A257838 Main diagonal of iterated partial sums array of Fibonacci numbers (starting with the first partial sums).

0, 1, 4, 16, 63, 247, 967, 3785, 14820, 58060, 227612, 892926, 3505386, 13770404, 54129602, 212904952, 837885495, 3299264407, 12997784803, 51230474669, 202014314769, 796928589755, 3145066003589, 12416625685891, 49037912997003, 193734379979677, 765632076098287, 3026670770970925, 11968378998073935

Offset: 0

Author

Luciano Ancora, May 10 2015

Keywords

• nonn
• easy

Comments

The array used here starts in row n=0 with the first partial sums of A000045. The array which starts with the Fibonacci numbers in row k=0 is shown in A136431. The diagonal of that array is given in A176085. - Wolfdieter Lang, Jun 03 2015

Examples

			This sequence is the main diagonal of the following array (see the comment and Example field of A136431):
0, 1, 2,  4,  7,  12, ...  A000071
0, 1, 3,  7, 14,  26, ...  A001924
0, 1, 4, 11, 25,  51, ...  A014162
0, 1, 5, 16, 41,  92, ...  A014166
0, 1, 6, 22, 63, 155, ...  A053739
0, 1, 7, 29, 92, 247, ...  A053295
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000045, A000071, A001924, A014162, A014166, A136431, A176085.

Programs

Mathematica

Table[DifferenceRoot[Function[{a, n},{(2*n + 4*n^2)*a[n] + (2 + 7*n + 15*n^2)*a[1 + n] + (8 - 6*n - 8*n^2)*a[2 + n] + (-2 + n + n^2)*a[3 + n] == 0, a[1] == 0, a[2] == 1, a[3] == 4, a[4] == 16}]][n], {n, 30}]

Maxima

a(n):=sum(binomial(2*n-k,n-k)*fib(k),k,0,n); /* Vladimir Kruchinin, Oct 09 2016 */

PARI

x='x+O('x^50); concat([0], Vec(-(4*x+sqrt(1-4*x)-1)/(8*x^2+sqrt(1-4*x)*(8*x-2)-2*x))) \\ G. C. Greubel, Apr 08 2017

Formula

a(n) = F^{n+1}(n), n >= 0, with the k-th iterated partial sum F^{k} of the Fibonacci number A000045. - Wolfdieter Lang, Jun 03 2015

Conjecture: n*(n-3)*a(n) +2*(-4*n^2+13*n-6)*a(n-1) +(15*n^2-53*n+48)*a(n-2) +2*(2*n-3)*(n-2)*a(n-3)=0. - R. J. Mathar, Dec 10 2015

G.f.: -(4*x+sqrt(1-4*x)-1)/(8*x^2+sqrt(1-4*x)*(8*x-2)-2*x). - Vladimir Kruchinin, Oct 09 2016

a(n) = Sum_{k=0..n} binomial(2*n-k,n-k)*F(k), where F(k) = A000045(k). - Vladimir Kruchinin, Oct 09 2016

a(n) ~ 2^(2*n+1)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 09 2016

Extensions

Name edited by Wolfdieter Lang, Jun 03 2015