A257067 Number of length 5 1..(n+1) arrays with every leading partial sum divisible by 2 or 3.
3, 20, 113, 243, 1024, 1636, 3866, 6599, 12387, 16807, 32768, 41744, 66291, 90598, 133205, 161051, 248832, 292932, 401910, 501113, 661703, 759375, 1048576, 1185856, 1507979, 1788296, 2222649, 2476099, 3200000, 3532100, 4287258, 4926235, 5889323
Offset: 1
Keywords
Examples
Some solutions for n=4 ..3....2....2....3....3....2....4....4....2....3....2....4....4....4....3....2 ..5....4....1....3....3....1....5....5....4....3....1....4....5....4....1....1 ..4....3....1....3....4....5....3....3....3....2....1....2....3....2....2....1 ..2....5....2....1....5....2....4....3....1....1....4....4....2....5....2....4 ..1....2....2....4....3....5....4....3....5....5....1....2....1....5....2....2
Links
- R. H. Hardin, Table of n, a(n) for n = 1..210
Crossrefs
Cf. A257062
Formula
Empirical: a(n) = a(n-2) +a(n-3) -a(n-5) +4*a(n-6) -4*a(n-8) -4*a(n-9) +4*a(n-11) -6*a(n-12) +6*a(n-14) +6*a(n-15) -6*a(n-17) +4*a(n-18) -4*a(n-20) -4*a(n-21) +4*a(n-23) -a(n-24) +a(n-26) +a(n-27) -a(n-29)
Empirical for n mod 6 = 0: a(n) = (32/243)*n^5 + (32/81)*n^4 + (4/9)*n^3 + (1/9)*n^2
Empirical for n mod 6 = 1: a(n) = (32/243)*n^5 + (128/243)*n^4 + (487/486)*n^3 + (853/972)*n^2 + (34/243)*n + (313/972)
Empirical for n mod 6 = 2: a(n) = (32/243)*n^5 + (112/243)*n^4 + (355/486)*n^3 + (145/486)*n^2 + (125/486)*n + (209/243)
Empirical for n mod 6 = 3: a(n) = (32/243)*n^5 + (16/27)*n^4 + (8/9)*n^3 + (8/9)*n^2 + (1/3)*n
Empirical for n mod 6 = 4: a(n) = (32/243)*n^5 + (80/243)*n^4 + (80/243)*n^3 + (40/243)*n^2 + (10/243)*n + (1/243)
Empirical for n mod 6 = 5: a(n) = (32/243)*n^5 + (160/243)*n^4 + (320/243)*n^3 + (320/243)*n^2 + (160/243)*n + (32/243)
Comments