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A376498 Array read by ascending antidiagonals: A(n, k) = 2^k*Sum_{j=1..n} cos((2*j - 1)*Pi/(2*n + 1))^k.

%I A376498 #47 Oct 26 2024 10:59:33
%S A376498 0,1,0,2,1,0,3,1,1,0,4,1,3,1,0,5,1,5,4,1,0,6,1,7,4,7,1,0,7,1,9,4,13,
%T A376498 11,1,0,8,1,11,4,19,16,18,1,0,9,1,13,4,25,16,38,29,1,0,10,1,15,4,31,
%U A376498 16,58,57,47,1,0,11,1,17,4,37,16,78,64,117,76,1,0
%N A376498 Array read by ascending antidiagonals: A(n, k) = 2^k*Sum_{j=1..n} cos((2*j - 1)*Pi/(2*n + 1))^k.
%C A376498 It is only a conjecture that the A(n, k) are always integers.
%F A376498 A(n + k, 2*k - 1) = A(k, 2*k-1) = 4^(k-1).
%F A376498 Let P_n(x) be the polynomial: Sum_{k=0..n} x^k*A180870(n, k). Let R_n(x) be the polynomial Product_{k=0..n} x-Roots(P_n, k)^m. A(n, k) = abs([x^1] R_n(x))/2^(m*(n-1)), for n > 0. - _Thomas Scheuerle_, Oct 07 2024
%e A376498 Array starts:
%e A376498 [0] 0, 0,  0, 0,  0,  0,   0,  0,   0,   0,    0,    0, ...  [A000004]
%e A376498 [1] 1, 1,  1, 1,  1,  1,   1,  1,   1,   1,    1,    1, ...  [A000012]
%e A376498 [2] 2, 1,  3, 4,  7, 11,  18, 29,  47,  76,  123,  199, ...  [A000032]
%e A376498 [3] 3, 1,  5, 4, 13, 16,  38, 57, 117, 193,  370,  639, ...  [A096975]
%e A376498 [4] 4, 1,  7, 4, 19, 16,  58, 64, 187, 247,  622,  925, ...  [A094649]
%e A376498 [5] 5, 1,  9, 4, 25, 16,  78, 64, 257, 256,  874, 1013, ...  [A189234]
%e A376498 [6] 6, 1, 11, 4, 31, 16,  98, 64, 327, 256, 1126, 1024, ...  [A216605]
%e A376498 [7] 7, 1, 13, 4, 37, 16, 118, 64, 397, 256, 1378, 1024, ...
%e A376498 [8] 8, 1, 15, 4, 43, 16, 138, 64, 467, 256, 1630, 1024, ...
%e A376498 [9] 9, 1, 17, 4, 49, 16, 158, 64, 537, 256, 1882, 1024, ...
%o A376498 (C++) double gen(int n, int m) {
%o A376498     double s = 0, d = 1;
%o A376498     for(int i = 1; i <= m; i++) d *= 2;
%o A376498     for(int i = 1; i <= n; i++) {
%o A376498         double v = 1;
%o A376498         for(int j = 1; j <= m; j++) v *= cos((2 * i - 1) * M_PI / (2 * n + 1));
%o A376498         s += v * d;
%o A376498     }
%o A376498     return s;
%o A376498 }
%o A376498 (PARI) A(n, k) = 2^k*sum(j=1, n, (cos((2*j-1)*Pi/(2*n+1)))^k, x=0)
%Y A376498 Rows: A000004 (n=0), A000012 (n=1), A000032 (n=2), A096975 (n=3), A094649 (n=4), A189234 (n=5), A216605 (n=6, with alternate signs).
%Y A376498 Columns: A001477 (k=0), A057427 (k=1).
%Y A376498 Cf. A180870.
%K A376498 nonn,tabl
%O A376498 0,4
%A A376498 _Cheng-Jun Li_, Sep 25 2024