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A246462 a(n) = Sum_{k=0..n} (2k+1)*C(n,k)^2*C(n+k,k)^2, where C(n,k) denotes the binomial coefficient n!/(k!*(n-k)!). 3
1, 13, 289, 7733, 223001, 6689045, 205569505, 6422252485, 203029535305, 6476057609045, 208013166524153, 6718923443380109, 218021269879802377, 7101635058978727909, 232072490781790669153, 7604916953685880646885 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

For any n > 0, n^3 divides Sum_{k=0..n-1} (2k+1)*a(k).

LINKS

Zhi-Wei Sun, Table of n, a(n) for n = 0..100

Zhi-Wei Sun, Two new kinds of numbers and their arithmetic properties, arXiv:1408.5381 [math.NT], 2014-2018.

FORMULA

Recurrence: n^3*(2*n-3)*(8*n^4 - 48*n^3 + 96*n^2 - 72*n + 13)*a(n) = (2*n-1)*(280*n^7 - 2096*n^6 + 5728*n^5 - 6536*n^4 + 1383*n^3 + 3160*n^2 - 2432*n + 552)*a(n-1) - (2*n-3)*(280*n^7 - 1824*n^6 + 4096*n^5 - 3384*n^4 - 345*n^3 + 2046*n^2 - 1100*n + 192)*a(n-2) + (n-2)^3*(2*n-1)*(8*n^4 - 16*n^3 + 8*n - 3)*a(n-3). - Vaclav Kotesovec, Aug 27 2014

a(n) ~ sqrt(24+17*sqrt(2)) * (17+12*sqrt(2))^n / (4*Pi^(3/2)*sqrt(n)). - Vaclav Kotesovec, Aug 27 2014

EXAMPLE

a(1) = 13 since Sum_{k=0..1} (2k+1)*C(1,k)^2*C(1+k,k)^2 = 1 + 3*2^2 = 13.

MAPLE

A246462:=n->add((2*k+1)*binomial(n, k)^2*binomial(n+k, k)^2, k=0..n): seq(A246462(n), n=0..20); # Wesley Ivan Hurt, Aug 27 2014

MATHEMATICA

a[n_]:=Sum[(2k+1)*Binomial[n, k]^2*Binomial[n+k, k]^2, {k, 0, n}]

Table[a[n], {n, 0, 15}]

CROSSREFS

Cf. A245769, A246065, A246138, A246459, A246460, A246461.

Sequence in context: A064750 A228753 A249864 * A023357 A165218 A317632

Adjacent sequences:  A246459 A246460 A246461 * A246463 A246464 A246465

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Aug 26 2014

STATUS

approved

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Last modified November 12 19:25 EST 2019. Contains 329078 sequences. (Running on oeis4.)