login
Search: a365672 -id:a365672
     Sort: relevance | references | number | modified | created      Format: long | short | data
Expansion of e.g.f. sqrt(sec(sqrt(2)*x)), showing coefficients of only the even powers of x.
+10
13
1, 1, 7, 139, 5473, 357721, 34988647, 4784061619, 871335013633, 203906055033841, 59618325600871687, 21297483077038703899, 9127322584507530151393, 4621897483978366951337161, 2730069675607609356178641127, 1860452328661957054823447670979, 1448802510679254790311316267306753
OFFSET
0,3
COMMENTS
Previous name was: Column 0 and row sums of symmetric triangle A126155.
This is the square root of the Euler numbers (A122045) with respect to the Cauchy type product as described by J. Singh (see link and the second Maple program) normalized by 2^n. A241885 shows the corresponding sqrt of the Bernoulli numbers. - Peter Luschny, May 07 2014
REFERENCES
H. S. Wall, Analytic Theory of Continued Fractions, Chelsea 1973, p. 366.
LINKS
Alain Connes, Caterina Consani and Henri Moscovici, Zeta zeros and prolate wave operators, arXiv:2310.18423 [math.NT], Oct 2023, p.31.
Denis S. Grebenkov, Vittoria Sposini, Ralf Metzler, Gleb Oshanin, and Flavio Seno, Exact distributions of the maximum and range of random diffusivity processes, New J. Phys. (2021) Vol. 23, 023014.
Jitender Singh, On an arithmetic convolution, arXiv:1402.0065 [math.NT], 2014.
FORMULA
a(n) = Sum_{k=0..n} A087736(n,k)*3^(n-k). - Philippe Deléham, Jul 17 2007
E.g.f.: Sum_{n>=0} a(n)*x^(2*n)/(2*n)! = sqrt(sec(sqrt(2)*x)). - David Callan, Jan 03 2011
E.g.f. satisfies: A(x) = exp( Integral Integral A(x)^4 dx dx ), where A(x) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)! and the constant of integration is zero. - Paul D. Hanna, May 30 2015
E.g.f. satisfies: A(x) = exp( Integral A(x)^2 * Integral 1/A(x)^2 dx dx ), where A(x) = Sum_{n>=0} a(n)*x^(2*n)/(2*n)! and the constant of integration is zero. - Paul D. Hanna, Jun 02 2015
G.f.: 1/(1-x/(1-6*x/(1-15*x/(1-28*x/(1-45*x/(1-66*x/(1-91*x/(1-... or 1/U(0) where U(k) = 1-x*(k+1)*(2*k+1)/U(k+1); (continued fraction). [See Wall.] - Sergei N. Gladkovskii, Oct 31 2011
G.f.: 1/U(0) where U(k) = 1 - (4*k+1)*(4*k+2)*x/(2 - (4*k+3)*(4*k+4)*x/ U(k+1)); (continued fraction, 2-step). - Sergei N. Gladkovskii, Oct 24 2012
G.f.: 1/G(0) where G(k) = 1 -x*(k+1)*(2*k+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013
G.f.: Q(0), where Q(k) = 1 - x*(2*k+1)*(k+1)/( x*(2*k+1)*(k+1) - 1/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, Oct 09 2013
a(n) ~ 2^(5*n+2) * n^(2*n) / (exp(2*n) * Pi^(2*n+1/2)). - Vaclav Kotesovec, Jul 13 2014
a(n) = (1/(4*n))*Sum_{k=1..n} binomial(2*n,2*k)*((2^(2*k)-1)*2^(3*k)*(-1)^((k-1))*Bernoulli(2*k)*a(n-k)), a(0)=1. - Vladimir Kruchinin, Feb 25 2015
a(n) = Sum_{k=1..n} a(n-k)*binomial(2*n,2*k)*(k/(2*n)-1)*(-2)^k, a(0)=1. - Tani Akinari, Sep 11 2023
For n > 0, a(n) = -Sum_{j=0..n} Sum_{k=0..floor(j/2)} (2*n+1)!*(2*k-j)^(2*n)/(n!*(2*j+1)*(n-j)!*k!*(j-k)!*(-2)^(n+j-1)). - Tani Akinari, Sep 28 2023
EXAMPLE
E.g.f.: A(x) = 1 + x^2/2! + 7*x^4/4! + 139*x^6/6! + 5473*x^8/8! + 357721*x^10/10! + ...
where the logarithm begins:
log(A(x)) = x^2/2! + 4*x^4/4! + 64*x^6/6! + 2176*x^8/8! + 126976*x^10/10! + 11321344*x^12/12! + ...
compare the logarithm to
A(x)^4 = 1 + 4*x^2/2! + 64*x^4/4! + 2176*x^6/6! + 126976*x^8/8! + 11321344*x^10/10! + ...
