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 A322790 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j. 7
 1, 1, 1, 1, 3, 1, 1, 17, 5, 1, 1, 99, 49, 7, 1, 1, 577, 485, 97, 9, 1, 1, 3363, 4801, 1351, 161, 11, 1, 1, 19601, 47525, 18817, 2889, 241, 13, 1, 1, 114243, 470449, 262087, 51841, 5291, 337, 15, 1, 1, 665857, 4656965, 3650401, 930249, 116161, 8749, 449, 17, 1 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 LINKS Seiichi Manyama, Antidiagonals n = 0..139, flattened Wikipedia, Chebyshev polynomials. FORMULA a(n) = 2 * A322699(n) + 1. A(n,k) + sqrt(A(n,k)^2 - 1) = (sqrt(n+1) + sqrt(n))^(2*k). A(n,k) - sqrt(A(n,k)^2 - 1) = (sqrt(n+1) - sqrt(n))^(2*k). A(n,0) = 1, A(n,1) = 2*n+1 and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) for k > 1. A(n,k) = T_{k}(2*n+1) where T_{k}(x) is a Chebyshev polynomial of the first kind. T_1(x) = x. So A(n,1) = 2*n+1. EXAMPLE Square array begins:    1,  1,   1,    1,      1,       1,         1, ...    1,  3,  17,   99,    577,    3363,     19601, ...    1,  5,  49,  485,   4801,   47525,    470449, ...    1,  7,  97, 1351,  18817,  262087,   3650401, ...    1,  9, 161, 2889,  51841,  930249,  16692641, ...    1, 11, 241, 5291, 116161, 2550251,  55989361, ...    1, 13, 337, 8749, 227137, 5896813, 153090001, ... MATHEMATICA A[0, k_] := 1; A[n_, k_] := Sum[Binomial[2 k, 2 j]*(n + 1)^(k - j)*n^j, {j, 0, k}]; Table[A[n - k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 26 2018 *) CROSSREFS Columns 0-3 give A000012, A005408, A069129(n+1), A322830. Rows 0-9 give A000012, A001541, A001079, A011943(n+1), A023039, A077422, A097308, A068203, A056771, A078986. Main diagonal gives A173174. A(n-1,n) gives A173148(n). Cf. A322699, A322747. Sequence in context: A135021 A087987 A290311 * A333560 A176293 A176339 Adjacent sequences:  A322787 A322788 A322789 * A322791 A322792 A322793 KEYWORD nonn,tabl AUTHOR Seiichi Manyama, Dec 26 2018 STATUS approved

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Last modified April 5 06:13 EDT 2020. Contains 333238 sequences. (Running on oeis4.)