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A100309 Modulo 2 binomial transform of 6^n. 6
1, 7, 37, 259, 1297, 9079, 47989, 335923, 1679617, 11757319, 62145829, 435020803, 2178463249, 15249242743, 80603140213, 564221981491, 2821109907457, 19747769352199, 104381066575909, 730667466031363, 3658979549971729 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

6^n may be retrieved through 6^n=sum{k=0..n, (-1)^A010060(n-k)*mod(binomial(n,k),2)A100309(k)}.

LINKS

Table of n, a(n) for n=0..20.

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)6^k}.

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

sum{n>=0}1/a(n)^r = prod{k>=0}(1 + 1/(6^(2^k)+1)^r),

sum{n>=0}(-1)^A000120(n)/a(n)^r = prod{k>=0}(1 - 1/(6^(2^k)+1)^r), where r>0 is a real number.

In particular,

sum{n>=0}1/a(n) = prod{k>=0}(1 + 1/(6^(2^k)+1)) = 1.1746508...;

sum{n>=0}(-1)^A000120(n)/a(n) = 5/6.

a(2^n)=6^(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)) = 35*(6^(2^(t-1)+1))/(6^(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) = 37 * a(4*n);

a(8*n+4) = 1297/37 * a(8*n+2);

a(16*n+8)= 1679617/47989 * a(16*n+6), etc. (End)

CROSSREFS

Cf. A001316, A001317, A038183, A100307, A100308, A100310, A100311.

Sequence in context: A096965 A159597 A217723 * A198410 A199192 A089677

Adjacent sequences:  A100306 A100307 A100308 * A100310 A100311 A100312

KEYWORD

easy,nonn

AUTHOR

Paul Barry, Dec 06 2004

STATUS

approved

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Last modified August 13 02:09 EDT 2020. Contains 336441 sequences. (Running on oeis4.)