A054446 Triangle of partial row sums of triangle A037027(n,m), n >= m >= 0 (Fibonacci convolution triangle).
1, 2, 1, 5, 3, 1, 12, 9, 4, 1, 29, 24, 14, 5, 1, 70, 62, 42, 20, 6, 1, 169, 156, 118, 67, 27, 7, 1, 408, 387, 316, 205, 100, 35, 8, 1, 985, 951, 821, 588, 332, 142, 44, 9, 1, 2378, 2323, 2088, 1614, 1020, 509, 194, 54, 10, 1, 5741, 5652, 5232, 4290, 2966, 1671, 747, 257
Offset: 0
Examples
{1}; {2,1}; {5,3,1}; {12,9,4,1};...
Fourth row polynomial (n=3): p(3,x)= 12+9*x+4*x^2+x^3
Formula
a(n, m)=sum(A037027(n, k), k=m..n), n >= m >= 0, a(n, m) := 0 if n
Column m recursion: a(n, m)= sum(a(j-1, m)*A037027(n-j, 0), j=m..n) + A037027(n, m), n >= m >= 0, a(n, m) := 0 if n
G.f. for column m: Pell(x)*(x*Fib(x))^m, m >= 0, with Fib(x) = g.f. A000045(n+1) and Pell(x) = g.f. A000129(n+1).
T(n,0) = 2*T(n-1,0) + T(n-2,0), T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k) for k>0, T(0,0) = 1, T(1,0) = 2, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 26 2014
A291231 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S).
10, 65, 360, 1831, 8830, 41104, 186710, 833401, 3672840, 16034303, 69506930, 299700192, 1287010850, 5509712833, 23531008200, 100312445063, 427025152550, 1815832379312, 7714875191470, 32756357939033, 139008007848360, 589672772732671, 2500620567692890
Offset: 0
Comments
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
See A291219 for a guide to related sequences.
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (10, -31, 20, 40, -20, -31, -10, -1)
Programs
Formula
G.f.: (10 - 35 x + 20 x^2 + 46 x^3 - 20 x^4 - 35 x^5 - 10 x^6)/(1 - 10 x + 31 x^2 - 20 x^3 - 40 x^4 + 20 x^5 + 31 x^6 + 10 x^7 + x^8).
a(n) = 10*a(n-1) - 31*a(n-2) + 20*a(n-3) + 40*a(n-4) - 20*a(n-5) -31*a(n-6) - 10*a(n-7) - a(n-8) for n >= 7.
A291264 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - 2 S)^2.
4, 12, 36, 104, 292, 804, 2180, 5840, 15492, 40764, 106532, 276792, 715556, 1841748, 4722180, 12066208, 30737924, 78088812, 197892388, 500374024, 1262618148, 3180066180, 7995639940, 20071580784, 50312160388, 125942854044, 314865132324, 786254598872
Offset: 0
Comments
Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
See A291219 for a guide to related sequences.
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (4, -2, -4, -1)
Programs
-
Mathematica
z = 60; s = x/(1 - x^2); p = (1 - 2 s)^2; Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *) u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291264 *) u/4 (A054447 *) -
PARI
Vec(4*(1 - x - x^2) / (1 - 2*x - x^2)^2 + O(x^40)) \\ Colin Barker, Aug 26 2017
Formula
G.f.: -((4 (-1 + x + x^2))/(-1 + 2 x + x^2)^2).
a(n) = 4*a(n-1) - 2*a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.
a(n) - 4*A054447(n) for n >= 0.
a(n) = ((1-sqrt(2))^(1+n)*(2-3*sqrt(2) + 2*n) + (1+sqrt(2))^(1+n)*(2+3*sqrt(2) + 2*n)) / 4. - Colin Barker, Aug 26 2017
Comments