A107946 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.

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

Offset: 1

Author

Paul D. Hanna, May 28 2005

Keywords

• nonn
• look

Comments

The partial sums is A107947. Terms at positions 2^k forms A107948.

Examples

			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}.
		

Links

• Ivan Neretin, Table of n, a(n) for n = 1..8192

Crossrefs

Cf. A107947, A107948.

Programs

Mathematica

Nest[Join[#, Accumulate@#] &, {1}, 7] (* Ivan Neretin, Jan 31 2018 *)

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]}