A107239 -id:A107239 - cp's OEIS frontend

A337285 a(n) = Sum_{i=1..n} (i-1)^2*T(i)^2, where T(i) = A000073(i) is the i-th tribonacci number.

0, 1, 5, 41, 297, 1522, 7606, 35830, 159734, 691175, 2911275, 11995471, 48573775, 193800376, 763577276, 2976338876, 11493413820, 44020618429, 167385941185, 632387189285, 2375420846885, 8876467428110, 33013780952786, 122261706093330, 451010242361106, 1657768413841731, 6073328651742855

Offset: 1

N. J. A. Sloane, Sep 12 2020

nonn

R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202. (Note that this paper uses an offset for the tribonacci numbers that is different from that used in A000073.)

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-9,-7,-56,96,108,252,-162,-114,-318,126,-16,136,-36,12,-21,3,-1,1).

Cf. A000073, A085697, A107239, A337282, A337283, A337284, A337286.

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x^2*(1 -2*x+15*x^2+62*x^3-97*x^4+96*x^5+73*x^6-64*x^7-57*x^8-194*x^9-127*x^10-138*x^11 -55*x^12-12*x^13-9*x^14-4*x^15)/((1-x)*(1+x+x^2-x^3)^3*(1-3*x-x^2 -x^3)^3) )); // G. C. Greubel, Nov 22 2021

T[n_]:= T[n]= If[n<2, 0, If[n==2, 1, T[n-1] +T[n-2] +T[n-3]]]; (* A000073 *)
A337285[n_]:= Sum[j^2*T[j+1]^2, {j,0,n-1}];
Table[A337285[n], {n, 40}] (* G. C. Greubel, Nov 22 2021 *)

@CachedFunction
def T(n): # A000073
    if (n<2): return 0
    elif (n==2): return 1
    else: return T(n-1) +T(n-2) +T(n-3)
def A337285(n): return sum( j^2*T(j+1)^2 for j in (0..n-1) )
[A337285(n) for n in (1..40)] # G. C. Greubel, Nov 22 2021

From G. C. Greubel, Nov 22 2021: (Start)
a(n) = A337286(n) - 2*A337283(n) + A107239(n).
a(n) = Sum_{j=0..n-1} j^2*A000073(j+1)^2.
G.f.: x^2*(1 -2*x +15*x^2 +62*x^3 -97*x^4 +96*x^5 +73*x^6 -64*x^7 -57*x^8 -194*x^9 -127*x^10 -138*x^11 -55*x^12 -12*x^13 -9*x^14 -4*x^15)/((1-x)*(1 +x +x^2 -x^3)^3*(1 -3*x -x^2 -x^3)^3). (End)

A140973 Numbers k such that arithmetic mean of squares of the first k tribonacci numbers is an integer.

Original entry on oeis.org

1, 2, 8, 15, 16, 18, 22, 32, 47, 48, 53, 58, 64, 70, 77, 78, 80, 94, 95, 96, 103, 106, 128, 138, 163, 199, 206, 256, 257, 266, 269, 311, 326, 330, 352, 358, 385, 397, 398, 401, 419, 421, 499, 512, 514, 538, 587, 599, 617, 622, 640, 672, 683, 757, 768, 770, 773

Offset: 1

Ctibor O. Zizka, Jul 27 2008

Can the arithmetic mean of the tribonacci numbers (T(0)+...+T(k-1)) / k be an integer?

The arithmetic means are integers for the first 1, 2, 47, 53, 94, 103, 106, ... (A141579) tribonacci numbers. - R. J. Mathar, Aug 04 2008

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000073, A141579.

Cf. A107239. - R. J. Mathar, Aug 04 2008

With[{m = 1000}, Position[Accumulate[LinearRecurrence[{1, 1, 1}, {0, 0, 1}, m]^2] / Range[m], ?IntegerQ] // Flatten] (* _Amiram Eldar, Jul 04 2025 *)

Numbers k such that (T(0)^2+ T(1)^2+ ... + T(k-1)^2) / k is an integer, where T(i) = i-th tribonacci number.

a(1)-a(2) inserted and a(32) onwards added by R. J. Mathar, Aug 04 2008

A343138 Array A(k, n) read by descending antidiagonals: A(k, n) = Sum_{m=0..n} F(k, m)^2, where F are the k-generalized Fibonacci numbers A092921.

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 2, 1, 0, 1, 4, 6, 2, 1, 0, 1, 5, 15, 6, 2, 1, 0, 1, 6, 40, 22, 6, 2, 1, 0, 1, 7, 104, 71, 22, 6, 2, 1, 0, 1, 8, 273, 240, 86, 22, 6, 2, 1, 0, 1, 9, 714, 816, 311, 86, 22, 6, 2, 1, 0, 1, 10, 1870, 2752, 1152, 342, 86, 22, 6, 2, 1, 0

Offset: 0

Peter Luschny, Apr 06 2021

			Array starts:
  n = 0  1  2  3   4   5    6     7     8     9      10
------------------------------------------------------------
[k=0] 0, 1, 1, 1,  1,  1,   1,    1,    1,     1,     1, ...  [A057427]
[k=1] 0, 1, 2, 3,  4,  5,   6,    7,    8,     9,    10, ...  [A001477]
[k=2] 0, 1, 2, 6, 15, 40, 104,  273,  714,  1870,  4895, ...  [A001654]
[k=3] 0, 1, 2, 6, 22, 71, 240,  816, 2752,  9313, 31514, ...  [A107239]
[k=4] 0, 1, 2, 6, 22, 86, 311, 1152, 4288, 15952, 59216, ...
[k=5] 0, 1, 2, 6, 22, 86, 342, 1303, 5024, 19424, 75120, ...
[k=6] 0, 1, 2, 6, 22, 86, 342, 1366, 5335, 20960, 82464, ...
[k=7] 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21591, 85600, ...
[k=8] 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21846, 86871, ...
[k=9] 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21846, 87382, ...
[...]
[ oo] 0, 1, 2, 6, 22, 86, 342, 1366, 5462, 21846, 87382, ...  [A047849]
Note that the first parameter in A(k, n) refers to rows, and the second parameter refers to columns, as always. The usual naming convention for the indices is not adhered to because the row sequences are the sums of the squares of the k-bonacci numbers.

