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Revision History for A001317 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
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Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.
(history; published version)

#292 by Joerg Arndt at Sun Jun 30 02:01:42 EDT 2024

STATUS

editing

proposed

#291 by Joerg Arndt at Sun Jun 30 02:01:26 EDT 2024

FORMULA

For n >= 1, m >= 2, a(2^(n + Floor(m/2)) - m - 2) = Round(2^2^(n + Floor(m/2))/a(m + 1)). - Alan Michael Gómez Calderón, Jun 29 2024

MATHEMATICA

fa[n_] := Nest[ BitXor[#, BitShiftLeft[#, 1]] &, 1, n]; Array[f, a, 42, 0] (* Joel Madigan (dochoncho(AT)gmail.com), Dec 03 2007 *)

f[n_] := FromDigits[ Table[ Mod[ Binomial[n, k], 2], {k, 0, n}], 2]; Array[f, 42, 0] (* Robert G. Wilson v *)

#290 by Alan Michael Gómez Calderón at Sat Jun 29 21:45:55 EDT 2024

FORMULA

For n >= 1, m >= 1, 2, a(2^(n + Floor(m/2)) - m - 2) = Round(2^2^(n + Floor(m/2))/a(m + 1)). - Alan Michael Gómez Calderón, Jun 29 2024

Discussion

Sun Jun 30

01:26

Kevin Ryde: But I believe you have a restricted, and moderately obscured, form of Vladimir Shevelev's "a(n+m) = a(n)*a(m)" and I'd recommend against for that reason.

01:28

Kevin Ryde: (What you're looking for in the 2^(n + Floor(m/2)) is any power 2^x > n and m ...)

01:31

Alan Michael Gómez Calderón: Hello Kevin Ryde, could you provide an example?

01:37

Kevin Ryde: a(9)*a(6) = a(15) since 9 = binary 1001 and 6 = binary 110 have no bits in common.  Their total is 9+6 = 15 = binary 1111 and a(15) = 2^(2^4) - 1.  (That pattern for a(2^k-1) is, umm, hiding among the formulas ...)

01:49

Alan Michael Gómez Calderón: I just want to show that my formula work like a(2^n) = A000215(n), but shifted by a 'm' factor.

02:01

Kevin Ryde: At least clean it up.  It works with more exponents than n + Floor(m/2).  (You're still a special case of Vladimir Shevelev ...)

#289 by Alan Michael Gómez Calderón at Sat Jun 29 21:34:31 EDT 2024

FORMULA

For n >= 1, m >= 1, a(2^(n + Floor(m/2)) - m - 12) = Round(2^2^(n + Floor(m/2))/a(m + 1)). - Alan Michael Gómez Calderón, Jun 29 2024

Discussion

Sun Jun 30

01:23

Kevin Ryde: Your Round() is exact with a "-1" in the numerator?

#288 by Alan Michael Gómez Calderón at Sat Jun 29 21:31:10 EDT 2024

FORMULA

for For n >= 1, m >= 1, a(2^(n + Floor(m/2)) - m - 1) = Round(2^2^(n + Floor(m/2) + 1)/a(m + 1)). - Alan Michael Gómez Calderón, Jun 29 2024

STATUS

proposed

editing

#287 by Alan Michael Gómez Calderón at Sat Jun 29 19:06:29 EDT 2024

STATUS

editing

proposed

Discussion

Sat Jun 29

20:39

Andrey Zabolotskiy: Sorry but I don't see how this works. E.g. for n=2,m=1 we apparently get a(3)=51, which is incorrect.

21:07

Alan Michael Gómez Calderón: ok Sorry, its a offset issue, I will solve it, Thanks

#286 by Alan Michael Gómez Calderón at Sat Jun 29 17:12:25 EDT 2024

FORMULA

for n >= 1, m >= 1, a(2^(n + Floor(m/2)) - m) = Round(2^2^(n + Floor(m/2) + 1)/a(m + 1)). _- _Alan Michael Gómez Calderón_, Jun 29 2024

#285 by Alan Michael Gómez Calderón at Sat Jun 29 17:05:07 EDT 2024

FORMULA

for n >= 1, m >= 1, a(2^(n + Floor(m/2)) - m) = Round(2^2^(n + Floor(m/2) + 1)/a(m + 1)). Alan Michael Gómez Calderón, Jun 29 2024

STATUS

approved

editing

Discussion

Sat Jun 29

17:06

Alan Michael Gómez Calderón: Hello dear editors, here I present to your consideration this formula. Have a good day.

#284 by Amiram Eldar at Mon Jun 10 01:53:53 EDT 2024

STATUS

reviewed

approved

#283 by Michel Marcus at Mon Jun 10 01:52:29 EDT 2024

STATUS

proposed

reviewed