A245235 - OEIS

%I #29 Dec 18 2024 09:27:33

%S 1,2,2,8,8,8,64,64,64,64,1024,1024,1024,1024,1024,32768,32768,32768,

%T 32768,32768,32768,2097152,2097152,2097152,2097152,2097152,2097152,

%U 2097152,268435456,268435456,268435456,268435456,268435456,268435456,268435456,268435456

%N Repeat 2^(n*(n+1)/2) n+1 times.

%C For a(n), the successive exponents of 2 are 0, 1, 1, 3, 3, 3,... = A057944(n).

%F a(n) = 2^n/A059268(n).

%F T(n, k) = 2^(n*(n+1)/2), 0 <= k <= n. - _Michel Marcus_, Jul 17 2014

%e n+1 times repeated 2^(n*(n+1)/2)= 1, 2, 8, 64, 1024,... = A139685(n).

%e By the formula: a(0)=1/1=1, a(1)=2/1=2, a(2)=4/2=2, a(3)=8/1=8, a(4)=16/2=8,...

%e As triangle:

%e    1,

%e    2,    2,

%e    8,    8,    8,

%e   64,   64,   64,   64,

%e 1024, 1024, 1024, 1024, 1024,

%e etc.

%e Row sums: 1, 4, 24, 256,... = A095340.

%t Table[2^(n*(n+1)/2), {n, 0, 7}, {n+1}] // Flatten (* _Jean-François Alcover_, Jul 15 2014 *)

%o (Python)

%o from math import isqrt

%o def A245235(n): return 1<<((m:=isqrt(n+1<<3)-1>>1)*(m+1)>>1) # _Chai Wah Wu_, Dec 17 2024

%Y Cf. A000079, A006125, A057944, A059268, A095340, A123903, A139685.

%K nonn,tabl

%O 0,2

%A _Paul Curtz_, Jul 14 2014