Greg Dresden and Yichen Wang, Sums and convolutions of k-bonacci and k-Lucas numbers, draft 2021.

Russell Jay Hendel, Sums of Squares: Methods for Proving Identity Families, arXiv:2103.16756 [math.NT], 2021

Cf. A092921, A057427, A001477, A001654, A107239, A047849, A343125.

F := (k, n) -> (F(k, n) := `if`(n<2, n, add(F(k, n-j), j = 1..min(k, n)))):
A := (k, n) -> add(F(k, m)^2, m = 0..n):
seq(seq(A(k, n-k), k=0..n), n = 0..11);
# The following two functions implement Russell Jay Hendel's formula for k >= 2:
T := (k, n) -> (n + 3)*(k - n) - 4:
H := (k, n) -> (2*add(j*add((m-k+1)*F(k, n+j)*F(k, n+m), m = j+1..k), j = 1..k-1)
- add(T(k, j-1)*F(k, n+j)^2, j = 1..k) + (k - 2))/(2*k - 2):
seq(lprint([k], seq(H(k, n), n = 0..11)), k=2..9); # Peter Luschny, Apr 07 2021

A343138[k_, len_] := Take[Accumulate[LinearRecurrence[PadLeft[{1}, k, 1], PadLeft[{1}, k], len + k]^2], -len - 2];
A343138[0, len_] := Table[Boole[n != 0], {n, 0, len}];
A343138[1, len_] := Table[n, {n, 0, len}];
(* Table: *) Table[A343138[k, 12], {k, 0, 9}]
(* Sequence / descending antidiagonals: *)
Table[Table[Take[A343138[j, 12], {k + 1 - j, k + 1 - j}], {j, 0, k}], {k, 0, 10}] // Flatten (* Georg Fischer, Apr 08 2021 *)

Russell Jay Hendel gives the following representation, valid for k >= 2:

A(n, k) = Sum_{m=0..n} F(k, m)^2 = (1/(2*k-2)) * (2*Sum_{j=1..k-1}(j*Sum_{m=j+1..k} (m-k+1) * F(k, n+j) * F(k, n+m)) - Sum_{j=1..k}(A343125(k, j-1) * F(k, n+j)^2) + (k - 2)). - Peter Luschny, Apr 07 2021

A228609 Partial sums of the cubes of the tribonacci sequence A000073.

Original entry on oeis.org

0, 1, 2, 10, 74, 417, 2614, 16438, 101622, 633063, 3941012, 24511836, 152535900, 949133883, 5905611508, 36746590964, 228646935796, 1422699232325, 8852413871022, 55082039340022, 342734883853750, 2132586518002125

Offset: 0

R. J. Mathar, Dec 18 2013

R. Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.

Michael De Vlieger, Table of n, a(n) for n = 0..1261
H. Ohtsuka, Advanced Problems and Solutions, Fib. Quart. 51 (2) (2013) 186, H-736.
Index entries for linear recurrences with constant coefficients, signature (5,5,25,-58,26,-42,54,-13,1,-3,1).

Cf. A000073, A107239.

CoefficientList[Series[x (-1 + 3 x + 11 x^3 - 5 x^4 + x^5 - 3 x^6 + x^7 + 5 x^2)/((x^3 - 5 x^2 + 7 x - 1) (x^6 + 4 x^5 + 11 x^4 + 12 x^3 + 11 x^2 + 4 x + 1) (x - 1)^2), {x, 0, 21}], x] (* Michael De Vlieger, Jan 12 2022 *)
Accumulate[LinearRecurrence[{1,1,1},{0,1,1},30]^3]  (* or *) LinearRecurrence[ {5,5,25,-58,26,-42,54,-13,1,-3,1},{0,1,2,10,74,417,2614,16438,101622,633063,3941012},30] (* Harvey P. Dale, Sep 11 2022 *)

T(n)=([0, 1, 0; 0, 0, 1; 1, 1, 1]^n)[1, 3]; \\ A000073
a(n) = sum(k=1, n, T(k)^3); \\ Michel Marcus, Jan 12 2022

a(n) = a(n-1) + (A000073(n))^3.

G.f.: x*(-1+3*x+11*x^3-5*x^4+x^5-3*x^6+x^7+5*x^2) / ( (x^3-5*x^2+7*x-1) *(x^6+4*x^5+11*x^4+12*x^3+11*x^2+4*x+1) *(x-1)^2 )

cp's OEIS Frontend

A337285 a(n) = Sum_{i=1..n} (i-1)^2*T(i)^2, where T(i) = A000073(i) is the i-th tribonacci number.

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A140973 Numbers k such that arithmetic mean of squares of the first k tribonacci numbers is an integer.

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A343138 Array A(k, n) read by descending antidiagonals: A(k, n) = Sum_{m=0..n} F(k, m)^2, where F are the k-generalized Fibonacci numbers A092921.

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A228609 Partial sums of the cubes of the tribonacci sequence A000073.

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