A130253 - OEIS

A130253

Number of Jacobsthal numbers (A001045) <=n.

1, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9

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OFFSET

0,2

COMMENTS

Partial sums of the Jacobsthal indicator sequence (A105348).

For n<>1, we have a(A001045(n))=n+1.

LINKS

G. C. Greubel, Table of n, a(n) for n = 0..10000

Dorin Andrica, Ovidiu Bagdasar, and George Cătălin Tųrcąs, On some new results for the generalised Lucas sequences, An. Şt. Univ. Ovidius Constanţa (Romania, 2021) Vol. 29, No. 1, 17-36.

FORMULA

a(n) = floor(log_2(3n+1)) + 1 = ceiling(log_2(3n+2)).

a(n) = A130249(n) + 1 = A130250(n+1).

G.f.: 1/(1-x)*(Sum_{k>=0} x^A001045(k)).

EXAMPLE

a(9)=5 because there are 5 Jacobsthal numbers <=9 (0,1,1,3 and 5).

MATHEMATICA

Table[1+Floor[Log[2, 3n+1]], {n, 0, 100}] (* Harvey P. Dale, Jul 03 2013 *)

PROG

(PARI) a(n)=logint(3*n+1, 2)+1 \\ Charles R Greathouse IV, Oct 03 2016

(Magma) [Ceiling(Log(3*n+2)/Log(2)): n in [0..30]]; // G. C. Greubel, Jan 08 2018

CROSSREFS