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A215172
a(0)=1, a(n) = a(n-1)*4^n + 2^n - 1. That is, add n 0's and n 1's to the binary representation of previous term.
2
1, 5, 83, 5319, 1361679, 1394359327, 5711295803455, 93573870443806847, 6132457173405325525247, 1607586853265165654490350079, 1685676992249374341322873324438527, 7070241751299519797307892876185811552255
OFFSET
0,2
COMMENTS
Binary representations:
a(0) 1
a(1) 101
a(2) 1010011
a(3) 1010011000111
a(4) 101001100011100001111
a(5) 1010011000111000011110000011111
a(6) 1010011000111000011110000011111000000111111
a(7) 101001100011100001111000001111100000011111100000001111111
a(8) 1010011000111000011110000011111000000111111000000011111110000000011111111
LINKS
FORMULA
a(0)=1, a(n) = a(n-1)*4^n + 2^n - 1.
MATHEMATICA
nxt[{n_, a_}]:={n+1, FromDigits[Join[IntegerDigits[a], PadRight[{}, n, 0], PadRight[ {}, n, 1]]]}; FromDigits[IntegerDigits[#], 2]&/@NestList[nxt, {1, 1}, 12][[All, 2]] (* Harvey P. Dale, Apr 30 2019 *)
PROG
(Python)
a = 1
for n in range(1, 13):
#print 'a('+str(n-1)+')', bin(a)[2:], a
print a,
a = a*(4**n) + 2**n - 1
CROSSREFS
Cf. A076131: add n 0's and one 1 to the binary representation of previous term.
Sequence in context: A250549 A035512 A054953 * A301811 A216146 A111834
KEYWORD
nonn,base
AUTHOR
Alex Ratushnyak, Aug 05 2012
STATUS
approved