%I #14 Oct 08 2024 03:14:05

%S 1,4,5,33,63,112,179,260,369,502,659,868,1098,1367,1703,2046,2457,

%T 2895,3392,3975,4591,5322,6069,6899,7807,8828,9836,10948,12290,13440,

%U 14838,16392,18046,19712,21259,23381,25382,27506,29679,32083

%N a(n) = (1/3^n) * Sum_{k=0..n^3} ( (binomial(n^3, k) * 2^k) (mod 3^n) ).

%H J. B. Roberts, <a href="https://doi.org/10.4153/CJM-1957-043-6">On Binomial Coefficient Residues</a>, Canadian Journal of Mathematics, Volume 9 1957, pp. 363 - 370.

%e The table of residues of coefficients of x^k in (1 + 2*x)^(n^3) modulo 3^n begins:

%e (1+2*x) (mod 3): [1, 2];

%e (1+2*x)^8 (mod 3^2): [1, 7, 4, 7, 4, 1, 1, 7, 4];

%e (1+2*x)^27 (mod 3^3): [1, 0, 0, 18, 0, 0, 9, 0, 0, 24, 0, 0, 18, 0, 0, 9, 0, 0, 3, 0, 0, 18, 0, 0, 9, 0, 0, 26];

%e (1+2*x)^64 (mod 3^4): [1, 47, 45, 78, 30, 72, 66, 6, 45, 11, 40, 54, 45, 36, 54, 36, 18, 54, 48, 15, 27, 63, 18, 27, 45, 63, 27, 46, 29, 72, 6, 21, 18, 3, 15, 72, 53, 43, 27, 9, 72, 27, 72, 36, 27, 33, 66, 54, 72, 9, 54, 63, 72, 54, 7, 32, 45, 51, 57, 72, 39, 33, 45, 44, 52];

%e (1+2*x)^125 (mod 3^5): [1, 7, 139, 220, 55, 232, 46, 106, 211, 140, 89, 182, 56, 95, 62, 140, 224, 128, 78, 60, 150, 42, 132, 168, 213, 195, 42, 65, 212, 44, 233, 119, 122, 47, 140, 242, 238, 127, 115, 214, 94, 46, 184, 100, 196, 171, 225, 198, 36, 9, 144, 9, 63, 36, 177, 24, 60, 222, 177, 159, 204, 132, 6, 12, 84, 210, 183, 228, 3, 174, 84, 48, 72, 18, 45, 207, 234, 99, 234, 180, 207, 95, 179, 83, 83, 203, 8, 158, 26, 38, 79, 229, 127, 145, 16, 13, 160, 13, 181, 165, 183, 93, 201, 111, 75, 30, 48, 201, 148, 64, 160, 214, 175, 208, 31, 82, 232, 17, 200, 95, 131, 53, 119, 71, 65, 176];

%e ...

%e where a(n) equals the sum of row n divided by 3^n:

%e a(1) = (1 + 2)/3 = 1;

%e a(2) = (1 + 7 + 4 + 7 + 4 + 1 + 1 + 7 + 4)/3^2 = 4;

%e a(3) = (1 + 0 + 0 + 18 + 0 + 0 + 9 + 0 + 0 + 24 + 0 + 0 + 18 + 0 + 0 + 9 + 0 + 0 + 3 + 0 + 0 + 18 + 0 + 0 + 9 + 0 + 0 + 26)/3^3 = 5;

%e a(4) = (1 + 47 + 45 + 78 + 30 + 72 + 66 + 6 + 45 + 11 + 40 + 54 + 45 + 36 + 54 + 36 + 18 + 54 + 48 + 15 + 27 + 63 + 18 + 27 + 45 + 63 + 27 + 46 + 29 + 72 + 6 + 21 + 18 + 3 + 15 + 72 + 53 + 43 + 27 + 9 + 72 + 27 + 72 + 36 + 27 + 33 + 66 + 54 + 72 + 9 + 54 + 63 + 72 + 54 + 7 + 32 + 45 + 51 + 57 + 72 + 39 + 33 + 45 + 44 + 52)/3^4 = 33;

%e ...

%o (PARI) {a(n) = sum(k=0,n^3, (binomial(n^3,k) * 2^k) % (3^n) )/3^n}

%o for(n=1,35,print1(a(n),", "))

%o (Python)

%o def A376536(n):

%o m, r, c, a, b, d = 3**n, n**3, 1, 2, n**3, 1

%o for k in range(1,r+1):

%o c += b//d*a%m

%o a = a*2%m

%o b *= r-k

%o d *= k+1

%o return c//m # _Chai Wah Wu_, Oct 07 2024

%Y Cf. A376531, A376532, A376533, A376534, A376535.

%K nonn

%O 1,2

%A _Paul D. Hanna_, Oct 06 2024