A126155 -id:A126155 - cp's OEIS frontend

A126156 Expansion of e.g.f. sqrt(sec(sqrt(2)*x)), showing coefficients of only the even powers of x.

1, 1, 7, 139, 5473, 357721, 34988647, 4784061619, 871335013633, 203906055033841, 59618325600871687, 21297483077038703899, 9127322584507530151393, 4621897483978366951337161, 2730069675607609356178641127, 1860452328661957054823447670979, 1448802510679254790311316267306753

Offset: 0

Paul D. Hanna, Dec 20 2006

nonn

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

			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! + ...

H. S. Wall, Analytic Theory of Continued Fractions, Chelsea 1973, p. 366.

G. C. Greubel, Table of n, a(n) for n = 0..200
Peter Bala, A triangle for calculating A126156
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.

Diagonals: A126157, A126158.

Cf. A126155, A365672.

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

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 *)

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 */

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 */

/* 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),", "))

{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),", "))

{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 */

def A126156(n): return A126155(n, 0)
print([A126156(n) for n in range(17)])  # Peter Luschny, Dec 14 2023

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

New name based on a comment of David Callan, Peter Luschny, May 07 2014

A126150 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 4 times first term.

Original entry on oeis.org

1, 1, 4, 1, 6, 24, 36, 24, 6, 96, 384, 636, 744, 636, 384, 96, 2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, 2976, 151416, 605664, 1042056, 1407024, 1650456, 1736064, 1650456, 1407024, 1042056, 605664, 151416, 11449296, 45797184

Offset: 0

Paul D. Hanna, Dec 19 2006

			Triangle begins:
  1;
  1, 4, 1;
  6, 24, 36, 24, 6;
  96, 384, 636, 744, 636, 384, 96;
  2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, 2976;
  151416, 605664, 1042056, 1407024, 1650456, 1736064, 1650456, 1407024, 1042056, 605664, 151416; ...
If we write the triangle like this:
  .......................... ....1;
  ................... ....1, ....4, ....1;
  ............ ....6, ...24, ...36, ...24, ....6;
  ..... ...96, ..384, ..636, ..744, ..636, ..384, ...96;
  .2976, 11904, 20256, 26304, 28536, 26304, 20256, 11904, .2976;
then the first term in each row is the sum of the previous row:
  2976 = 96 + 384 + 636 + 744 + 636 + 384 + 96
the next term is 4 times the first:
  11904 = 4*2976,
and the remaining terms in each row are obtained by the rule illustrated by:
  20256 = 2*11904 - 2976 - 6*96;
  26304 = 2*20256 - 11904 - 6*384;
  28536 = 2*26304 - 20256 - 6*636;
  26304 = 2*28536 - 26304 - 6*744;
  20256 = 2*26304 - 28536 - 6*636;
  11904 = 2*20256 - 26304 - 6*384;
  2976 = 2*11904 - 20256 - 6*96.
An alternate recurrence is illustrated by:
  11904 = 2976 + 3*(96 + 384 + 636 + 744 + 636 + 384 + 96);
  20256 = 11904 + 3*(384 + 636 + 744 + 636 + 384);
  26304 = 20256 + 3*(636 + 744 + 636);
  28536 = 26304 + 3*(744);
and then for k>n, T(n,k) = T(n,2n-k).

Cf. A126151 (column 0); diagonals: A126152, A126153; A126154; variants: A008301, A125053, A126155.

PARI
```
T(n,k)=local(p=3);if(2*n
				
```

/* Alternate Recurrence: */ T(n,k)=local(p=3);if(2*n

Sum_{k=0..2n} (-1)^k*C(2n,k)*T(n,k) = (-6)^n.

cp's OEIS Frontend

A126157 Main diagonal and central terms of symmetric triangle A126155.

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A126158 Secondary diagonal of symmetric triangle A126155: a(n) = A126155(n+1,n).

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A126159 a(n) = Sum_{k=0..n} C(2n,k)*A126155(n,k).

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A126156 Expansion of e.g.f. sqrt(sec(sqrt(2)*x)), showing coefficients of only the even powers of x.

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A126150 Symmetric triangle, read by rows of 2*n+1 terms, similar to triangle A008301. Second term 4 times first term.

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