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A071922
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Unimodal analog of binomial coefficient, such that A071921(n,m) = a(n+m-1,n) for all (n,m) different from (0,0), arranged in a Pascal-like triangle.
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6
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1, 1, 1, 1, 2, 1, 1, 3, 4, 1, 1, 4, 9, 7, 1, 1, 5, 16, 22, 11, 1, 1, 6, 25, 50, 46, 16, 1, 1, 7, 36, 95, 130, 86, 22, 1, 1, 8, 49, 161, 295, 296, 148, 29, 1, 1, 9, 64, 252, 581, 791, 610, 239, 37, 1, 1, 10, 81, 372, 1036, 1792, 1897, 1163, 367, 46, 1, 1, 11, 100, 525, 1716, 3612
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OFFSET
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0,5
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COMMENTS
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Also, number of n-length k-ary words avoiding the pattern 1'-2-1". - Ralf Stephan, Apr 28 2004
The matrix inverse starts
1;
-1, 1;
1, -2, 1;
-2, 5, -4, 1;
8, -21, 19, -7, 1;
-56, 148, -137, 55, -11, 1;
608, -1608, 1493, -608, 130, -16, 1;
-9440, 24968, -23190, 9461, -2044, 266, -22, 1;
198272, -524416, 487088, -198761, 42997, -5642, 490, -29, 1; - R. J. Mathar, Mar 15 2013
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LINKS
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FORMULA
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a(n, m) = Sum_{k=0..n-m} binomial(2*k+m-1, 2*k).
Sum_{m=0..n} a(n, m) = 1 + Fibonacci(2*n).
Sum_{m=0..n} (-1)^m*a(n, m) = 1 if 3 divides n, 0 otherwise.
G.f. for k-th row: 1/(1-x)^(2k-1) + Sum_{j=1..k-1} x/(1-x)^(2j). - Ralf Stephan, Apr 28 2004
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EXAMPLE
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Triangle begins
1;
1, 1;
1, 2, 1;
1, 3, 4, 1;
1, 4, 9, 7, 1;
1, 5, 16, 22, 11, 1;
1, 6, 25, 50, 46, 16, 1;
1, 7, 36, 95, 130, 86, 22, 1;
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MAPLE
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add( binomial(2*j+k-1, 2*j), j=0..n-k) ;
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MATHEMATICA
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a[n_, m_]:= Sum[Binomial[2k+m-1, 2k], {k, 0, n-m}]; Flatten[ Table[ a[n, m], {n, 0, 11}, {m, 0, n}]]
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PROG
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(PARI) a(n, k) = sum(j=0, n-k, binomial(2*j+k-1, 2*j));
for(n=0, 11, for(k=0, n, print1(a(n, k), ", "))) \\ G. C. Greubel, Aug 26 2019
(Magma) [&+[Binomial(2*j+k-1, 2*j): j in [0..n-k]]: k in [0..n], n in [0..11]]; // G. C. Greubel, Aug 26 2019
(Sage) [[sum(binomial(2*j+k-1, 2*j) for j in (0..n-k)) for k in (0..n)] for n in (0..11)] # G. C. Greubel, Aug 26 2019
(GAP) Flat(List([0..11], n-> List([0..n], k-> Sum([0..n-k], j-> Binomial( 2*j+k-1, 2*j) )))); # G. C. Greubel, Aug 26 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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Michele Dondi (bik.mido(AT)tiscalinet.it), Jun 14 2002
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EXTENSIONS
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STATUS
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approved
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