Timeline for What is this sequence counting?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 2, 2019 at 8:03 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
added 19 characters in body
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| Sep 2, 2019 at 7:36 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
added 24 characters in body
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| Sep 2, 2019 at 7:31 | comment | added | Bjørn Kjos-Hanssen | @TheTwistedSector sorry I think I fixed it now | |
| Sep 2, 2019 at 7:27 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
from wrong to right
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| Sep 2, 2019 at 7:18 | history | undeleted | Bjørn Kjos-Hanssen | ||
| Sep 2, 2019 at 6:10 | history | deleted | Bjørn Kjos-Hanssen | via Vote | |
| Sep 2, 2019 at 3:33 | comment | added | TheTwistedSector |
Here's the Mathematica code. add2[lst_] := Table[lst + 2 IdentityMatrix[Length[lst]][[i]], {i, 1, Length[lst]}]; cBoole[f_] := 1 - Boole[f]; isSpecial[lst_] := cBoole@MemberQ[cBoole@MemberQ[#, 1] & /@ add2[lst], 1]; isForm[lst_] := If[EvenQ[Total[lst]] && Count[lst, 1] == 2 && Count[lst, 2] == Length[lst] - 2, 0, 1]; which[n_] := (isForm[#] & /@ Partitions[n])*(isSpecial[#] & /@ Partitions[n]); bruteCount[n_] := Total[which[n]]; Table[bruteCount[n], {n, 1, 15}]
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| Sep 2, 2019 at 3:32 | comment | added | TheTwistedSector | Thanks for the answer. I implemented this on the computer and I seem to get the sequence 0, 0, 1, 1, 3, 4, 7, 10, 15, 21, 30, 41, 56, 76, 101,.. i.e. there is agreement till the 10. It is quite possible I am making a mistake though. | |
| Sep 1, 2019 at 7:06 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
added 81 characters in body
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| Sep 1, 2019 at 7:00 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |