A164364 -id:A164364 - cp's OEIS frontend

A164363 The number of 1's in the n-th stage of A164349.

1, 1, 2, 3, 6, 12, 23, 46, 91, 182, 364, 727, 1454, 2908, 5816, 11631, 23262, 46523, 93046, 186092, 372183, 744366, 1488732, 2977463, 5954926, 11909851, 23819702, 47639404, 95278807, 190557614, 381115228, 762230456, 1524460911, 3048921822, 6097843643

Offset: 1

Jack W Grahl, Aug 14 2009

nonn

A164349 is generated as follows. Start with the string 01, and at each stage copy the previous string twice and remove the last symbol.

Since the number of symbols in the whole string is 2^n + 1, A164363 + A164362 = 2^n + 1.

			01 -> 010 -> 01001 -> 010010100 -> 01001010001001010 etc.
So the number of 1's in the n-th stage is the sequence 1, 1, 2, 3, 6 etc.

Paul Tek, Table of n, a(n) for n = 1..3324

Cf. A164349, A164362, A164364.

t = Nest[ Most@ Flatten@ {#, #} &, {0, 1}, 25]; Table[ Count[ Take[t, 2^n + 1], 1], {n, 0, 25}] (* Robert G. Wilson v, Aug 17 2009 *)

A recurrence is given in terms of A164364(n) = A164349(2^n):

a(n+1) = 2*a(n) - A164364(n).

a(23)-a(26) from Robert G. Wilson v, Aug 17 2009

A164362 The number of 0's in the n-th stage of A164349.

Original entry on oeis.org

1, 2, 3, 6, 11, 21, 42, 83, 166, 331, 661, 1322, 2643, 5285, 10569, 21138, 42275, 84550, 169099, 338197, 676394, 1352787, 2705573, 5411146, 10822291, 21644582, 43289163, 86578325, 173156650, 346313299, 692626597, 1385253193, 2770506386, 5541012771, 11082025542

Offset: 1

Jack W Grahl, Aug 14 2009

nonn

A164349 is generated as follows. Start with the string 01, and at each stage copy the previous string twice and remove the last symbol.

Since the number of symbols in the whole string is 2^n + 1, A164362 + A164363 = 2^n + 1.

			01 -> 010 -> 01001 -> 010010100 -> 01001010001001010 etc.
So the number of 0's in the n-th stage is the sequence 1, 2, 3, 6, 11 etc.

Paul Tek, Table of n, a(n) for n = 1..3323
Paul Tek, PERL program for this sequence

Cf. A164349, A164363, A164364.

t = Nest[ Most@ Flatten@ {#, #} &, {0, 1}, 25]; Table[ Count[ Take[t, 2^n + 1], 0], {n, 0, 25}] (* Robert G. Wilson v, Aug 17 2009 *)

A recurrence is given in terms of A164364(n) = A164349(2^n).

a(n+1) = 2*a(n) + A164364(n) - 1.

a(23)-a(26) from Robert G. Wilson v, Aug 17 2009

cp's OEIS Frontend

A164363 The number of 1's in the n-th stage of A164349.

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