MAPLE
A126156 := proc(n)
sqrt(sec(sqrt(2)*z)) ;
coeftayl(%, z=0, 2*n) ;
%*(2*n)! ;
end;
seq(A126156(n), n=0..10) ; # Sergei N. Gladkovskii, Oct 31 2011
g := proc(f, n) option remember; local g0, m; g0 := sqrt(f(0));
if n=0 then g0 else if n=1 then 0 else add(binomial(n, m)*g(f, m)* g(f, n-m), m=1..n-1) fi; (f(n)-%)/(2*g0) fi end:
a := n -> (-2)^n*g(euler, 2*n);
seq(a(n), n=0..14); # Peter Luschny, May 07 2014
# Alternative: an algorithm as described by Peter Bala, see also A365672:
T := proc(n, k) option remember; if k = 0 then 1 else if k = n then
T(n, k-1) else (n - k + 1) * (2 * (n - k) + 1) * T(n, k - 1) + T(n - 1, k)
fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..14); # Peter Luschny, Sep 29 2023
MATHEMATICA
a[n_] := SeriesCoefficient[ Sqrt[ Sec[ Sqrt[2]*x]], {x, 0, 2 n}]*(2*n)!; Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Nov 29 2013, after Sergei N. Gladkovskii *)
PROG
(Maxima)
a(n):=if n=0 then 1 else 1/(4*n)*sum(binomial(2*n, 2*k)*((2^(2*k)-1)*2^(3*k)*(-1)^((k-1))*bern(2*k)*a(n-k)), k, 1, n); /* Vladimir Kruchinin, Feb 25 2015 */
(Maxima)
a[n]:=if n=0 then 1 else sum(a[n-k]*binomial(2*n, 2*k)*(k/(2*n)-1)*(-2)^k, k, 1, n);
makelist(a[n], n, 0, 30); /* Tani Akinari, Sep 11 2023 */
(PARI) /* E.g.f. A(x) = exp( Integral^2 A(x)^4 dx^2 ): */
{a(n)=local(A=1+x*O(x)); for(i=1, n, A=exp(intformal(intformal(A^4 + x*O(x^(2*n))))) ); (2*n)!*polcoeff(A, 2*n, x)}
for(n=0, 20, print1(a(n), ", "))
(PARI) {a(n) = local(A=1+x); for(i=1, n, A = exp( intformal( A^2 * intformal( 1/A^2 + x*O(x^n)) ) ) ); n!*polcoeff(A, n)}
for(n=0, 20, print1(a(2*n), ", "))
(PARI) {a(n)=-(n<1)-sum(j=0, n, sum(k=0, j/2, (2*n+1)!*(2*k-j)^(2*n)/(n!*(2*j+1)*(n-j)!*k!*(j-k)!*(-2)^(n+j-1))))}; /* Tani Akinari, Sep 28 2023 */
(SageMath)
def A126156(n): return A126155(n, 0)
print([A126156(n) for n in range(17)]) # Peter Luschny, Dec 14 2023
CROSSREFS
Diagonals: A126157, A126158.
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Dec 20 2006
EXTENSIONS
New name based on a comment of David Callan, Peter Luschny, May 07 2014
STATUS
approved
Array A(n, k) read by ascending antidiagonals. Polygonal number weighted generalized Catalan sequences.
+10
6
1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 4, 1, 1, 1, 4, 15, 8, 1, 1, 1, 5, 34, 105, 16, 1, 1, 1, 6, 61, 496, 945, 32, 1, 1, 1, 7, 96, 1385, 11056, 10395, 64, 1, 1, 1, 8, 139, 2976, 50521, 349504, 135135, 128, 1, 1, 1, 9, 190, 5473, 151416, 2702765, 14873104, 2027025, 256, 1
OFFSET
0,9
COMMENTS
Using polygonal numbers as weights, a recursion for triangles is defined, whose main diagonals represents a family of sequences, which include, among others, the powers of 2, the double factorial of odd numbers, the reduced tangent numbers, and the Euler numbers.
