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A086880
a(n) = floor(Sum_{k=0..infinity} k^n/(k!)^2); related to generalized Bell numbers.
6
2, 1, 2, 3, 7, 17, 45, 128, 391, 1287, 4524, 16889, 66657, 276982, 1207598, 5507362, 26203307, 129757596, 667358910, 3558097578, 19632277761, 111930731957, 658482495614, 3992062349412, 24911272290567, 159833355923362
OFFSET
0,1
COMMENTS
Define B(n) = Sum_{k=0..infinity} k^n/(k!)^2, then there exists a complex linear relation: B(3) = B(2) + B(1); B(4) = 2*B(3); B(5) = 2*B(4) + B(2); B(6) = 5*B(4) + 3*B(2); B(7) = 7*B(5) + B(3); B(12) = B(11) + 11*B(10); ...
LINKS
Toshiki Matsusaka, Taichi Miyazaki, and Shunta Yara, On finite analogues of DobiƄski's formula and of Euler's constant via Gregory polynomials, arXiv:2604.01578 [math.NT], 2026. See p. 3.
FORMULA
Sum_{k>=0} k^n/(k!)^2 = A000994(n)*BesselI(0, 2) + A000995(n)*BesselI(1, 2), using Bessel function values BesselI(0, 2)=2.2795853023..., BesselI(1, 2) = 1.5906368546... (A096789) and where A000994 and A000995 shift 2 places left under binomial transform: A000994={1, 0, 1, 1, 2, 5, 13, 36, 109, 359, 1266, 4731, ...} A000995={0, 1, 0, 1, 2, 4, 10, 29, 90, 295, 1030, 3838, ...}.
EXAMPLE
a(5) = floor(1^5/(1!)^2 + 2^5/(2!)^2 + 3^5/(3!)^2 + 4^5/(4!)^2 +...)
MATHEMATICA
Table[Floor[Sum[k^n/(k!)^2, {k, 0, Infinity}]], {n, 0, 20}] (* Vaclav Kotesovec, Jul 31 2014 *)
Flatten[{2, 1, Table[Floor[HypergeometricPFQ[ConstantArray[2, n-2], ConstantArray[1, n-1], 1]], {n, 2, 20}]}] (* Vaclav Kotesovec, May 23 2015 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 16 2003
STATUS
approved