A142987 - OEIS

A142987

a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1) + (n + 1)*(n + 2)*a(n).

1, 10, 106, 1180, 13920, 174600, 2330640, 33084000, 498646080, 7964020800, 134491276800, 2396163513600, 44942274316800, 885524502643200, 18293122632960000, 395457106963968000, 8930300425804800000

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OFFSET

1,2

COMMENTS

This is the case m = 5 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n), which arises when accelerating the convergence of a certain series for the constant log(2). See A142983 for remarks on the general case.

REFERENCES

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

FORMULA

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = (2*n^5 + 10*n^3 + 3*n)/15 = A069038(n).

Recurrence: a(1) = 1, a(2) = 10, a(n+2) = 10*a(n+1) + (n + 1)*(n + 2)*a(n).

The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 10 and b(2) = 102.

Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(10 + 1*2/(10 + 2*3/(10 + 3*4/(10 + .. +n*(n - 1)/(10))))), for n >= 2.

The behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p(k)*p(k+1)) = 1/(10 + 1*2/(10 + 2*3/(10 + 3*4/(10 + ... + n*(n - 1)/(10 + ...))))) = 10*log(2) - 41/6, where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).

MAPLE

p := n -> (2*n^5+10*n^3+3*n)/15: a := n -> n!*p(n+1)*sum ((-1)^(k+1)/(p(k)*p(k+1)), k = 1..n): seq(a(n), n = 1..20);

MATHEMATICA

RecurrenceTable[{a[1]==1, a[2]==10, a[n+2]==10a[n+1]+(n+1)(n+2)a[n]}, a, {n, 20}] (* Harvey P. Dale, Mar 23 2021 *)

PROG

(Haskell)

a142987 n = a142987_list !! (n-1)

a142987_list = 1 : 10 : zipWith (+)

(map (* 10) $ tail a142987_list)

(zipWith (*) (drop 2 a002378_list) a142987_list)

-- Reinhard Zumkeller, Jul 17 2015

CROSSREFS