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Search: id:A034867
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| A034867 |
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Triangle of odd-numbered terms in rows of Pascal's triangle. |
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+0
8
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| 1, 2, 3, 1, 4, 4, 5, 10, 1, 6, 20, 6, 7, 35, 21, 1, 8, 56, 56, 8, 9, 84, 126, 36, 1, 10, 120, 252, 120, 10, 11, 165, 462, 330, 55, 1, 12, 220, 792, 792, 220, 12, 13, 286, 1287, 1716, 715, 78, 1, 14, 364, 2002, 3432, 2002, 364, 14, 15, 455, 3003, 6435, 5005, 1365, 105, 1
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Also triangle of numbers of n-sequences of 0,1 with k subsequences of consecutive 01 because this number is C(n+1,2*k+1). - Roger Cuculiere (cuculier(AT)imaginet.fr), Nov 16 2002
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REFERENCES
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A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 136.
L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.
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FORMULA
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T(n, k) = C(n+1, 2k+1) = Sum[i=k..n-k, C(i, k)*C(n-i, k) ].
E.g.f.: 1+(exp(x)*sinh(x*sqrt(y)))/sqrt(y). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Mar 20 2005
G.f.=1/[(1-z)^2-tz^2]. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 01 2005
T(n, k) = Sum_{ j = 0, . . ., n} A034839(j, k). - Philippe DELEHAM, May 18 2005
Pell(n+1) = A000129(n+1) = sum(k=0,...,n) T(n,k) * 2^k = (1/n!)sum(k=0,...,n) A131980(n,k) * 2^k . - Tom Copeland (tcjpn(AT)msn.com), Nov 30 2007
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EXAMPLE
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Triangle starts:
1;
2;
3,1;
4,4;
5,10,1;
6,20,6;
T(5,2)=6 because only the following binary sequences of length 5 contain 2 subsequences 01: 00101,01001,01010,01011,01101, and 10101.
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MAPLE
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seq(seq(binomial(n+1, 2*k+1), k=0..floor(n/2)), n=0..14); (Deutsch)
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CROSSREFS
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Cf. A007318, A034839.
Sequence in context: A100035 A090244 A096180 this_sequence A055446 A104706 A094137
Adjacent sequences: A034864 A034865 A034866 this_sequence A034868 A034869 A034870
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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njas
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EXTENSIONS
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More terms from Emeric Deutsch (deutsch(AT)duke.poly.edu), Apr 01 2005
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