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A064405
Number of even entries (A048967) minus the number of odd entries (A001316) in row n of Pascal's triangle (A007318).
6
-1, -2, -1, -4, 1, -2, -1, -8, 5, 2, 3, -4, 5, -2, -1, -16, 13, 10, 11, 4, 13, 6, 7, -8, 17, 10, 11, -4, 13, -2, -1, -32, 29, 26, 27, 20, 29, 22, 23, 8, 33, 26, 27, 12, 29, 14, 15, -16, 41, 34, 35, 20, 37, 22, 23, -8, 41, 26, 27, -4, 29, -2, -1, -64, 61, 58, 59, 52, 61, 54, 55, 40, 65, 58, 59, 44, 61, 46, 47, 16, 73, 66, 67, 52, 69, 54
OFFSET
0,2
LINKS
FORMULA
a(n) = Sum_{k=0..n} (-1)^binomial(n, k); a(2^n) = 2^n-3; a(2^n+1)=2^n-6; more generally there's a sequence z(k) such that for any k>=0 and for 2^n >k, a(2^n+k) = 2^n+z(k); for k=0, 1, 2, 3, 4, 5, 6, 7, 8... z(k) = -3, -6, -5, -12, -3, -10, -9, -24, 1, ... - Benoit Cloitre, Oct 18 2002
a(2n) = a(n) + n, a(2n+1) = 2a(n). - Ralf Stephan, Mar 05 2004
a(n) = -Sum_{k=0..n} moebius(binomial(n, k) mod 2). - Paul Barry, Apr 29 2005
a(2^n-1) = -2^n. - Seiichi Manyama, Aug 24 2022
MATHEMATICA
Table[ n + 1 - 2Sum[ Mod[ Binomial[ n, k ], 2 ], {k, 0, n} ], {n, 0, 100} ]
PROG
(PARI) a(n)=sum(i=0, n, (-1)^binomial(n, i))
(PARI) a(n)=if(n<1, -1, if(n%2==0, a(n/2)+n/2, 2*a((n-1)/2)))
CROSSREFS
KEYWORD
easy,sign,look
AUTHOR
Robert G. Wilson v, Sep 29 2001
STATUS
approved