A292632
a(n) = n! * [x^n] exp((n+2)*x)*(BesselI(0,2*x) - BesselI(1,2*x)).
Original entry on oeis.org
1, 2, 10, 77, 798, 10392, 162996, 2991340, 62893270, 1490758022, 39334017996, 1143492521437, 36318168041260, 1251270023475864, 46481870133666792, 1852054390616046345, 78792796381529620710, 3564894013016856836190, 170921756533520140861020, 8657018996674423681277455, 461881087606113071895396420
Offset: 0
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Table[n!*SeriesCoefficient[E^((n+2)*x)*(BesselI[0,2*x] - BesselI[1,2*x]),{x,0,n}], {n,0,20}] (* Vaclav Kotesovec, Sep 20 2017 *)
Join[{1}, Table[Sum[Binomial[n, j] * CatalanNumber[j] * n^(n-j), {j, 0, n}], {n, 1, 20}]] (* Vaclav Kotesovec, Nov 23 2021 *)
A349640
a(n) = Sum_{k=0..n} binomial(n,k) * A000108(k) * k!.
Original entry on oeis.org
1, 2, 7, 46, 485, 7066, 130987, 2946182, 77923561, 2369742130, 81467904431, 3124302688222, 132237820201357, 6123150708289226, 307903794151741075, 16709463201832993846, 973385368533058021457, 60583668821975488285282, 4012342371757905842648791, 281735471040327667890013070
Offset: 0
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gf := exp(x)*(1 - sqrt(1 - 4*x))/(2*x): ser := series(gf, x, 24):
seq(n!*coeff(ser, x, n), n = 0..19);
# Alternative:
a := n -> `if`(n < 4, [1, 2, 7, 46][n + 1], ((4*n^2 - 12*n + 8)*a(n - 3) - (8*n^2 - 13*n + 5)*a(n - 2) + 4*n^2*a(n - 1))/(n + 1)):
seq(a(n), n = 0..19); # Peter Luschny, Nov 23 2021
# Alternative
seq(simplify(hypergeom([-n, 1/2, 1], [2], -4)), n = 0..19); # Peter Bala, Mar 13 2025
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Table[Sum[Binomial[n, j]*CatalanNumber[j]*j!, {j, 0, n}], {n, 0, 20}]
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a(n) = sum(k=0, n, binomial(n,k) * (binomial(2*k,k)/(k+1)) * k!); \\ Michel Marcus, Nov 23 2021
A349639
a(n) = Sum_{k=0..n} binomial(n,k) * A000108(k) * k^k.
Original entry on oeis.org
1, 2, 11, 163, 4177, 150606, 7002679, 399296682, 26997867705, 2112814307980, 187919721166951, 18727570061711897, 2067435790679136937, 250474099952311886236, 33043529154916822685459, 4715582224589290429430011, 723854564711343436767660481, 118933484485939500023357177356
Offset: 0
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Table[1+Sum[Binomial[n, j]*CatalanNumber[j]*j^j, {j, 1, n}], {n, 0, 20}]
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a(n) = sum(k=0, n, binomial(n,k) * (binomial(2*k,k)/(k+1)) * k^k); \\ Michel Marcus, Nov 23 2021
Showing 1-3 of 3 results.
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