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A245195
a(n) = 2^A014081(n).
4
1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 4, 4, 4, 8, 8, 8, 16, 32, 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16, 2, 2, 2, 4, 2
OFFSET
0,4
COMMENTS
This sequence provides a bridge between A245180 (and, presumably, A160239) and A014081.
See A245196 for more about this class of sequences.
Run length transform of A011782: 1,1,2,4,8,16,32,64,... - Chai Wah Wu, Oct 19 2016
LINKS
Chai Wah Wu and Robert Israel, Table of n, a(n) for n = 0..10000
Robert Israel, Proof that A277560 is the same as A245195 [This will be modified to reflect the fact that the two sequences have now been merged]
FORMULA
The entries may be arranged into blocks of sizes 1,2,4,8,...:
B_0: 1,
B_1: 1, 2,
B_2: 1, 1, 2, 4,
B_3: 1, 1, 1, 2, 2, 2, 4, 8,
B_4: 1, 1, 1, 2, 1, 1, 2, 4, 2, 2, 2, 4, 4, 4, 8, 16,
B_5: 1, 1, 1, 2, 1, 1, 2, 4, 1, 1, 1, 2, 2, 2, 4, 8, 2, 2, 2, 4, 2, 2, 4, 8, 4, 4, 4, 8, 8, 8, 16, 32,
...
Consider the block B_{k-1} containing terms a(2^(k-1)), a(2^(k-1)+1), ..., a(2^k-1). It is convenient to index the terms working backwards from the next, 2^k-th, term. For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
(if j=0) a(2^k-2^r) = 2^(k-r-1),
(if j>0) a(2^k-2^r+j) = 2^(k-r-1)*a(j).
a(n) = A162510(A005940(1+n)). - Antti Karttunen, Oct 29 2016
From Robert Israel, Nov 02 2016: (Start)
a(2*k) = a(k).
a(4*k+1) = a(k).
a(4*k+3) = 2*a(2*k+1).
G.f. g(x) satisfies g(x) = x + (2*x+1)*g(x^2) - x*g(x^4). (End)
Also, a(n) = Sum_{k=0..floor(n/2)} ((binomial(n,2k)*binomial(n,k)) mod 2). - Chai Wah Wu, Oct 19 2016 and Robert Israel, Nov 04 2016. For proof, see the article by Chai Wah Wu, Sums of products of binomial coefficients mod 2 and run length transforms of sequences, arXiv:1610.06166, or the Robert Israel link.
MAPLE
# This Maple program applies more generally to a sequence where the recurrence across a block is as follows. The parameters to be set are the sequence G(0), G(1), G(2), ... (the final terms in the blocks), and the multiplier m.
# For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
# (if j=0) a(2^k-2^r) = G(k-r-1),
# (if j>0) a(2^k-2^r+j) = m*G(k-r-1)*a(j).
# Since Maple gives its lists an offset of 1, it is necessary to add 1 to the arguments of G.
# For the present sequence, G(n)=2^n and m=1.
G:=[seq(2^n, n=0..30)];
m:=1;
f:=proc(n) option remember; global m, G; local k, r, j, np;
if n <= 2 then G[0+1] elif n=3 then G[1+1]
elif n=4 then G[0+1] elif n=5 then m*G[0+1] elif n=6 then G[1+1] elif n=7 then G[2+1]
else
k:=1+floor(log[2](n)); np:=2^k-n;
if np=1 then r:=0; j:=0; else r:=1+floor(log[2](np-1)); j:=2^r-np; fi;
if j=0 then G[k-r-1+1]; else m*G[k-r-1+1]*f(j); fi;
fi;
end;
[seq(f(n), n=1..520)]:
# Setting G(n) = A083424(n) and m = 8 gives A245180. Setting G(n) = 2^n and m = 2 gives A048896.
A245195:=n->add(binomial(n, 2*k)*binomial(n, k) mod 2, k=0..floor(n/2)): seq(A245195(n), n=0..200); # Wesley Ivan Hurt, Nov 01 2016
MATHEMATICA
Table[Sum[Mod[Binomial[n, 2 k] Binomial[n, k], 2], {k, 0, n}], {n, 0, 85}] (* Michael De Vlieger, Oct 21 2016 *)
PROG
(PARI) a(n) = 2^hammingweight(bitand(n, n>>1)) \\ Charles R Greathouse IV, Jul 16 2016
(PARI) a(n) = sum(k=0, n, binomial(n, 2*k)*binomial(n, k) % 2); \\ Michel Marcus, Oct 21 2016
(Python)
from __future__ import division
def A277560(n):
return sum(int(not (~n & 2*k) | (~n & k)) for k in range(n//2+1))
(Python)
def A245195(n): return 1<<(n&(n>>1)).bit_count() # Chai Wah Wu, Feb 11 2023
KEYWORD
nonn,tabf
AUTHOR
N. J. A. Sloane, Jul 24 2014
EXTENSIONS
Changed offset to 0, merged former entry A277560 from Chai Wah Wu (Oct 19 2016) with this sequence. - N. J. A. Sloane, Nov 05 2016
STATUS
approved