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 A181496 Number of degree n polynomials p(x,y) with all coefficients 0 or 1 such that x+y=1 implies p(x,y)=1. 1
 1, 1, 2, 5, 14, 43, 140, 474, 1648, 5839, 20980, 76202, 279168, 1029962, 3822336, 14256228, 53400912 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS The polynomials are of the form P = (x+y-1) * Q + 1. LINKS Table of n, a(n) for n=0..16. Murray S. Klamkin, Problem #3, Missouri State Math Dept. Challenge Archive, 2005/06 discusses the generalization to m variables. EXAMPLE For degree 0, the only solution is 1, hence a(0) = 1. For degree 1, the only solution is x + y, hence a(1) = 1. For degree 2 there are x^2 + x*y + y and x + x*y + y^2 (and no more), hence a(2) = 2. For n = 3 the a(3) = 5 polynomials are x^3 + x^2 y + x y + y, x^2 y + x y^2 + x^2 + y, x^2 y + x y^2 + x^2 + x y + y^2, x y^2 + y^3 + x y + x, x^2 y + x y^2 +y^2 + x. PROG (PARI) padbin(n, len) = {b = binary(n); while(length(b) < len, b = concat(0, b); ); b; } a(n) = {if (n == 0, return (1)); kill(x); kill(y); poln = vector(n+1, i, x^(i-1)*y^(n-i+1)); polm = [1]; for (i = 1, n-1, polm = concat(polm, vector(i+1, j, x^(j-1)*y^(i-j+1))); ); nbpol = 0; nbn = #poln; nbm = #polm; for (i = 1, 2^nbn, bi = padbin(i, nbn); poli = sum(ki = 1, nbn, bi[ki]*poln[ki]); for (j = 0, 2^nbm, bj = padbin(j, nbm); pol = poli + sum(kj = 1, nbm, bj[kj]*polm[kj]); if (subst(pol, y, 1-x) == 1, nbpol++; ); ); ); nbpol; } \\ Michel Marcus, Aug 25 2013 CROSSREFS Cf. A122743. Sequence in context: A366042 A149880 A066351 * A276989 A272461 A213264 Adjacent sequences: A181493 A181494 A181495 * A181497 A181498 A181499 KEYWORD nonn,more AUTHOR Hagen von Eitzen, Oct 24 2010 EXTENSIONS a(13) from Hagen von Eitzen, Oct 25 2010 a(14)-a(16) from Artem Romanov, Jul 25 2019 STATUS approved

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Last modified August 13 07:12 EDT 2024. Contains 375113 sequences. (Running on oeis4.)