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A245235 Repeat 2^(n*(n+1)/2) n+1 times.

%I A245235 #29 Dec 18 2024 09:27:33
%S A245235 1,2,2,8,8,8,64,64,64,64,1024,1024,1024,1024,1024,32768,32768,32768,
%T A245235 32768,32768,32768,2097152,2097152,2097152,2097152,2097152,2097152,
%U A245235 2097152,268435456,268435456,268435456,268435456,268435456,268435456,268435456,268435456
%N A245235 Repeat 2^(n*(n+1)/2) n+1 times.
%C A245235 For a(n), the successive exponents of 2 are 0, 1, 1, 3, 3, 3,... = A057944(n).
%F A245235 a(n) = 2^n/A059268(n).
%F A245235 T(n, k) = 2^(n*(n+1)/2), 0 <= k <= n. - _Michel Marcus_, Jul 17 2014
%e A245235 n+1 times repeated 2^(n*(n+1)/2)= 1, 2, 8, 64, 1024,... = A139685(n).
%e A245235 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 A245235 As triangle:
%e A245235    1,
%e A245235    2,    2,
%e A245235    8,    8,    8,
%e A245235   64,   64,   64,   64,
%e A245235 1024, 1024, 1024, 1024, 1024,
%e A245235 etc.
%e A245235 Row sums: 1, 4, 24, 256,... = A095340.
%t A245235 Table[2^(n*(n+1)/2), {n, 0, 7}, {n+1}] // Flatten (* _Jean-François Alcover_, Jul 15 2014 *)
%o A245235 (Python)
%o A245235 from math import isqrt
%o A245235 def A245235(n): return 1<<((m:=isqrt(n+1<<3)-1>>1)*(m+1)>>1) # _Chai Wah Wu_, Dec 17 2024
%Y A245235 Cf. A000079, A006125, A057944, A059268, A095340, A123903, A139685.
%K A245235 nonn,tabl
%O A245235 0,2
%A A245235 _Paul Curtz_, Jul 14 2014