A100308 Modulo 2 binomial transform of 5^n.

1, 6, 26, 156, 626, 3756, 16276, 97656, 390626, 2343756, 10156276, 60937656, 244531876, 1467191256, 6357828776, 38146972656, 152587890626, 915527343756, 3967285156276, 23803710937656, 95520019531876, 573120117191256

Offset: 0

Author

Paul Barry, Dec 06 2004

Keywords

• easy
• nonn

Comments

5^n may be retrieved through 5^n = Sum_{k=0..n} (-1)^A010060(n-k) * (binomial(n,k) mod 2)*a(k).

Links

• Robert Israel, Table of n, a(n) for n = 0..1429
• 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, A038183, A100307, A100309, A100310, A100311.

Programs

Magma

[(&+[5^k*(Binomial(n, k) mod 2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Feb 03 2023

Maple

f:= proc(n) local L,M;
   L:= convert(n,base,2);
   mul(1+5^(2^(k-1)), k = select(t -> L[t]=1, [$1..nops(L)]));
end proc:
map(f, [$0..30]); # Robert Israel, Aug 26 2018

Mathematica

a[n_]:= Sum[Mod[Binomial[n, k], 2] 5^k, {k, 0, n}];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Sep 19 2018 *)

Python

def A100308(n): return sum((bool(~n&n-k)^1)*5**k for k in range(n+1)) # Chai Wah Wu, May 02 2023

SageMath

def A100308(n): return sum(5^k*(binomial(n, k)%2) for k in range(n+1))
[A100308(n) for n in range(41)] # G. C. Greubel, Feb 03 2023

Formula

a(n) = Sum_{k=0..n} (binomial(n, k) mod 2)*5^k.

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

Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(5^(2^k)+1)^r),

Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(5^(2^k)+1)^r), where r > 0 is a real number.

In particular,

Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(5^(2^k)+1)) = 1.2134769...;

Sum_{n>=0} (-1)^A000120(n)/a(n) = 0.8.

a(2^n) = 5^(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)) = 24*(5^(2^(t-1)+1))/(5^(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) = 26 * a(4*n),

a(8*n+4) = (313/13) * a(8*n+2),

a(16*n+8) = (195313/8138) * a(16*n+6), etc. (End)