A337283
a(n) = Sum_{i=0..n} i*T(i)^2, where T(i) = A000073(i) is the i-th tribonacci number.
Original entry on oeis.org
0, 0, 2, 5, 21, 101, 395, 1578, 6186, 23610, 89220, 333431, 1234343, 4536551, 16567157, 60172532, 217524468, 783111476, 2809027334, 10043413545, 35805255545, 127314522569, 451629771519, 1598650868766, 5647706073630, 19916305738030, 70117445671624, 246478579433947, 865201260035147
Offset: 0
- Raphael Schumacher, Explicit formulas for sums involving the squares of the first n Tribonacci numbers, Fib. Q., 58:3 (2020), 194-202.
- Colin Barker, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (5,-2,-2,-35,3,0,48,-11,7,-14,2,-1,1).
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R:=PowerSeriesRing(Integers(), 30); [0,0] cat Coefficients(R!( x^2*(1-x^2)*(2-7*x+7*x^2+3*x^3+9*x^4+7*x^5+x^6+x^7+x^8)/((1-x)*(1+x+x^2-x^3)*(1-3*x-x^2-x^3))^2 )); // G. C. Greubel, Nov 20 2021
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T[n_]:= T[n]= If[n<2, 0, If[n==2, 1, T[n-1] +T[n-2] +T[n-3]]]; (* A000073 *)
a[n_]:= a[n]= Sum[j*T[j]^2, {j,0,n}];
Table[a[n], {n,0,30}] (* G. C. Greubel, Nov 20 2021 *)
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concat([0,0], Vec(x^2*(1+x)*(2 -7*x +7*x^2 +3*x^3 +9*x^4 +7*x^5 +x^6 + x^7 +x^8)/((1-x)*(1 +x +x^2 -x^3)^2*(1 -3*x -x^2 -x^3)^2) + O(x^30))) \\ Colin Barker, Sep 19 2020
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@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 A337283(n): return sum( j*T(j)^2 for j in (0..n) )
[A337283(n) for n in (0..40)] # G. C. Greubel, Nov 20 2021
A337284
a(n) = Sum_{i=1..n} (i-1)*T(i)^2, where T(i) = A000073(i) is the i-th tribonacci number.
Original entry on oeis.org
0, 1, 3, 15, 79, 324, 1338, 5370, 20858, 79907, 301917, 1127753, 4175945, 15347222, 56045572, 203563012, 735880196, 2649245173, 9502874215, 33976624115, 121128306995, 430701953720, 1527852568478, 5408197139806, 19106052817630, 67376379676855, 237205619596129, 833831061604429, 2926954896983117
Offset: 1
- 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).
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R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( x^2*(1-2*x+2*x^2+12*x^3+8*x^5+2*x^6+4*x^7+3*x^8+2*x^9)/((1-x)*(1-2*x-3*x^2-6*x^3+x^4+x^6)^2) )); // G. C. Greubel, Nov 22 2021
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T[n_]:= T[n]= If[n<2, 0, If[n==2, 1, T[n-1] +T[n-2] +T[n-3]]];
a[n_]:= a[n]= Sum[(j-1)*T[j]^2, {j,0,n}];
Table[a[n], {n,40}] (* G. C. Greubel, Nov 22 2021 *)
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@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 A337284(n): return sum( (j-1)*T(j)^2 for j in (0..n) )
[A337284(n) for n in (1..40)] # G. C. Greubel, Nov 22 2021
A141583
Squares of tribonacci numbers A000213.
Original entry on oeis.org
1, 1, 1, 9, 25, 81, 289, 961, 3249, 11025, 37249, 126025, 426409, 1442401, 4879681, 16507969, 55845729, 188925025, 639128961, 2162157001, 7314525625, 24744863025, 83711270241, 283193201281, 958035736849, 3241011678961
Offset: 0
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Eric Weisstein's World of Mathematics, Ladder Graph
- Eric Weisstein's World of Mathematics, Total Dominating Set
- Index entries for linear recurrences with constant coefficients, signature (2,3,6,-1,0,-1).
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I:=[1,1,1,9,25,81]; [n le 6 select I[n] else 2*Self(n-1) + 3*Self(n-2) + 6*Self(n-3) - Self(n-4) - Self(n-6): n in [1..30]]; // Vincenzo Librandi, Dec 13 2012
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CoefficientList[Series[(1+x)^2*(1-3*x+x^2-x^3)/((1+x+x^2-x^3)*(1-3*x-x^2-x^3)), {x, 0, 40}], x] (* Vincenzo Librandi, Dec 13 2012 *)
Table[RootSum[-1 - # - #^2 + #^3 &, 2 #^n - 4 #^(n + 1) + 3 #^(n + 2) &]^2/121, {n, 0, 20}] (* Eric W. Weisstein, Apr 10 2018 *)
LinearRecurrence[{2,3,6,-1,0,-1}, {1,1,9,25,81,289}, {0, 20}] (* Eric W. Weisstein, Apr 10 2018 *)
LinearRecurrence[{1,1,1},{1,1,1},40]^2 (* Harvey P. Dale, Aug 01 2021 *)
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@CachedFunction
def T(n): # A000213
if (n<3): return 1
else: return T(n-1) +T(n-2) +T(n-3)
def A141583(n): return T(n)^2
[A141583(n) for n in (0..40)] # G. C. Greubel, Nov 22 2021
A337285
a(n) = Sum_{i=1..n} (i-1)^2*T(i)^2, where T(i) = A000073(i) is the i-th tribonacci number.
