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A079547
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((n^6-(n-1)^6)-(n^2-(n-1)^2))/60.
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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; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| 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: ((n^5-(n-1)^5)-(n^3-(n-1)^3))/24, giving A006322 for n>1; ((n^4-(n-1)^4)-(n^2-(n-1)^2))/12, giving A000330; ((n^3-(n-1)^3)-(n^1-(n-1)^1))/6, giving A000217; ((n^2-(n-1)^2)-(n^1-(n-1)^1))/2, giving n; ((n^2-(n-1)^2)-(n^0-(n-1)^0))/1, giving 2*n-1. In those examples, x is equal to 1,2,6,12,24, and 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.rs), Jul 01 2003
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LINKS
| Bruno Berselli, Table of n, a(n) for n = 1..1000
Index to sequences with linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
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FORMULA
| a(n+1) = sum(i=1..n) (i^2 + i^4)/2 = n*(2*n+1)*(n+1)*(3*n^2+3*n+4)/60. - Vladeta Jovovic, Mar 17 2006
G.f.: x^2*(x+1)*(1+4*x+x^2)/(1-x)^6. - Bruno Berselli, Feb 13 2012
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MATHEMATICA
| Table[((n^6 - (n - 1)^6) - (n^2 - (n - 1)^2))/60, {n, 27}] (* Bruno Berselli, Feb 13 2012 *)
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CROSSREFS
| Cf. A006322, A000330, A000217, A047969, A003215.
Cf. A083200, A088889-A088894 (similar sequences).
Sequence in context: A099532 A041226 A042503 * A034264 A051946 A201150
Adjacent sequences: A079544 A079545 A079546 * A079548 A079549 A079550
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KEYWORD
| nonn,easy,changed
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AUTHOR
| Xavier Acloque, Jan 22 2003
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