 2^n and 2^(n+1), both excluded. - Gionata Neri, Jun 16 2015

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Paul Barry, On the Gap-sum and Gap-product Sequences of Integer Sequences, arXiv:2104.05593 [math.CO], 2021.

Index entries for linear recurrences with constant coefficients, signature (6,-8).

Formula

a(n) = 3*A006516(n).

From Bruno Berselli, Sep 19 2011: (Start)

G.f.: 3*x/((1-2*x)*(1-4*x)).

a(n+2) = A061561(4n-2). (End)

E.g.f.: (3/2)*(exp(4*x) - exp(2*x)). - Stefano Spezia, Nov 10 2019

Mathematica

Table[3*2^(n - 1)*(2^n - 1), {n, 30}]
LinearRecurrence[{6, -8}, {3, 18}, 30] (* Harvey P. Dale, Feb 11 2018 *)

Prog

(Magma) [3*2^(n-1)*(2^n-1): n in [1..24]];  // Bruno Berselli, Sep 19 2011
(PARI) a(n)=3*2^(n-1)*(2^n-1) \\ Charles R Greathouse IV, Jun 08 2015
(Python) b = list(range(0, 2**20-1)); a = [sum(b[2**i-1:2**(i+1)-1]) for i in range(1, 20)] ## Johan Claes, Nov 10 2019

Crossrefs

Cf. A006516.

Sequence in context: A078904 A099012 A122069 * A119424 A301996 A218924

Adjacent sequences: A103894 A103895 A103896 * A103898 A103899 A103900

Keyword

nonn,easy

Author

Zak Seidov, Mar 30 2005

Status

approved