∴
© "Tous droits réservés" - 2019 par Cédric Christian Bernard Gagneux né le 19/07/64.
∴
IXXX) LES FONCTIONS TRIGONOMETRIQUES EQUIVALENTES A LA FONCTION MODULO.
∴
n mod(x)=x/2+x*arctan(tan(pi*((n+1)/x-1/2)))/pi+((n+1)/x+1/2-arctan(tan(pi*((n+1)/x-1/2)))/pi-((n)/x+1/2-arctan(tan(pi*((n)/x-1/2)))/pi))*(x/2-x*arctan(tan(pi*((n)/x-1/2)))/pi)*x-1
Pour illustrer les formules d'équivalences précédentes, 3 exemples de formules exactes et proposées pour publication sur le suite de "l'Encyclopédie des Séquences en Lignes", O.E.I.S., mais refusées à la publication:
Soit la séquence répertorié sur le site O.E.I.S correspondante au numéro A063524 et intitulée, "Characteristic function of 1": {0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}.
a(n)=(((2+n-(n/2+n*arctan(tan(Pi*(2/n-1/2)))/Pi+(1/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi-(1/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi))*(n/2-n*arctan(tan(Pi*(1/n-1/2)))/Pi)*n))/n-2)/2-(1+2*arctan(tan(Pi*(((2+n-(n/2+n*arctan(tan(Pi*(2/n-1/2)))/Pi+((2-1)/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi-(1/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi))*(n/2-n*arctan(tan(Pi*(1/n-1/2)))/Pi)*n))/n-2+1)/2-1/2)))/Pi+((((n+3-((n+1)/2+(n+1)*arctan(tan(Pi*(2/(n+1)-1/2)))/Pi+(1/(n+1)+1/2-arctan(tan(Pi*(1/(n+1)-1/2)))/Pi-(1/(n+1)+1/2-arctan(tan(Pi*(1/(n+1)-1/2)))/Pi))*((n+1)/2-(n+1)*arctan(tan(Pi*(1/(n+1)-1/2)))/Pi)*(n+1)))/(n+1)-2+1)-1)/2+1/2-arctan(tan(Pi*(((2+(n+1)-((n+1)/2+(n+1)*arctan(tan(Pi*(2/(n+1)-1/2)))/Pi+((2-1)/(n+1)+1/2-arctan(tan(Pi*(1/(n+1)-1/2)))/Pi-(1/(n+1)+1/2-arctan(tan(Pi*(1/(n+1)-1/2)))/Pi))*((n+1)/2-(n+1)*arctan(tan(Pi*(1/(n+1)-1/2)))/Pi)*(n+1)))/(n+1)-2)/2-1/2)))/Pi-(((2+n-(n/2+n*arctan(tan(Pi*(2/n-1/2)))/Pi+(1/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi-(1/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi))*(n/2-n*arctan(tan(Pi*(1/n-1/2)))/Pi)*n))/n-2)/2+1/2-arctan(tan(Pi*(((2+n-(n/2+n*arctan(Tan(Pi*(2/n-1/2)))/Pi+(1/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi-(1/n+1/2-arctan(Tan(Pi*(1/n-1/2)))/Pi))*(n/2-n*arctan(tan(Pi*(1/n-1/2)))/Pi)*n))/n-2)/2-1/2)))/Pi))*(1-2*arctan(tan(Pi*(((2+n-(n/2+n*arctan(tan(Pi*(2/n-1/2)))/Pi+(1/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi-(1/n+1/2-arctan(tan(Pi*(1/n-1/2)))/Pi))*(n/2-n*arctan(tan(Pi*(1/n-1/2)))/Pi)*n))/n-2)/2-1/2)))/Pi)*2-1)/2)/2.. - Cédric Christian Bernard Gagneux_, May 28 2019.
Soit la séquence répertorié sur le site O.E.I.S correspondante au numéro A079998, intitulée, "The characteristic function of the multiples of five": {1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1}.
a(n)=((n+1+5-(5/2+5*arctan(tan(Pi*((n+1+1)/5-1/2)))/Pi+(((n+2+1)-1)/5+1/2-arctan(tan(Pi*((n+2)/5-1/2)))/Pi-((n+1)/5+1/2-arctan(tan(Pi*((n+1)/5-1/2)))/Pi))*(5/2-5*arctan(tan(Pi*((n+1)/5-1/2)))/Pi)*5-1))/5)-(5+n+1-(5/2+5*arctan(tan(Pi*((n+1)/5-1/2)))/Pi+((n+2-1)/5+1/2-arctan(tan(Pi*((n+2-1)/5-1/2)))/Pi-((n+1-1)/5+1/2-arctan(tan(Pi*((n+1-1)/5-1/2)))/Pi))*(5/2-5*arctan(tan(Pi*((n+1-1)/5-1/2)))/Pi)*5))/5 –n-1+(n+1)*((n+1+1-((n+1)/2+(n+1)*arctan(tan(Pi*((n+1+2)/(n+1)-1/2)))/Pi+(((n+2+1)-1)/(n+1)+1/2-arctan(tan(Pi*((n+2)/(n+1)-1/2)))/Pi-((n+1+1)/(n+1)+1/2-arctan(tan(Pi*((n+1+1)/(n+1)-1/2)))/Pi))*((n+1)/2-(n+1)*arctan(tan(Pi*((n+1+1)/(n+1)-1/2)))/Pi)*(n+1)-1))/(n+1)). - Cédric Christian Bernard Gagneux_, May 28 2019.
Soit la séquence répertorié sur le site O.E.I.S correspondante au numéro A027868, intitulée, "Number of trailing zeros in n!; highest power of 5 dividing n!": {0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 19}.
a(n)=(6+n-(5/2+5*arctan(tan(Pi*((n+1)/5-1/2)))/Pi+((n+1)/5+1/2-arctan(tan(Pi*((n+1)/5-1/2)))/Pi-(n/5+1/2-arctan(tan(Pi*(n/5-1/2)))/Pi))*(5/2-5*arctan(tan(Pi*(n/5-1/2)))/Pi)*5))/5-1. - Cédric Christian Bernard Gagneux_, May 28 2019.
∴
∀∈⌊⌋⌈⌉ ₀₁₂₃₄₅₆₇₈₉ ₊ ₋ ₌₍₎ ₐ ₑ ₒ ₓ ₔ ₕ ₖ ₗ ₘ ₙ ₚ ₛ ₜ ⱼ