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A101624
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Stern-Jacobsthal numbers.
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9
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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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
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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