A100311 Modulo 2 binomial transform of 8^n.

1, 9, 65, 585, 4097, 36873, 266305, 2396745, 16777217, 150994953, 1090519105, 9814671945, 68736258049, 618626322441, 4467856773185, 40210710958665, 281474976710657, 2533274790395913, 18295873486192705, 164662861375734345

Offset: 0

Author

Paul Barry, Dec 06 2004

Keywords

• easy
• nonn

Comments

8^n may be retrieved through 8^n = Sum_{k=0..n} (-1)^A010060(n-k) * (binomial(n,k) mod 2) * A100311(k).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Vladimir Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, arXiv:1011.6083 [math.NT], 2010-2012; J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.

Crossrefs

Cf. A001316, A001317, A010060, A038183, A047999.

Cf. A100307, A100308, A100309, A100310.

Programs

Magma

[(&+[(Binomial(n,k) mod 2)*8^k: k in [0..n]]): n in [0..40]]; // G. C. Greubel, Jan 25 2023

Mathematica

A100311[n_]:= A100311[n]= Sum[Mod[Binomial[n, k], 2]*8^k, {k, 0, n}];
Table[A100311[n], {n, 0, 30}] (* G. C. Greubel, Jan 25 2023 *)

Python

a=1
for i in range(33):
    print(a, end=", ")
    a = (a*8) ^ a
# Alex Ratushnyak, Apr 22 2012

Python

def A100311(n): return sum((bool(~n&n-k)^1)<<3*k for k in range(n+1)) # Chai Wah Wu, May 02 2023

SageMath

def A100311(n): return sum( (binomial(n, k)%2)*8^k for k in range(n+1))
[A100311(n) for n in range(41)] # G. C. Greubel, Jan 25 2023

Formula

a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*8^k.

Conjecture: a(0)=1, a(n+1) = (a(n)*8) XOR a(n), where XOR is the bitwise exclusive-or operator. - Alex Ratushnyak, Apr 22 2012

From Vladimir Shevelev, Dec 26-27 2013: (Start)

Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(8^(2^k)+1)^r),

Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(8^(2^k)+1)^r), where r>0 is a real number.

In particular,

Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(8^(2^k)+1)) = 1.1284805...;

Sum_{n>=0} (-1)^A000120(n)/a(n) = 7/8.

a(2^n) = 8^(2^n) + 1, n >= 0.

Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations:

a(2^t*n+2^(t-1)) = 63*(8^(2^(t-1)+1))/(8^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t >= 2.

In particular, for t=2,3,4, we have the following formulas:

a(4*n+2) = 65 * a(4*n);

a(8*n+4) = 4097/65 * a(8*n+2);

a(16*n+8) = (16777217/266305) * a(16*n+6), etc. (End)