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A101855
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a(n) = n*(n+1)*(n+2)*(n+4)*(n+23)/120.
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1
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6, 30, 91, 216, 441, 812, 1386, 2232, 3432, 5082, 7293, 10192, 13923, 18648, 24548, 31824, 40698, 51414, 64239, 79464, 97405, 118404, 142830, 171080, 203580, 240786, 283185, 331296, 385671, 446896, 515592, 592416, 678062, 773262, 878787, 995448
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OFFSET
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1,1
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COMMENTS
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6th partial summation within series as series accumulate n times from an initial sequence of Euler Triangle's row 3: 1,4,1. 6th row of the array shown in A101853. Partial sums of A101854.
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LINKS
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FORMULA
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a(1)=6, a(2)=30, a(3)=91, a(4)=216, a(5)=441, a(6)=812, a(n)=6*a(n-1)- 15*a(n-2)+ 20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Harvey P. Dale, Feb 07 2013
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MATHEMATICA
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Table[n(n+1)(n+2)(n+4)(n+23)/120, {n, 40}] (* or *) LinearRecurrence[ {6, -15, 20, -15, 6, -1}, {6, 30, 91, 216, 441, 812}, 40](* Harvey P. Dale, Feb 07 2013 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 18 2004
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STATUS
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approved
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