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Monthly Problem 10791, second expression.
2

%I #42 Sep 08 2022 08:45:27

%S 1,1,2,9,61,554,6565,96677,1716730,36072181,881242577,24674241834,

%T 783024550969,27896201305769,1106485798248706,48517267642373105,

%U 2337333266369553253,123040664089658462650,7043260281573138384701,436533086101058798529933

%N Monthly Problem 10791, second expression.

%H G. C. Greubel, <a href="/A121870/b121870.txt">Table of n, a(n) for n = 0..325</a>

%H A. Fekete and G. Martin, <a href="http://www.jstor.org/stable/2695545">Problem 10791: Squared Series Yielding Integers</a>, Amer. Math. Monthly, 108 (No. 2, 2001), 177-178.

%F a(n) = A121867(n)^2 + A121868(n)^2.

%F From _Gary W. Adamson_, Jul 22 2011: (Start)

%F sqrt(a(n)) = upper left term in M^n as to the modulus of a polar term; M = an infinite square production matrix in which a column of (i, i, i, ...) is appended to the right of Pascal's triangle, as follows (with i = sqrt(-1)):

%F 1, i, 0, 0, 0, ...

%F 1, 1, i, 0, 0, ...

%F 1, 2, 1, i, 0, ...

%F 1, 3, 3, 1, i, ...

%F ... (End)

%F a(n) = |B_n(i)|^2, where B_n(x) is the n-th Bell polynomial, i = sqrt(-1) is the imaginary unit. - _Vladimir Reshetnikov_, Oct 15 2017

%F a(n) ~ (n*exp(-1 + Re(LambertW(i*n)) / Abs(LambertW(i*n))^2) / Abs(LambertW(i*n)))^(2*n) / Abs(1 + LambertW(i*n)), where i is the imaginary unit. - _Vaclav Kotesovec_, Jul 28 2021

%p A121870a:= proc(a) local i, t:

%p i:=1: t:=0: for i to 100 do t:=t+evalf((i^(a-1))*(I)^i/(i)!): od:

%p RETURN(round(abs(t^2))):

%p end: a:= A121870a(n);

%p # _Russell Walsmith_, Apr 18 2008

%p # Alternate:

%p seq(abs(BellB(n,I))^2, n=0..30); # _Robert Israel_, Oct 15 2017

%t Table[Abs[BellB[n, I]]^2, {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 15 2017 *)

%o (PARI) a(n) = abs( (sum(k=0,n, I^k*stirling(n,k,2)))^2 );

%o vector(25, n, a(n-1)) \\ _G. C. Greubel_, Oct 08 2019

%o (Magma) C<I>:= ComplexField(); a:= func< n | Round(Abs( (&+[I^k*StirlingSecond(n,k): k in [0..n]])^2 )) >;

%o [a(n): n in [0..25]]; // _G. C. Greubel_, Oct 08 2019

%o (Sage) [abs( sum(I^k*stirling_number2(n,k) for k in (0..n))^2 ) for n in (0..25)] # _G. C. Greubel_, Oct 08 2019

%o (GAP) List([0..25], n-> (Sum([0..Int(n/2)], k-> Stirling2(n,2*k)*(-1)^(k)) )^2 + (Sum([0..Int(n/2)], k-> (-1)^k*Stirling2(n,2*k+1)))^2 ); # _G. C. Greubel_, Oct 08 2019

%Y Cf. A024429, A024430, A121867, A121868, A121869.

%K nonn

%O 0,3

%A _N. J. A. Sloane_, Sep 05 2006