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<DIV>As noted in the posting(s) by Rob Johnson in that thread, the resulting sequence is the tetrahedral numbers A000292, plus 1 when n = 2 or 3 (mod 4):</DIV>
<DIV> </DIV>
<DIV>1, 5, 11, 20, 35, 57, 85, 120, 165, 221, 287, 364, 455, 561, 681, 816, 969, 1141, 1331, 1540, 1771, 2025, 2301, 2600, 2925, 3277, 3655, 4060, 4495, 4961, 5457, 5984, 6545, 7141, 7771, 8436, 9139, 9881, 10661, 11480, 12341, 13245, 14191, 15180, 16215, 17297, 18425, 19600, 20825, 22101<BR></DIV>
<DIV>I'll submit this sequence.</DIV>
<DIV> </DIV>
<DIV>Franklin T. Adams-Watters<BR> <BR>-----Original Message-----<BR>From: Leroy Quet <qq-quet@mindspring.com><BR>To: seqfan@ext.jussieu.fr<BR>Sent: Mon, 15 May 06 09:34:29 -0600<BR>Subject: Minimizing the sum(permutation*inverse)<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_22b36aab-e8fc-47fd-a3bc-e3aaec53e96d><PRE><TT>Regarding the sci.math thread:
<A href="http://groups.google.com/group/sci.math/browse_thread/thread/6cf265b0d0e2f1" target=_blank>http://groups.google.com/group/sci.math/browse_thread/thread/6cf265b0d0e2f1</A>57
Let {b(k)} be a permutation of {1,2,3,...,n}.
Let {c(k)} be the inverse-permutation of {b(k)}.
(ie. b(c(j)) =j, for every j.)
What is the minimum possible sum:
sum{k=1 to n} b(k) * c(k) ?
For example, if b is:
[1,2,3],
[2,1,3],
[3,2,1],
each give a sum of 14 (since each of these permutations is its own
inverse).
But
[2,3,1],
[3,1,2],
(which are inverses of each other)
both give a sum of 11.
I get (possibly erroneously) that the sequences of minimum sums begins:
1, 5, 11, 20, 35,...
Could someone please calculate/submit this sequence (unless it is already
in the EIS, of course).
thanks,
Leroy Quet
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