A165149
a(n) = (3*9^n - 5^n)/2.
Original entry on oeis.org
1, 11, 109, 1031, 9529, 87011, 789349, 7135391, 64374769, 580154171, 5225293789, 47047175351, 423522234409, 3812188390931, 34312136924629, 308821439352911, 2779453989332449, 25015391079773291, 225140045596865869, 2026268039766324071, 18236450504869572889, 164128245278689437251
Offset: 0
A165147
a(n) = (3*7^n-3^n)/2.
Original entry on oeis.org
1, 9, 69, 501, 3561, 25089, 176109, 1234221, 8643921, 60520569, 423683349, 2965901541, 20761665081, 145332718449, 1017332217789, 7121335090461, 49849374331041, 348945706410729, 2442620203155429, 17098342196928981
Offset: 0
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[ (3*7^n-3^n)/2: n in [0..19] ];
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LinearRecurrence[{10, -21}, {1, 9}, 25] (* Paolo Xausa, Apr 22 2024 *)
Table[(3*7^n-3^n)/2,{n,0,20}] (* Harvey P. Dale, Aug 05 2025 *)
A181107
Triangle read by rows: T(n,k) is the number of 2 X 2 matrices over Z(n) having determinant congruent to k mod n, 1 <= n, 0 <= k <= n-1.
Original entry on oeis.org
1, 10, 6, 33, 24, 24, 88, 48, 72, 48, 145, 120, 120, 120, 120, 330, 144, 240, 198, 240, 144, 385, 336, 336, 336, 336, 336, 336, 736, 384, 576, 384, 672, 384, 576, 384, 945, 648, 648, 864, 648, 648, 864, 648, 648, 1450, 720, 1200, 720, 1200, 870, 1200, 720, 1200, 720
Offset: 1
From _Andrew Howroyd_, Jul 16 2018: (Start)
Triangle begins:
1;
10, 6;
33, 24, 24;
88, 48, 72, 48;
145, 120, 120, 120, 120;
330, 144, 240, 198, 240, 144;
385, 336, 336, 336, 336, 336, 336;
736, 384, 576, 384, 672, 384, 576, 384;
945, 648, 648, 864, 648, 648, 864, 648, 648;
... (End)
- Erdos Pal, Rows n=1..100 of triangle, flattened
- Richard P. Brent and Brendan D. McKay, Determinants and ranks of random matrices over Z_m, Discrete Mathematics 66 (1987) pp. 35-49.
- A. K. Gupta, Generalized hidden hexagon squares, The Fibonacci Quarterly, Vol 12, Number 1, Feb.1974, pp. 45-46.
- S. Hitotumatu, D. Sato, Star of David theorem (I), The Fibonacci Quarterly, Vol 13, Number 1, Feb.1975, p. 70.
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(* computing T(p^e,k) ; p=prime, 1<=e, 0<=k
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S(p,e)={my(u=vector(p^e)); my(t=(p-1)*p^(e-1)); u[1] = p^e + e*t; for(j=1, p^e-1, u[j+1] = t*(1+valuation(j, p))); vector(#u, k, sum(j=0, #u-1, u[j + 1]*u[(j+k-1) % #u + 1]))}
T(n)={my(f=factor(n), v=vector(n,i,1)); for(i=1, #f~, my(r=S(f[i,1], f[i,2])); for(j=0, #v-1, v[j + 1] *= r[j % #r + 1])); v}
for(n=1, 10, print(T(n))); \\ Andrew Howroyd, Jul 16 2018
Showing 1-3 of 3 results.
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