This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A248464 #8 Jul 23 2025 11:44:07 %S A248464 16,72,460,1512,4272,9684,20236,37868,67140,111104,177024,269892, %T A248464 400040,574140,806808,1107132,1493952,1978984,2586340,3330852,4242444, %U A248464 5338788,6656352,8217136,10064140,12222784,14744480,17659176,21025952,24879392 %N A248464 Number of length 3+2 0..n arrays with no three consecutive terms having the sum of any two elements equal to twice the third. %C A248464 Row 3 of A248461 %H A248464 R. H. Hardin, <a href="/A248464/b248464.txt">Table of n, a(n) for n = 1..210</a> %F A248464 Empirical: a(n) = a(n-1) +a(n-3) -2*a(n-7) +a(n-8) -2*a(n-9) +a(n-10) -a(n-11) +2*a(n-12) +2*a(n-15) -a(n-16) +a(n-17) -2*a(n-18) +a(n-19) -2*a(n-20) +a(n-24) +a(n-26) -a(n-27) %F A248464 Also a polynomial of degree 5 plus a quadratic quasipolynomial with period 840, the first 12 being: %F A248464 Empirical for n mod 840 = 0: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n %F A248464 Empirical for n mod 840 = 1: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (66/5)*n - (33/10) %F A248464 Empirical for n mod 840 = 2: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (113/15)*n - (152/15) %F A248464 Empirical for n mod 840 = 3: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (26/5)*n - (1/2) %F A248464 Empirical for n mod 840 = 4: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - (12/5) %F A248464 Empirical for n mod 840 = 5: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (158/15)*n + (1/6) %F A248464 Empirical for n mod 840 = 6: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - (24/5) %F A248464 Empirical for n mod 840 = 7: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (26/5)*n + (67/10) %F A248464 Empirical for n mod 840 = 8: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (113/15)*n - (4/3) %F A248464 Empirical for n mod 840 = 9: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (66/5)*n - (9/10) %F A248464 Empirical for n mod 840 = 10: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (197/30)*n^2 + (51/5)*n - 8 %F A248464 Empirical for n mod 840 = 11: a(n) = n^5 + (1/2)*n^4 + (20/3)*n^3 - (31/15)*n^2 + (38/15)*n + (41/30) %e A248464 Some solutions for n=6 %e A248464 ..2....2....3....1....4....2....2....1....2....4....0....1....3....3....6....5 %e A248464 ..3....5....2....3....1....6....6....4....0....4....2....6....6....0....0....4 %e A248464 ..5....6....0....1....4....1....2....4....0....1....5....5....6....5....2....4 %e A248464 ..2....0....5....6....4....6....5....0....3....5....3....0....2....5....2....0 %e A248464 ..4....5....3....2....1....4....2....6....5....5....6....5....2....6....3....5 %K A248464 nonn %O A248464 1,1 %A A248464 _R. H. Hardin_, Oct 06 2014