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Search: id:A079547
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| A079547 |
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a(n)=[(n^6-(n-1)^6)-(n^2-(n-1)^2}]/60 |
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+0
12
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| 0, 1, 11, 56, 192, 517, 1183, 2408, 4488, 7809, 12859, 20240, 30680, 45045, 64351, 89776, 122672, 164577, 217227, 282568, 362768, 460229, 577599, 717784, 883960, 1079585, 1308411
(list; graph; listen)
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OFFSET
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1,3
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COMMENT
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Polynexus numbers of order 6.
A polynexus (subtractive) function is composed of two or more subtracted nexus numbers divided by an integer x. The general form of the formula is a(n)=[(n^p-(n-1)^p)-(n^q-(n-1)^q}]/x, where n, p, q and x are integers.
Already known: a(n)=[(n^5-(n-1)^5)-(n^3-(n-1)^3}]/24, giving A006322; a(n)=[(n^4-(n-1)^4)-(n^2-(n-1)^2}]/12, giving Sum(n^2); a(n)=[(n^3-(n-1)^3)-(n^1-(n-1)^1}]/6, giving Sum(n); a(n)=[(n^2-(n-1)^2)-(n^1-(n-1)^1}]/2, giving n; a(n)=[(n^2-(n-1)^2)-(n^0-(n-1)^0}]/1, giving 2*n-1... x equals, in those examples, 1,2,6,12,24. 3 is also possible.
Also number of monotone n-weightings of complete bipartite digraph K(3,2) if offset were 0; cf. A085464-A085465. - Goran Kilibarda, Vladeta Jovovic (vladeta(AT)Eunet.yu), Jul 01 2003
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FORMULA
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a(n) = SUM[i=1..n] (i^2 + i^4 )/2 = n*(2*n+1)*(n+1)*(3*n^2+3*n+4)/60 = A079547(n+1). - Vladeta Jovovic, Mar 17 2006
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CROSSREFS
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Cf. A006322, A000330, A000217, A047969, A003215.
Adjacent sequences: A079544 A079545 A079546 this_sequence A079548 A079549 A079550
Sequence in context: A099532 A041226 A042503 this_sequence A034264 A051946 A114030
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KEYWORD
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nonn
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AUTHOR
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Xavier Acloque, Jan 22 2003
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