A100310 Modulo 2 binomial transform of 7^n.
1, 8, 50, 400, 2402, 19216, 120100, 960800, 5764802, 46118416, 288240100, 2305920800, 13847054404, 110776435232, 692352720200, 5538821761600, 33232930569602, 265863444556816, 1661646528480100, 13293172227840800
Offset: 0
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.
Programs
-
Magma
[(&+[7^k*(Binomial(n,k) mod 2): k in [0..n]]): n in [0..40]]; // G. C. Greubel, Feb 01 2023
-
Mathematica
a[n_]:= a[n]= Sum[7^k*Mod[Binomial[n,k], 2], {k,0,n}]; Table[a[n], {n,0,40}] (* G. C. Greubel, Feb 01 2023 *) -
Python
def A100310(n): return sum((bool(~n&n-k)^1)*7**k for k in range(n+1)) # Chai Wah Wu, May 03 2023
-
SageMath
def A100310(n): return sum(7^k*(binomial(n, k)%2) for k in range(n+1)) [A100310(n) for n in range(41)] # G. C. Greubel, Feb 01 2023
Formula
a(n) = Sum_{k=0..n} mod(binomial(n, k), 2)*7^k.
From Vladimir Shevelev, Dec 26-27 2013: (Start)
Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(7^(2^k)+1)^r),
Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(7^(2^k)+1)^r), where r>0 is a real number.
In particular,
Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(7^(2^k)+1)) = 1.1479779...;
Sum_{n>=0} (-1)^A000120(n)/a(n) = 6/7.
a(2^n) = 7^(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)) = 48*(7^(2^(t-1)+1))/(7^(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) = 50 * a(4*n);
a(8*n+4) = 1201/25 * a(8*n+2);
a(16*n+8)= 2882401/60050 * a(16*n+6), etc. (End)
Comments