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Search: id:A111808
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| A111808 |
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Left half of trinomial triangle (A027907), triangle read by rows. |
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+0
22
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| 1, 1, 1, 1, 2, 3, 1, 3, 6, 7, 1, 4, 10, 16, 19, 1, 5, 15, 30, 45, 51, 1, 6, 21, 50, 90, 126, 141, 1, 7, 28, 77, 161, 266, 357, 393, 1, 8, 36, 112, 266, 504, 784, 1016, 1107, 1, 9, 45, 156, 414, 882, 1554, 2304, 2907, 3139, 1, 10, 55, 210, 615, 1452, 2850, 4740, 6765, 8350
(list; table; graph; listen)
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OFFSET
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1,5
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COMMENT
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Consider a doubly infinite chessboard with squares labeled (i,j), i in Z, j in Z; number of king-paths of length j from (0,0) to (i,j), 0 <= i <= j, is T(j,i-j). - Harrie Grondijs (hgrondijs(AT)epo.org), May 27 2005. Cf. A026300, A114929, A114972.
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REFERENCES
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Harrie Grondijs, Neverending Quest of Type C, Volume B - the endgame study-as-struggle.
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LINKS
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Eric Weisstein's World of Mathematics, Trinomial Triangle
Eric Weisstein's World of Mathematics, Trinomial Coefficient
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FORMULA
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(1 + x + x^2)^n = Sum(T(n,k)*x^k: 0<=k<=n) + Sum(T(n,k)*x^(2*n-k): 0<=k<n);
T(n, k) = A027907(n, k) = Sum(binomial(n, n-k+2*i) * binomial(n-k+2*i, i): 0<=i<k/2), 0<=k<=n.
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CROSSREFS
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Row sums give A027914; central terms give A027908;
T(n, 0)=0; T(n, 1)=n for n>1; T(n, 2)=A000217(n) for n>1;
T(n, 3) = A005581(n) for n>2;
T(n, 4) = A005712(n) for n>3;
T(n, 5) = A000574(n) for n>4;
T(n, 6) = A005714(n) for n>5;
T(n, 7) = A005715(n) for n>6;
T(n, 8) = A005716(n) for n>7;
T(n, 9) = A064054(n-5) for n>8;
T(n, n-5) = A098470(n) for n>4;
T(n, n-4) = A014533(n-3) for n>3;
T(n, n-3) = A014532(n-2) for n>2;
T(n, n-2) = A014531(n-1) for n>1;
T(n, n-1) = A005717(n) for n>0;
T(n, n) = central terms of A027907 = A002426(n).
Sequence in context: A118981 A117938 A101912 this_sequence A081422 A027555 A059481
Adjacent sequences: A111805 A111806 A111807 this_sequence A111809 A111810 A111811
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KEYWORD
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nonn,tabl
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AUTHOR
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Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 17 2005
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