cp's OEIS Frontend

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.

A248436 Number of length 3+2 0..n arrays with every three consecutive terms having the sum of some two elements equal to twice the third.

This page as a plain text file.
%I A248436 #10 Nov 08 2018 10:46:29
%S A248436 2,9,24,69,118,185,252,357,470,593,716,881,1046,1217,1400,1621,1842,
%T A248436 2081,2320,2593,2874,3161,3448,3785,4126,4473,4828,5213,5598,6005,
%U A248436 6412,6857,7310,7769,8232,8737,9242,9753,10272,10833,11394,11973,12552,13161,13782
%N A248436 Number of length 3+2 0..n arrays with every three consecutive terms having the sum of some two elements equal to twice the third.
%H A248436 R. H. Hardin, <a href="/A248436/b248436.txt">Table of n, a(n) for n = 1..210</a>
%F A248436 Empirical: a(n) = a(n-1) - a(n-2) + a(n-3) - a(n-4) + 2*a(n-5) - a(n-6) + a(n-7) - a(n-11) + a(n-12) - 2*a(n-13) + a(n-14) - a(n-15) + a(n-16) - a(n-17) + a(n-18).
%F A248436 Also a quadratic polynomial plus a constant quasipolynomial with period 840, the first 12 being:
%F A248436 Empirical for n mod 840 = 0: a(n) = (106/15)*n^2 - (178/15)*n + 1
%F A248436 Empirical for n mod 840 = 1: a(n) = (106/15)*n^2 - (178/15)*n + (34/5)
%F A248436 Empirical for n mod 840 = 2: a(n) = (106/15)*n^2 - (178/15)*n + (67/15)
%F A248436 Empirical for n mod 840 = 3: a(n) = (106/15)*n^2 - (178/15)*n - 4
%F A248436 Empirical for n mod 840 = 4: a(n) = (106/15)*n^2 - (178/15)*n + (17/5)
%F A248436 Empirical for n mod 840 = 5: a(n) = (106/15)*n^2 - (178/15)*n + (2/3)
%F A248436 Empirical for n mod 840 = 6: a(n) = (106/15)*n^2 - (178/15)*n + (9/5)
%F A248436 Empirical for n mod 840 = 7: a(n) = (106/15)*n^2 - (178/15)*n - (56/5)
%F A248436 Empirical for n mod 840 = 8: a(n) = (106/15)*n^2 - (178/15)*n - (1/3)
%F A248436 Empirical for n mod 840 = 9: a(n) = (106/15)*n^2 - (178/15)*n + (22/5)
%F A248436 Empirical for n mod 840 = 10: a(n) = (106/15)*n^2 - (178/15)*n + 5
%F A248436 Empirical for n mod 840 = 11: a(n) = (106/15)*n^2 - (178/15)*n - (128/15).
%F A248436 Empirical g.f.: x*(2 + 7*x + 17*x^2 + 52*x^3 + 66*x^4 + 117*x^5 + 124*x^6 + 200*x^7 + 175*x^8 + 222*x^9 + 167*x^10 + 210*x^11 + 118*x^12 + 113*x^13 + 52*x^14 + 54*x^15 - x^16 + x^17) / ((1 - x)^3*(1 + x)*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)*(1 + x^4)*(1 + x + x^2 + x^3 + x^4)). - _Colin Barker_, Nov 08 2018
%e A248436 Some solutions for n=6:
%e A248436 ..0....1....1....3....4....3....3....1....4....3....2....4....4....4....6....3
%e A248436 ..0....3....5....5....3....1....4....3....3....4....6....3....5....3....2....1
%e A248436 ..0....5....3....4....5....2....2....5....2....2....4....2....3....2....4....2
%e A248436 ..0....4....1....6....4....3....0....1....1....3....5....1....1....4....0....3
%e A248436 ..0....6....2....2....3....4....1....3....3....1....6....0....5....3....2....1
%Y A248436 Row 3 of A248433.
%K A248436 nonn
%O A248436 1,1
%A A248436 _R. H. Hardin_, Oct 06 2014