A001025 Powers of 16: a(n) = 16^n.
1, 16, 256, 4096, 65536, 1048576, 16777216, 268435456, 4294967296, 68719476736, 1099511627776, 17592186044416, 281474976710656, 4503599627370496, 72057594037927936, 1152921504606846976, 18446744073709551616, 295147905179352825856, 4722366482869645213696, 75557863725914323419136, 1208925819614629174706176
Offset: 0
References
- N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Muniru A Asiru, Table of n, a(n) for n = 0..820 (terms n = 0..100 from T. D. Noe)
- P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
- INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 280
- Tanya Khovanova, Recursive Sequences
- Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
- Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
- Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
- Index entries for linear recurrences with constant coefficients, signature (16).
Programs
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GAP
List([0..20],n->16^n); # Muniru A Asiru, Nov 07 2018
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Haskell
a001025 = (16 ^) a001025_list = iterate (* 16) 1 -- Reinhard Zumkeller, Nov 07 2012
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Maple
A001025:=-1/(-1+16*z); # Simon Plouffe in his 1992 dissertation
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Mathematica
Table[4^(2*n), {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Mar 01 2009 *)
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Maxima
A001025(n):=16^n$ makelist(A001025(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
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PARI
a(n)=1<<(4*n) \\ Charles R Greathouse IV, Feb 01 2012
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Python
print([16**n for n in range(20)]) # Stefano Spezia, Nov 10 2018
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Sage
[lucas_number1(n,16,0) for n in range(1, 18)] # Zerinvary Lajos, Apr 29 2009
Formula
G.f.: 1/(1-16*x).
E.g.f.: exp(16*x).
From Muniru A Asiru, Nov 07 2018: (Start)
a(n) = 16^n.
a(0) = 1, a(n) = 16*a(n-1). (End)
a(n) = 4^A005843(n) = 2^A008586(n) = A000302(n)^2 = A000079(n)*A001018(n). - Muniru A Asiru, Nov 10 2018
a(n) = ( Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k) ) * ( Sum_{k = 0..n} (-1)^k/(2*k + 1)*binomial(2*n + 1, n - k) ). - Peter Bala, Feb 12 2019
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