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A081498
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Consider the triangle in which the n-th row starts with n, contains n terms and the difference of successive terms is 1,2,3,... up to n-1. Sequence gives row sums.
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2
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1, 3, 5, 6, 5, 1, -7, -20, -39, -65, -99, -142, -195, -259, -335, -424, -527, -645, -779, -930, -1099, -1287, -1495, -1724, -1975, -2249, -2547, -2870, -3219, -3595, -3999, -4432, -4895, -5389, -5915, -6474, -7067, -7695, -8359, -9060, -9799, -10577, -11395, -12254, -13155, -14099, -15087, -16120
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OFFSET
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1,2
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COMMENTS
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The triangle whose row sums are being considered is:
1;
2, 1;
3, 2, 0;
4, 3, 1, -2;
5, 4, 2, -1, -5;
6, 5, 3, 0, -4, -9;
7, 6, 4, 1, -3, -8, -14;
The leading diagonal is given by A080956(n-1) = n*(3-n)/2.
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LINKS
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FORMULA
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a(n) = n^2 - binomial(n+1, n-2). - C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004
a(n) = binomial(n,2)+binomial(n,1)-binomial(n,3). - Zerinvary Lajos, Jul 23 2006
G.f.: x * (1 - x - x^2) / (1 - x)^4.
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EXAMPLE
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G.f. = x * (1 + 3*x + 5*x^2 + 6*x^3 + 5*x^4 + x^5 - 7*x^6 - 20*x^7 - 39*x^8 - 65*x^9 + ...).
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MAPLE
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seq(n^2-binomial(n+1, n-2), n=1..50); # C. Ronaldo
[seq(binomial(n, 2)+binomial(n, 1)-binomial(n, 3), n=1..49)]; # Zerinvary Lajos, Jul 23 2006
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MATHEMATICA
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LinearRecurrence[{4, -6, 4, -1}, {1, 3, 5, 6}, 50] (* G. C. Greubel, Mar 06 2019 *)
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PROG
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(PARI) {a(n) = if( n< 0, n = -2 - n; polcoeff( (1 + x - x^2) / (1 - x)^4 + x * O(x^n), n), polcoeff( (1 - x - x^2) / (1 - x)^4 + x * O(x^n), n))} /* Michael Somos, Jul 04 2012 */
(PARI) vector(50, n, n*(1+6*n-n^2)/6) \\ G. C. Greubel, Mar 06 2019
(GAP) List([1..50], n->n^2-Binomial(n+1, n-2)); # Muniru A Asiru, Mar 05 2019
(Magma) [n*(1+6*n-n^2)/6: n in [1..50]]; // G. C. Greubel, Mar 06 2019
(Sage) [n*(1+6*n-n^2)/6 for n in (1..50)] # G. C. Greubel, Mar 06 2019
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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EXTENSIONS
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More terms from C. Ronaldo (aga_new_ac(AT)hotmail.com), Dec 20 2004
Offset changed to 1 at the suggestion of Michel Marcus, Mar 05 2019
Formulas and programs addapted for offset 1 by Michel Marcus, Mar 05 2019
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STATUS
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approved
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