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A100311 Modulo 2 binomial transform of 8^n. 8
1, 9, 65, 585, 4097, 36873, 266305, 2396745, 16777217, 150994953, 1090519105, 9814671945, 68736258049, 618626322441, 4467856773185, 40210710958665, 281474976710657, 2533274790395913, 18295873486192705, 164662861375734345 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

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

LINKS

Table of n, a(n) for n=0..19.

V. Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.

FORMULA

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

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

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

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

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

In particular,

sum{n>=0}1/a(n) = prod{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)

PROG

(Python)

a=1

for i in range(33):

. print a,

. a = (a*8) ^ a

# from Alex Ratushnyak, Apr 22 2012

CROSSREFS

Cf. A001316, A001317, A038183, A100307, A100308, A100309, A100310.

Sequence in context: A154996 A128195 A103459 * A259242 A120286 A152581

Adjacent sequences:  A100308 A100309 A100310 * A100312 A100313 A100314

KEYWORD

easy,nonn

AUTHOR

Paul Barry, Dec 06 2004

STATUS

approved

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Last modified August 13 00:27 EDT 2020. Contains 336441 sequences. (Running on oeis4.)