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Search: id:A117963
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| A117963 |
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Diagonal sums of a Legendre-binomial triangle for p=3. |
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+0
3
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| 1, 1, 2, 0, 2, 2, 1, 3, 4, -2, 2, 0, 2, 2, 4, 0, 4, 4, -1, 3, 2, 2, 4, 6, 1, 7, 8, -6, 2, -4, 4, 0, 4, -2, 2, 0, 2, 2, 4, 0, 4, 4, 2, 6, 8, -4, 4, 0, 4, 4, 8, 0, 8, 8, -5, 3, -2, 4, 2, 6, -1, 5, 4, 0, 4, 4, 2, 6, 8, 2, 10, 12, -5, 7, 2, 6, 8, 14, 1, 15, 16, -14, 2, -12, 8, -4, 4, -6, -2, -8, 8, 0, 8, -4, 4, 0, 4, 4, 8, -6, 2
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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a(n)=a(3n+2)/a(2). Diagonal sums of A117947. A117963 mod 2 is A117964.
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FORMULA
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a(n)=sum{k=0..floor(n/2), L(C(n-k,k)/3)} where L(j/p) is the Legendre symbol of j and p.
G.f. satisfies: A(x) = A(x^3)*(1 - 4*x^3 - x^6)/(1 - x - x^2). a(n) == fibonacci(n+1) (mod 3); a(n) == a(n-1) + a(n-2) (mod 3). - Paul D. Hanna (pauldhanna(AT)juno.com), Jul 11 2006
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PROGRAM
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(PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, #binary(n), A=subst(A, x, x^3+x*O(x^n))*(1-4*x^3-x^6)/(1-x-x^2+x*O(x^n))); polcoeff(A, n, x)} - Paul D. Hanna (pauldhanna(AT)juno.com), Jul 11 2006
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CROSSREFS
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Sequence in context: A059451 A083817 A029273 this_sequence A112803 A124242 A112274
Adjacent sequences: A117960 A117961 A117962 this_sequence A117964 A117965 A117966
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KEYWORD
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easy,sign
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AUTHOR
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Paul Barry (pbarry(AT)wit.ie), Apr 05 2006
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