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Search: id:A107946
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| A107946 |
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Start with S(0)={1}, then S(k+1) equals the concatenation of S(k) with the partial sums of S(k); the limit gives this sequence. |
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5
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| 1, 1, 1, 2, 1, 2, 3, 5, 1, 2, 3, 5, 6, 8, 11, 16, 1, 2, 3, 5, 6, 8, 11, 16, 17, 19, 22, 27, 33, 41, 52, 68, 1, 2, 3, 5, 6, 8, 11, 16, 17, 19, 22, 27, 33, 41, 52, 68, 69, 71, 74, 79, 85, 93, 104, 120, 137, 156, 178, 205, 238, 279, 331, 399, 1, 2, 3, 5, 6, 8, 11, 16, 17, 19, 22, 27, 33
(list; graph; listen)
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OFFSET
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1,4
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COMMENT
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The partial sums is A107947. Terms at positions 2^k forms A107948.
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EXAMPLE
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Concatenate the initial 2^3 terms: {1,1,1,2,1,2,3,5} to the partial sums {1,2,3,5,6,8,11,16}
to obtain the initial 2^4 terms: {1,1,1,2,1,2,3,5, 1,2,3,5,6,8,11,16}.
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PROGRAM
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(PARI) {a(n)=local(A=[1, 1], B=[1]); for(i=1, #binary(n)-1, B=concat(B, vector(#B, k, polcoeff(Ser(A)/(1-x), #B+k-1))); A=concat(A, B); ); A[n]}
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CROSSREFS
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Cf. A107947, A107948.
Sequence in context: A002730 A081664 A117673 this_sequence A054502 A059346 A076492
Adjacent sequences: A107943 A107944 A107945 this_sequence A107947 A107948 A107949
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KEYWORD
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nonn
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), May 28 2005
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