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Search: id:A129995
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| A129995 |
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a(n) = (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*(n^4 + 4)/4!. |
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+0
14
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| 1, 5, 165, 4675, 65325, 543456, 3155425, 14146210, 52259625, 166192975, 469090061, 1201490445, 2839166005, 6268589250, 13060542825, 25881747316, 49095506065, 89615392425, 158091087925, 270522770375, 450420100221, 731644012660, 1162094343345, 1808433948150
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OFFSET
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0,2
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COMMENT
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Following my conjecture, computations by Peter J. C. Moses, mediation by Clark Kimberling and helpful comments from George E. Andrews, it is now known that a(n) = (n^1 + 1)*(n^2 + 2)*(n^3 + 3)*...*(n^k + k)/k! is an integer-valued sequence if and only if k belongs to {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 17, 18, 19, 20, 21}.
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FORMULA
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G.f.: (1-6x+165x^2+2970x^3+22480x^4+55969x^5+51511x^6+16490x^7+1595x^8+25x^9)/(1-x)^11. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 26 2007
G.f.: -(1 + x*(-6 + x*(165 + x*(2970 + x*(22480 + x*(55969 + x*(51511 + 5*x*(3298 + x*(319 + 5*x))))))))) / (x - 1)^11. - Peter J. C. Moses, Aug 29 2007
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MAPLE
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p:=proc(n, i) mul( n^j+j, j=1..i)/i!; end; [seq(p(n, 4), n=0..30)];
seq((n+1)*(n^2+2)*(n^3+3)*(n^4+4)/factorial(4), n = 0 .. 20) - Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 26 2007
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CROSSREFS
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Cf. A000027 (k=1), A064808 (k=2), A131509 (k=3), this sequence (k=4), A131675 (k=5), ..., A131680 (k=10).
See A131685 for a generalization.
Sequence in context: A136368 A117068 A128068 this_sequence A047940 A116629 A139986
Adjacent sequences: A129992 A129993 A129994 this_sequence A129996 A129997 A129998
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KEYWORD
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nonn
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AUTHOR
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Alexander Povolotsky (pevnev(AT)juno.com), Aug 19 2007, Aug 25 2007
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