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A121544
Sum of all proper base 4 numbers with n digits (those not beginning with 0).
1
6, 114, 1896, 30624, 491136, 7862784, 125822976, 2013241344, 32212156416, 515395682304, 8246335635456, 131941389041664, 2111062300164096, 33776997104615424, 540431954881806336, 8646911282940739584, 138350580546379186176, 2213609288819376390144
OFFSET
1,1
COMMENTS
Sum of the first 3 * 4^(n-1) integers starting with 4^(n-1).
Sum of the integers from 4^(n-1) to 4^n -1.
First differences of A026337.
FORMULA
a(n) = 3 * 4^(n-1) * (4^(n-1) + 4^n - 1)/2.
G.f.: 6*x*(1-x) / ((1-4*x)*(1-16*x)). - Colin Barker, Apr 30 2013
From G. C. Greubel, Nov 07 2024: (Start)
a(n) = (3/5)*binomial(5*4^(n-1), 2).
E.g.f.: (3/32)*(-1 - 4*exp(4*x) + 5*exp(16*x)). (End)
EXAMPLE
a(1) = 6 = 1 + 2 + 3.
a(2) = 114 = 10_4 + 11_4 + 12_4 + 13_4 + 20_4 + 21_4 + 22_4 + 23_4 + 30_4 + 31_4 + 32_4 + 33_4 = (4+5+6+7+8+9+10+11+12+13+14+15)_10.
MATHEMATICA
Table[3*4^(n-1)*(5*4^(n-1) - 1)/2, {n, 20}] (* James C. McMahon, Oct 19 2024 *)
PROG
(Magma) [3*Binomial(5*4^(n-1), 2)/5: n in [1..20]]; // G. C. Greubel, Nov 07 2024
(SageMath)
def A121544(n): return 3*binomial(5*4^(n-1), 2)//5
[A121544(n) for n in range(1, 21)] # G. C. Greubel, Nov 07 2024
CROSSREFS
KEYWORD
easy,nonn,base,changed
AUTHOR
Jonathan Vos Post, Sep 08 2006
EXTENSIONS
More terms from Colin Barker, Apr 30 2013
Edited by Michel Marcus, Apr 15 2024
STATUS
approved