Original entry on oeis.org
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
- 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).
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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
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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 *)
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@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
A337286
a(n) = Sum_{i=0..n} i^2*T(i)^2, where T(i) = A000073(i) is the i-th tribonacci number.
Original entry on oeis.org
0, 0, 4, 13, 77, 477, 2241, 10522, 47386, 204202, 860302, 3546623, 14357567, 57286271, 225714755, 879795380, 3397426356, 13012405492, 49478890936, 186932228945, 702169068945, 2623863676449, 9758799153349, 36140284390030, 133317609306766, 490032600916766, 1795262239190210, 6557012850772931
Offset: 0
- 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 = 0..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).
-
R:=PowerSeriesRing(Integers(), 40); [0,0] cat Coefficients(R!( x^2*(4-15*x+22*x^2+83*x^3-90*x^4+11*x^5-128*x^6-207*x^7-224*x^8-233*x^9-162*x^10- 147*x^11-58*x^12-3*x^13-4*x^14-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
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T[n_]:= T[n]= If[n<2, 0, If[n==2, 1, T[n-1] +T[n-2] +T[n-3]]]; (* A000073 *)
A337286[n_]:= Sum[j^2*T[j]^2, {j,0,n}];
Table[A337286[n], {n, 0, 50}] (* G. C. Greubel, Nov 22 2021 *)
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@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 A337286(n): return sum( j^2*T(j)^2 for j in (0..n) )
[A337286(n) for n in (0..40)] # G. C. Greubel, Nov 22 2021
A343125
Triangle T(k, n) = (n+3)*(k-n) - 4, k >= 2, 1 <= n <= k-1, read by rows.
Original entry on oeis.org
0, 4, 1, 8, 6, 2, 12, 11, 8, 3, 16, 16, 14, 10, 4, 20, 21, 20, 17, 12, 5, 24, 26, 26, 24, 20, 14, 6, 28, 31, 32, 31, 28, 23, 16, 7, 32, 36, 38, 38, 36, 32, 26, 18, 8, 36, 41, 44, 45, 44, 41, 36, 29, 20, 9, 40, 46, 50, 52, 52, 50, 46, 40, 32, 22, 10
Offset: 2
Triangle T(k, n) begins:
k \ n| 1 2 3 4 5 6 7 8 9 10 11
------+----------------------------------
2 | 0
3 | 4 1
4 | 8 6 2
5 | 12 11 8 3
6 | 16 16 14 10 4
7 | 20 21 20 17 12 5
8 | 24 26 26 24 20 14 6
9 | 28 31 32 31 28 23 16 7
10 | 32 36 38 38 36 32 26 18 8
11 | 36 41 44 45 44 41 36 29 20 9
12 | 40 46 50 52 52 50 46 40 32 22 10
.
The following are the closed formulas for k = 3, 4 for A(k, n) = Sum_{m=0..n} F(k, m)^2, with F(k, n) = A092921(k, n), the k-generalized Fibonacci numbers, and A(k, n) = A343138(k, n), the sum of squares of F(k, n). These formulas are derived from the closed formula in the formula section. Of course further simplifications are possible. For k = 2, T(2, 1) = 0 so illustrations start with k = 3.
k | Formula
--+--------------------------------------------------------
3 | Sum_{m=0..n} F(3,m)^2 = (1/4)*(2*F(3,n)*F(3,n+2) + 4*F(3,n+1)*F(3,n+2) - (k - 2)*F(3,n)^2 - T(3,1)*F(3,n+1)^2 - T(3,2)*F(3,n+2)^2 + 1).
4 | Sum_{m=0..n} F(3,m)^2 = (1/6)*(-2*F(4,n)*F(4,n+1) + 2*F(4,n)*F(4,n+3) + 4*F(4,n+1)*F(4,n+3) + 6*F(4,n+2)*F(4,n+3) - (k-2)*F(4,n)^2 - T(4,1)*F(4,n+1)^2 - T(4, 2)*F(4,n+2)^2 - T(4,3)*F(4,n+3)^2 + 2).
- Raphael Schumacher, How to Sum the Squares of the Tetranacci Numbers and the Fibonacci m-step Numbers, Fibonacci Quarterly, 57, (2019), 168-175.
- Raphael Schumacher, Explicit Formulas for Sums Involving the Squares of the First n Tribonacci Numbers, Fibonacci Quarterly, 58 (2020), 194-202.
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T := (k, n) -> (n + 3)*(k - n) - 4:
seq(print(seq(T(k, n), n=1..k-1)), k = 2..12); # Peter Luschny, Apr 02 2021
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Table[(n + 3) (k - n) - 4, {k, 2, 12}, {n, k - 1}] // Flatten (* Michael De Vlieger, Apr 06 2021 *)
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T(k,n)=(n + 3)*(k - n) - 4
for(k = 2,12,for(n = 1,k - 1, print1(T(k,n),", ")))
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flatten([[(n+3)*(k-n) -4 for n in (1..k-1)] for k in (2..15)]) # G. C. Greubel, Nov 22 2021
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