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A050068 a(n) = a(n-1) + a(m) for n >= 3, where m = 2*n - 3 - 2^(p+1) and p is the unique integer such that 2^p < n - 1 <= 2^(p+1), starting with a(1) = 1 and a(2) = 3. 10
1, 3, 4, 5, 9, 10, 14, 23, 37, 38, 42, 51, 65, 102, 144, 209, 353, 354, 358, 367, 381, 418, 460, 525, 669, 1022, 1380, 1761, 2221, 2890, 4270, 6491, 10761, 10762, 10766, 10775, 10789, 10826, 10868, 10933, 11077, 11430, 11788 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

The author of the Mathematica program below uses the initial conditions a(1) = 1, a(2) = 3, and a(3) = 4. This is not necessary. We get the same sequence by using the initial conditions a(1) = 1 and a(2) = 3. - Petros Hadjicostas, Nov 15 2019

LINKS

Ivan Neretin, Table of n, a(n) for n = 1..8193

MAPLE

a := proc(n) option remember; `if`(n < 3, [1, 3][n],

        a(n - 1) + a(-2^ceil(log[2](n - 1)) + 2*n - 3))

     end proc:

seq(a(n), n = 1..40); # Petros Hadjicostas, Nov 15 2019

MATHEMATICA

Fold[Append[#1, #1[[-1]] + #1[[#2]]] &, {1, 3, 4}, Flatten@Table[2 k - 1, {n, 5}, {k, 2^n}]] (* Ivan Neretin, Sep 07 2015 *)

CROSSREFS

Same as A050036 and A050052 except for the second term.

Cf. similar sequences with different initial conditions: A050024 (1,1,1), A050028 (1,1,2), A050032 (1,1,3), A050036 (1,1,4), A050040 (1,2,1), A050044 (1,2,2), A050048 (1,2,3), A050052 (1,2,4), A050056 (1,3,1), A050060 (1,3,2), A050064 (1,3,3).

Sequence in context: A228941 A236211 A279616 * A250444 A327178 A190211

Adjacent sequences:  A050065 A050066 A050067 * A050069 A050070 A050071

KEYWORD

nonn

AUTHOR

Clark Kimberling

EXTENSIONS

Name edited by Petros Hadjicostas, Nov 15 2019

STATUS

approved

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Last modified August 18 02:34 EDT 2022. Contains 356204 sequences. (Running on oeis4.)