[math-fun] Last 4 digits of S sum up to a square
Hello Math-fun, S = 1, 2, 4, 9,... If we want S to be the lexicographically earliest seq of distinct positive terms, we cannot have another start (like 1, 2, 3, X). The seq is finite — but how many terms can we plug in? I think we could extend S with 12: S = 1, 2, 4, 9, 12,... (the last 4 digits have sum 16 again — a square). Instead of 12 we could have extended S with 21, 30 or 999. This idea opens the way to a few variants, of course (sum k digits instead of 4, get a prime instead of a square), etc. Best, É. Catapulté de mon aPhone
Seq S is infinite, sorry; just extend it with a (new) term ending in —0001. And BTW, the start 0, 1, 2, 6, 7, 12 is not in the OEIS either. Best, É. Catapulté de mon aPhone
Le 18 juil. 2020 à 08:16, Éric Angelini <eric.angelini@skynet.be> a écrit :
Hello Math-fun, S = 1, 2, 4, 9,... If we want S to be the lexicographically earliest seq of distinct positive terms, we cannot have another start (like 1, 2, 3, X). The seq is finite — but how many terms can we plug in? I think we could extend S with 12: S = 1, 2, 4, 9, 12,... (the last 4 digits have sum 16 again — a square). Instead of 12 we could have extended S with 21, 30 or 999. This idea opens the way to a few variants, of course (sum k digits instead of 4, get a prime instead of a square), etc. Best, É. Catapulté de mon aPhone
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EA: "... the start 0, 1, 2, 6, 7, 12 is not in the OEIS either ..." I don't like the fact that 0+1+2 is not a square. I would start it 0, 1, 3, 5, 7, 13, ... When S has exhausted its supply of integers less than 1000 (994 at position 1193), subsequent terms are determined by the final four digits of the number itself. This is fairly predictable and somewhat boring. Here is a graph: http://chesswanks.com/seq/last4digits
boring
... indeed Hans — thanks for the computation! à+ É. Catapulté de mon aPhone
Le 18 juil. 2020 à 23:13, Hans Havermann <gladhobo@bell.net> a écrit :
EA: "... the start 0, 1, 2, 6, 7, 12 is not in the OEIS either ..."
I don't like the fact that 0+1+2 is not a square. I would start it 0, 1, 3, 5, 7, 13, ...
When S has exhausted its supply of integers less than 1000 (994 at position 1193), subsequent terms are determined by the final four digits of the number itself. This is fairly predictable and somewhat boring. Here is a graph:
http://chesswanks.com/seq/last4digits
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participants (2)
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Hans Havermann -
Éric Angelini