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Search: id:A131823
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| A131823 |
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Triangle read by rows of [2^(n+1) - (n+1)] terms where row n has the g.f.: Product_{i=0..n-1} (1 + x^(2^i))^(n-i) for n>0, starting with a '1' in row 0. |
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+0
2
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| 1, 1, 1, 1, 2, 2, 2, 1, 1, 3, 5, 7, 8, 8, 8, 8, 7, 5, 3, 1, 1, 4, 9, 16, 24, 32, 40, 48, 55, 60, 63, 64, 64, 64, 64, 64, 63, 60, 55, 48, 40, 32, 24, 16, 9, 4, 1, 1, 5, 14, 30, 54, 86, 126, 174, 229, 289, 352, 416, 480, 544, 608, 672, 735, 795, 850, 898, 938, 970, 994, 1010, 1019
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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Generating rule: Start with a single '1' in row 0; let S(n) denote the initial [2^(n+1) - (n+1)] terms of the partial sums of row n; generate row n+1 by concatenating the following: S(n), 2^[n*(n-1)/2] repeated (n-1) times, and the terms of S(n) when read in reverse order.
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FORMULA
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Row sums are 2^(n*(n+1)/2) for n>=0.
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EXAMPLE
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Triangle begins:
1;
1,1;
1,2, 2, 2,1;
1,3,5,7,8, 8,8, 8,7,5,3,1;
1,4,9,16,24,32,40,48,55,60,63,64, 64,64,64, 64,63,60,55,48,40,32,24,16,9,4,1; ...
Illustrate the row g.f.s by:
(1+x)^2*(1+x^2) = g.f. of row 2: [1,2,2,2,1];
(1+x)^3*(1+x^2)^2*(1+x^4) = g.f. of row 3: [1,3,5,7,8,8,8,8,7,5,3,1];
(1+x)^4*(1+x^2)^3*(1+x^4)^2*(1+x^8) = g.f. of row 4.
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PROGRAM
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(PARI) {T(n, k)=local(A=[1]); if(n==0, 1, for(i=0, n-1, A=concat(Vec((Polrev(A)+O(x^(#A+i)))/(1-x)), Vec(O(x^(#A))+Pol(Vec(Ser(A)/(1-x)))))); A[k+1])}
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CROSSREFS
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Cf. A131824 (main diagonal).
Sequence in context: A087698 A101677 A128084 this_sequence A089722 A079562 A129320
Adjacent sequences: A131820 A131821 A131822 this_sequence A131824 A131825 A131826
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Jul 19 2007
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