A377000 -id:A377000 - cp's OEIS frontend

A152086 a(n) = Sum_{k=1..n-1} k*A110971(n,k).

1, 3, 8, 21, 52, 126, 296, 685, 1556, 3498, 7768, 17122, 37416, 81308, 175568, 377469, 807604, 1721970, 3657464, 7746838, 16357496, 34459428, 72407728, 151851986, 317777032, 663908196, 1384524656, 2883208740, 5994736336, 12448784824, 25816193952, 53479331357, 110652549620

Offset: 2

N. J. A. Sloane, Sep 20 2009

nonn

Paolo Xausa, Table of n, a(n) for n = 2..1000
M. A. Michels and U. Knauer, The congruence classes of paths and cycles, Discr. Math., 309 (2009), 5352-5359. See p. 5356.

Cf. A110971, A102699.

Main diagonal of A377000.

A110971[n_] := (n+1)*2^(n-2) - If[OddQ[n], (n-1/2)*Binomial[n-1, (n-1)/2], 2*(n-1)*Binomial[n-2, (n-2)/2]];
Array[A110971, 50, 2] (* Paolo Xausa, Oct 13 2024 *)

from math import comb
def A152086(n): return ((n+1<>1)>>1 if n&1 else (n-1)*comb(n-2,n-2>>1)<<1)) # Chai Wah Wu, Oct 28 2024

a(n) = A102699(n)/2. - Paolo Xausa, Oct 13 2024, from N. J. A. Sloane formula in A102699.

More terms from Paolo Xausa, Oct 13 2024

A228879 a(n+2) = 3*a(n), starting 4,7.

Original entry on oeis.org

4, 7, 12, 21, 36, 63, 108, 189, 324, 567, 972, 1701, 2916, 5103, 8748, 15309, 26244, 45927, 78732, 137781, 236196, 413343, 708588, 1240029, 2125764, 3720087, 6377292, 11160261, 19131876, 33480783, 57395628, 100442349, 172186884, 301327047, 516560652

Offset: 0

Richard R. Forberg, Sep 06 2013

Successive terms are the square roots of expressions of prior terms. The same recursive formula (see below) can be applied using any term of A001353 as the initializing value to produce the family of sequences, as given in the array A227418. This sequence uses A001353(2) = 4, and is the third row of that array.
a(4n) are the squares of A008776(n).
Binomial transform of a(n) is A021006.
First differences of a(n) = A083658 (without initial two terms).
2nd differences of a(n) = A068911 (with initial term).
a(n-1) is the number of n-digit base 10 numbers where all the digits are even numbers, and each digit differs by 2 from the previous and the next digit. - Graeme McRae, Jun 09 2014

Paolo Xausa, Table of n, a(n) for n = 0..1000
Twitter / MathQuizzes, Puzzle relating to this sequence
Index entries for linear recurrences with constant coefficients, signature (0,3).

Cf. A001353, A008776, A021006, A068911, A083658, A227418.

Row n = 5 of A377000.

LinearRecurrence[{0, 3}, {4, 7}, 50] (* Paolo Xausa, Oct 14 2024 *)

Vec(-(7*x+4)/(3*x^2-1) + O(x^100)) \\ Colin Barker, Jun 09 2014

a(n) = sqrt(3*a(n-1)^2 + (-3)^(n-1)), a(0) = 4.
This divisibility relation also applies: a(n+3) = a(n+2)*a(n+1)/a(n).
G.f.: -(7*x+4) / (3*x^2-1). - Colin Barker, Jun 09 2014
From Stefano Spezia, Mar 20 2022: (Start)
a(n) = 3^((n-1)/2)*(4*sqrt(3) + 7 + (-1)^n*(4*sqrt(3) - 7))/2.
E.g.f.: 4*cosh(sqrt(3)*x) + 7*sinh(sqrt(3)*x)/sqrt(3). (End)

More terms from Colin Barker, Jun 09 2014

cp's OEIS Frontend

A152086 a(n) = Sum_{k=1..n-1} k*A110971(n,k).

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