| Displaying 1-4 of 4 results found.
page
1
0, 2, -1, 5, -2, 8, -5, 21, -8, 34, -21, 81, -34, 128, -81, 337, -128, 546, -337, 1301, -546, 2056, -1301, 5381, -2056, 8706, -5381, 20737, -8706, 32768, -20737, 86273, -32768, 139778, -86273, 333061, -139778, 526344, -333061, 1377557, -526344, 2228770
PROG
(Haskell)
a257971 n = a257971_list !! n
a257971_list = zipWith (-) (tail a006921_list) a006921_list
(Python)
def A257971(n): return sum(int(not r & ~(n+2-r))*2**(n//2+1-r) for r in range(n//2+2)) if n & 1 else -sum(int(not r & ~(n-1-r))*2**(n//2-1-r) for r in range(n//2)) # Chai Wah Wu, Jun 20 2022
1, 3, 7, 13, 29, 55, 115, 209, 465, 883, 1847, 3357, 7437, 14087, 29443, 53505, 119041, 226051, 472839, 859405, 1903901, 3606327, 7537523, 13697489, 30474449, 57868403, 121045047, 220004381, 487391245, 923205639, 1929576451, 3506503681, 7801470977, 14814478339, 30987976711, 56321966093
PROG
(Haskell)
(Python)
def A260022(n): return sum(int(not r & ~(2*n-r))*2**(n-r) for r in range(n+1)) # Chai Wah Wu, Jun 20 2022
Stern-Jacobsthal numbers.
+10
9
1, 1, 3, 1, 7, 5, 11, 1, 23, 21, 59, 17, 103, 69, 139, 1, 279, 277, 827, 273, 1895, 1349, 2955, 257, 5655, 5141, 14395, 4113, 24679, 16453, 32907, 1, 65815, 65813, 197435, 65809, 460647, 329029, 723851, 65793, 1512983, 1381397, 3881019, 1118225
COMMENTS
The Stern diatomic sequence A002487 could be called the Stern-Fibonacci sequence, since it is given by A002487(n) = Sum_{k=0..floor(n/2)} (binomial(n-k,k) mod 2), where F(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k,k). Now a(n) = Sum_{k=0..floor(n/2)} (binomial(n-k,k) mod 2)*2^k, where J(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k,k)*2^k, with J(n) = A001045(n), the Jacobsthal numbers. - Paul Barry, Sep 16 2015 These numbers seem to encode Stern (0, 1)-polynomials in their binary expansion. See Dilcher & Ericksen paper, especially Table 1 on page 79, page 5 in PDF. See A125184 ( A260443) for another kind of Stern-polynomials, and also A177219 for a reference to maybe a third kind. - Antti Karttunen, Nov 01 2016
FORMULA
a(n) = Sum_{k=0..floor(n/2)} (binomial(n-k, k) mod 2)*2^k.
a(2^n-1)=1, a(2*n) = 2*a(n-1) + a(n+1) = A099902(n); a(2*n+1) = A101625(n+1). a(n) = Sum_{k=0..n} (binomial(k, n-k) mod 2)*2^(n-k). - Paul Barry, May 10 2005 a(0)=1, a(1)=1, a(n) = a(n-1) XOR (a(n-2)*2), where XOR is the bitwise exclusive-OR operator. - Alex Ratushnyak, Apr 14 2012
PROG
(Python)
prpr = 1
prev = 1
print("1, 1", end=", ")
for i in range(99):
current = (prev)^(prpr*2)
print(current, end=", ")
prpr = prev
prev = current
(Python)
def A101624(n): return sum(int(not k & ~(n-k))*2**k for k in range(n//2+1)) # Chai Wah Wu, Jun 20 2022 (Haskell)
a101624 = sum . zipWith (*) a000079_list . map (flip mod 2) . a011973_row
Lucas sequence U_n(x,1) over the field GF(2).
+10
8
0, 1, 2, 5, 8, 21, 34, 81, 128, 337, 546, 1301, 2056, 5381, 8706, 20737, 32768, 86273, 139778, 333061, 526344, 1377557, 2228770, 5308753, 8388736, 22085713, 35782690, 85262357, 134742024, 352649221, 570556418, 1359020033, 2147483648
COMMENTS
The Lucas sequence U_n(x,1) over the field GF(2)={0,1} is: 0, 1, x, x^2+1, x^3, x^4+x^2+1, x^5+x, ... Numerical values are obtained evaluating these 01-polynomials at x=2 over the integers.
The counterpart sequence is V_n(x,1) = x*U_n(x,1) that implies identities like U_{2n}(x,1) = x*U_n(x,1)^2. - Max Alekseyev, Nov 19 2009
FORMULA
For n>1, a(n) = (2*a(n-1)) XOR a(n-2).
MATHEMATICA
a[0] = 0; a[1] = 1; a[n_] := a[n] = BitXor[2 a[n - 1], a[n - 2]]; Table[a@ n, {n, 0, 32}] (* Michael De Vlieger, Dec 11 2015 *)
PROG
(PARI) { a=0; b=1; for(n=1, 50, c=bitxor(2*b, a); a=b; b=c; print1(c, ", "); ) }
(Python)
def A168081(n): return sum(int(not r & ~(2*n-1-r))*2**(n-1-r) for r in range(n)) # Chai Wah Wu, Jun 20 2022
Search completed in 0.012 seconds
| 