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A262927
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a(n+9) = a(n) + 10*(n+4) + 9. a(0)=0, a(1)=1, a(2)=3, a(3)=6, a(4)=10, a(5)=15, a(6)=23, a(7)=30, a(8)=39.
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1
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0, 1, 3, 6, 10, 15, 23, 30, 39, 49, 60, 72, 85, 99, 114, 132, 149, 168, 188, 209, 231, 254, 278, 303, 331, 358, 387, 417, 448, 480, 513, 547, 582, 620, 657, 696, 736, 777, 819, 862, 906, 951, 999, 1046, 1095, 1145, 1196, 1248, 1301, 1355, 1410, 1468, 1525
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OFFSET
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0,3
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COMMENTS
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The main (or principal) sequence for the 11 steps recurrence is 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 30, 33, 36, ..., the partial sums of A054898.
a(n) mod 9 is a sequence of period 90.
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,0,0,0,1,-2,1).
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FORMULA
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a(n) = 2*a(n-1) - a(n-2) + a(n-9) - 2*a(n-10) + a(n-11), n>10.
G.f.: -x*(x^8+2*x^7-x^6+3*x^5+x^4+x^3+x^2+x+1) / ((x-1)^3*(x^2+x+1)*(x^6+x^3+1)). - Colin Barker, Oct 04 2015
a(n) = (5n^2 + 4n)/9 + O(1), or more precisely (5n^2 + 4n + 3)/9 <= a(n) <= (5n^2 + 4n - 10)/9. - Charles R Greathouse IV, Oct 16 2015
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MATHEMATICA
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LinearRecurrence[{2, -1, 0, 0, 0, 0, 0, 0, 1, -2, 1}, {0, 1, 3, 6, 10, 15, 23, 30, 39, 49, 60}, 60] (* Vincenzo Librandi, Oct 06 2015 *)
RecurrenceTable[{a[n+9] == a[n] + 10*(n+4) + 9, a[0]=0, a[1]=1, a[2]=3, a[3]=6, a[4]=10, a[5]=15, a[6]=23, a[7]=30, a[8]=39}, a, {n, 0, 1000}] (* G. C. Greubel, Oct 16 2015 *)
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PROG
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(PARI) a(n) = numerator(((2*n)^2+4)/4)\9 + numerator(((2*n+1)^2+4)/4)\9;
(PARI) concat(0, Vec(-x*(x^8+2*x^7-x^6+3*x^5+x^4+x^3+x^2+x+1)/((x-1)^3*(x^2+x+1)*(x^6+x^3+1)) + O(x^100))) \\ Colin Barker, Oct 04 2015
(Magma) I:=[0, 1, 3, 6, 10, 15, 23, 30, 39]; [n le 9 select I[n] else (Self(n-9)+10*(n-6)+9): n in [1..60]]; // Vincenzo Librandi, Oct 06 2015
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CROSSREFS
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KEYWORD
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nonn,easy,less
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AUTHOR
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STATUS
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approved
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