Apart from the edge cases k = 0 and k = n the recursion is T(n, k) = w(n, k) * T(n, k - 1) + T(n - 1, k). T(n, 0) = 1 and T(n, n) = T(n, n-1) if n > 0.
The weights w(n, k) identical to 1 yield the recursion of the Catalan triangle A009766 (with main diagonal the Catalan numbers). Here the polygonal numbers are used as weights in the form w(n, k) = p(s, n - k + 1), where the parameter s is the number of sides of the polygon and p(s, n) = ((s-2) * n^2 - (s-4) * n) / 2, see A317302.
EXAMPLE
Array A(n, k) starts: (polygon|diagonal|triangle)
[0] 1, 1, 1, 1, 1, 1, 1, ... A258837 A000012
[1] 1, 1, 2, 4, 8, 16, 32, ... A080956 A011782
[2] 1, 1, 3, 15, 105, 945, 10395, ... A001477 A001147 A001498
[3] 1, 1, 4, 34, 496, 11056, 349504, ... A000217 A002105 A365674
[4] 1, 1, 5, 61, 1385, 50521, 2702765, ... A000290 A000364 A060058
[5] 1, 1, 6, 96, 2976, 151416, 11449296, ... A000326 A126151 A366138
[6] 1, 1, 7, 139, 5473, 357721, 34988647, ... A000384 A126156 A365672
[7] 1, 1, 8, 190, 9080, 725320, 87067520, ... A000566 A366150 A366149
[8] 1, 1, 9, 249, 14001, 1322001, 188106489, ... A000567
MAPLE
poly := (s, n) -> ((s - 2) * n^2 - (s - 4) * n) / 2:
T := proc(s, n, k) option remember; if k = 0 then 1 else if k = n then T(s, n, k-1) else poly(s, n - k + 1) * T(s, n, k - 1) + T(s, n - 1, k) fi fi end:
for n from 0 to 8 do A := (n, k) -> T(n, k, k): seq(A(n, k), k = 0..9) od;
# Alternative, using continued fractions:
A := proc(p, L) local CF, poly, k, m, P, ser;
poly := (s, n) -> ((s - 2)*n^2 - (s - 4)*n)/2;
CF := 1 + x;
for k from 1 to L do
m := L - k + 1;
P := poly(p, m);
CF := 1/(1 - P*x*CF)
od;
ser := series(CF, x, L);
seq(coeff(ser, x, m), m = 0..L-1)
end:
for p from 0 to 8 do lprint(A(p, 8)) od;
MATHEMATICA
poly[s_, n_] := ((s - 2) * n^2 - (s - 4) * n) / 2;
T[s_, n_, k_] := T[s, n, k] = If[k == 0, 1, If[k == n, T[s, n, k - 1], poly[s, n - k + 1] * T[s, n, k - 1] + T[s, n - 1, k]]];
A[n_, k_] := T[n, k, k];
Table[A[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 27 2023, from first Maple program *)
PROG
(Python)
from functools import cache
@cache
def T(s, n, k):
if k == 0: return 1
if k == n: return T(s, n, k - 1)
p = (n - k + 1) * ((s - 2) * (n - k + 1) - (s - 4)) // 2
return p * T(s, n, k - 1) + T(s, n - 1, k)
def A(n, k): return T(n, k, k)
for n in range(9): print([A(n, k) for k in range(9)])
(PARI)
A(p, n) = {
my(CF = 1 + x,
poly(s, n) = ((s - 2)*n^2 - (s - 4)*n)/2,
m, P
);
for(k = 1, n,
m = n - k + 1;
P = poly(p, m);
CF = 1/(1 - P*x*CF)
);
Vec(CF + O(x^(n)))
}
for(p = 0, 8, print(A(p, 8)))
\\ Michel Marcus and Peter Luschny, Oct 02 2023
CROSSREFS
Cf. A009766, A366137 (central diagonal), A317302 (table of polygonal numbers).
KEYWORD
nonn,tabl
AUTHOR
Peter Luschny, Sep 30 2023
STATUS
approved

Search completed in 0.005 seconds