A152800 Irregular triangle read by rows: the q-analog of the Euler numbers; expansion of the arithmetic inverse of the q-cosine of x.
1, 1, 0, 1, 2, 1, 1, 0, 0, 1, 3, 5, 8, 10, 10, 9, 7, 5, 2, 1, 0, 0, 0, 1, 4, 10, 21, 36, 55, 78, 101, 122, 138, 145, 143, 134, 117, 95, 72, 50, 32, 18, 9, 3, 1, 0, 0, 0, 0, 1, 5, 16, 41, 87, 164, 283, 452, 679, 967, 1311, 1700, 2118, 2540, 2937, 3282, 3546, 3706, 3751, 3676, 3487
Offset: 0
Examples
Nonzero coefficients in row n range from x^(n-1) to x^(2n(n-1)) for n>0. Triangle begins: 1; 1; 0,1,2,1,1; 0,0,1,3,5,8,10,10,9,7,5,2,1; 0,0,0,1,4,10,21,36,55,78,101,122,138,145,143,134,117,95,72,50,32,18,9,3,1; 0,0,0,0,1,5,16,41,87,164,283,452,679,967,1311,1700,2118,2540,2937,3282,3546,3706,3751,3676,3487,3202,2842,2436,2014,1602,1223,894,622,409,253,145,76,35,14,4,1; ... Explicit expansion of g.f.: 1/cos_q(x,q) = 1 + x^2/faq(2,q) + x^4*(q + 2*q^2 + q^3 + q^4)/faq(4,q) + x^6*(q^2 + 3*q^3 + 5*q^4 + 8*q^5 + 10*q^6 + 10*q^7 + 9*q^8 + 7*q^9 + 5*q^10 + 2*q^11 + q^12)/faq(6,q) + x^8*(q^3 + 4*q^4 + 10*q^5 + 21*q^6 + 36*q^7 + 55*q^8 + 78*q^9 + 101*q^10 + 122*q^11 + 138*q^12 + 145*q^13 + 143*q^14 + 134*q^15 + 117*q^16 + 95*q^17 + 72*q^18 + 50*q^19 + 32*q^20 + 18*q^21 + 9*q^22 + 3*q^23 + q^24)/faq(8,q) +...
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..2255, as a flattened triangle of rows 0..15
- M. M. Graev, Einstein equations for invariant metrics on flag spaces and their Newton polytopes, Transactions of the Moscow Mathematical Society, 2014, pp. 13-68. Original publication: Trudy Moskovskogo Matematicheskogo Obshchestva, tom 75 (2014), vypusk 1.
- Eric Weisstein, q-Cosine Function from MathWorld.
- Eric Weisstein, q-Factorial from MathWorld.
Crossrefs
Programs
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PARI
{T(n,k)=polcoeff(polcoeff(1/sum(m=0,n,(-1)^m*x^(2*m)/prod(j=1,2*m,(q^j-1)/(q-1))+x*O(x^(2*n+1))),2*n,x)*prod(j=1,2*n,(q^j-1)/(q-1)),k,q)} for(n=0,8,for(k=0,2*n*(n-1),print1(T(n,k),", "));print(""))
Formula
G.f.: 1/cos_q(x,q) = Sum_{n>=0} Sum_{k=0..2n(n-1)} T(n,k)*q^k*x^(2n)/faq(2n,q).
G.f.: 1/cos(x) = Sum_{n>=1} Sum_{k=0..2n(n-1)} T(n,k)*x^(2n)/(2n)!.
Sum_{k=0..2n(n-1)} T(n,k) = A000364(n).
Sum_{k=0..2n(n-1)} T(n,k)*(-1)^k = 1 for n>=0.
Sum_{k=0..2n(n-1)} T(n,k)*I^k = (-1)^[n/2] for n>=0 where I^2=-1.
Sum_{k=0..2n(n-1)} T(n,k)*exp(2*Pi*I*k/n) = 1 for n>0.
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