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%I #11 Sep 08 2018 16:08:40
%S 1,1,2,1,3,5,1,4,9,7,1,5,13,13,12,1,6,17,19,23,15,1,7,21,25,34,29,22,
%T 1,8,25,31,45,43,43,26,1,9,29,37,56,57,64,51,35,1,10,33,43,67,71,85,
%U 76,69,40,1,11,37,49,78,85,106,101,103,79,51
%N Triangle read by rows, antidiagonals of an array formed from variants of A001318, generalized pentagonal numbers.
%H Andrew Howroyd, <a href="/A175009/b175009.txt">Table of n, a(n) for n = 1..1275</a> (first 50 rows)
%F Let row 1 of the array = A001318 starting with offset 1: (1, 2, 5, 7, 12,...)
%F For rows k>1, begin with A026741 starting (1, 3, 2, 5, 3, 7, 4, 9, 5, 11,...)
%F = generator Q. Then k-th row = partial sums of (1,...(k * Q)).
%F T(n,k) = 1 + (n-k+1)*(binomial(k+1, 2) - 1 - binomial(floor(k/2)+1, 2)). - _Andrew Howroyd_, Sep 08 2018
%e First few rows of the array:
%e 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, ...
%e 1, 3, 9, 13, 23, 29, 43, 51, 69, 79, ...
%e 1, 4, 13, 19, 34, 43, 64, 76, 103, 118, ...
%e 1, 5, 17, 25, 45, 57, 85, 101, 137, 157, ...
%e 1, 6, 21, 31, 56, 71, 106, 126, 171, 196, ...
%e ...
%e Example: row 3 is generated from 3 * (1, 3, 2, 5, 3, 7, ...) = (3, 9, 6, 15,...)
%e Preface with a 1 getting (1, 3, 9, 6, 15, ...) then take partial sums, = (1, 4, 13, 19, 34, 43, 64, ...).
%e ...
%e First few rows of the triangle:
%e 1;
%e 1, 2
%e 1, 3, 5;
%e 1, 4, 9, 7;
%e 1, 5, 13, 13, 12;
%e 1, 6, 17, 29, 23, 15;
%e 1, 7, 21, 25, 34, 29, 22;
%e 1, 8, 25, 31, 45, 43, 43, 26;
%e 1, 9, 29, 37, 56, 57, 64, 51, 35;
%e 1, 10, 33, 43, 67, 71, 85, 76, 69, 40;
%e 1, 11, 37, 49, 78, 85, 106, 101, 103, 79, 51;
%e 1, 12, 41, 55, 89, 99, 127, 126, 137, 118, 101, 57;
%e 1, 13, 45, 61, 100, 113, 148, 151, 171, 157, 151, 113, 70;
%e 1, 14, 49, 67, 111, 127, 169, 176, 205, 196, 201, 169, 139, 77;
%e ...
%o (PARI) T(n,k)=if(k<=n, 1 + (n-k+1)*(binomial(k+1, 2) - 1 - binomial(k\2+1, 2)), 0) \\ _Andrew Howroyd_, Sep 08 2018
%Y Row sums are A175006.
%Y Cf. A001318, A026741.
%K nonn,tabl
%O 1,3
%A _Gary W. Adamson_, Apr 03 2010
%E a(22) corrected by _Andrew Howroyd_, Sep 08 2018
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