1,2

Comments from T. D. Noe, Jul 27 2007: (Start) This is similar to A00123 and A005704, which both have a recursion a(n)=a(n-1)+a([n/k]), where k is 2 and 3, respectively. Those sequences count "partitions of k*n into powers of k". For the present sequence, k=phi. Does A006336(n) count the partitions of n*phi into powers of phi?

Answering my own question: If the recursion starts with a(0)=1, then I think we obtain "number of partitions of n*phi into powers of phi" (see A131882).

Here we need negative powers of phi also: letting p=phi and q=1/phi, we have

n=0: 0*p = {} for 1 partition,

n=1: 1*p = p = 1+q for 2 partitions,

n=2: 2*p = p+p = 1+p+q = 1+1+q+q = p^2+q for 4 partitions, etc.

So the present sequence, which starts with a(1)=1, counts 1/2 of the "number of partitions of n*phi into powers of phi". (End)

T. D. Noe, Table of n, a(n) for n=1..1000

Max Alekseyev, Proof of Paul Hanna's formula

Comment from Paul D. Hanna, Jul 22 2007: It seems that A006336 can be generated by a rule using the golden ratio phi: a(n) = a(n-1) + a([n/Phi]) for n>1 with a(1)=1 where phi = (sqrt(5)+1)/2, I.e. the number of even terms up to position n-1 equals n-1 - [n/Phi] for n>1 where Phi = (sqrt(5)+1)/2. (This is true - see the Alekseyev link.)

a(n)=a(n-1)+a(A060143(n)) for n>1; subsequence of A134409; A134408 and A134409 give first and second differences; A001950(n)=Min(m:A134409(m)=a(n)). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 24 2007

a[n_Integer] := a[n] = Block[{c, k}, c = 0; k = 1; While[k < n, If[ EvenQ[ a[k] ], c++ ]; k++ ]; Return[a[n - 1] + a[n - 1 - c] ] ]; a[1] = 1; a[2] = 2; Table[ a[n], {n, 0, 60} ]

(PARI) A006336(N=99) = local(a=vector(N, i, 1), e=0); for(n=2, #a, e+=0==(a[n]=a[n-1]+a[n-1-e])%2); a \\ M. F. Hasler (Maximilian.Hasler(AT)gmail.com), Jul 23 2007

Cf. A007604.

Adjacent sequences: A006333 A006334 A006335 this_sequence A006337 A006338 A006339

Sequence in context: A071424 A008762 A101018 this_sequence A070228 A006304 A039847

nonn,easy,nice

D. R. Hofstadter

More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Mar 07 2001