A316938
Triangle read by rows formed using Pascal's rule except that n-th row begins and ends with Fibonacci(n+4).
Original entry on oeis.org
3, 5, 5, 8, 10, 8, 13, 18, 18, 13, 21, 31, 36, 31, 21, 34, 52, 67, 67, 52, 34, 55, 86, 119, 134, 119, 86, 55, 89, 141, 205, 253, 253, 205, 141, 89, 144, 230, 346, 458, 506, 458, 346, 230, 144, 233, 374, 576, 804, 964, 964, 804, 576, 374, 233, 377, 607, 950, 1380, 1768, 1928, 1768, 1380, 950, 607, 377
Offset: 0
Triangle begins:
3;
5, 5;
8, 10, 8;
13, 18, 18, 13;
21, 31, 36, 31, 21;
34, 52, 67, 67, 52, 34;
55, 86, 119, 134, 119, 86, 55;
89, 141, 205, 253, 253, 205, 141, 89;
144, 230, 346, 458, 506, 458, 346, 230, 144;
...
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t={}; Do[r={}; Do[If[k==0||k==n, m=Fibonacci[n+4], m=t[[n,k]]+t[[n,k+1]]]; r=AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t
A316528
a(n) = 3*a(n-1) - a(n-2) - 2*a(n-3) for n > 2, a(0)=1, a(1)=4, a(2)=10.
Original entry on oeis.org
1, 4, 10, 24, 54, 118, 252, 530, 1102, 2272, 4654, 9486, 19260, 38986, 78726, 158672, 319318, 641830, 1288828, 2586018, 5185566, 10393024, 20821470, 41700254, 83493244, 167136538, 334515862, 669424560, 1339484742, 2679997942, 5361659964, 10726012466, 21456381550
Offset: 0
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a:=[1,4,10];; for n in [4..35] do a[n]:=3*a[n-1]-a[n-2]-2*a[n-3]; od; a; # Muniru A Asiru, Jul 14 2018
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I:=[1,4,10]; [n le 3 select I[n] else 3*Self(n-1)-Self(n-2)-2*Self(n-3): n in [1..40]];
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seq(coeff(series((1+x-x^2)/(1-3*x+x^2+2*x^3), x,n+1),x,n),n=0..35); # Muniru A Asiru, Jul 14 2018
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RecurrenceTable[{a[n] == 3 a[n - 1] - a[n - 2] - 2 a[n - 3], a[0] == 1, a[1] == 4, a[2] == 10}, a, {n, 0, 40}]
Table[5 2^n - 2 Fibonacci[n + 3], {n, 0, 40}] (* Bruno Berselli, Jul 16 2018 *)
LinearRecurrence[{3,-1,-2},{1,4,10},40] (* Harvey P. Dale, Jul 18 2020 *)
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Vec((1 + x - x^2)/((1 - 2*x)*(1 - x - x^2)) + O(x^40)) \\ Colin Barker, Jul 23 2018
A347584
Triangle formed by Pascal's rule, except that the n-th row begins and ends with the n-th Lucas number.
Original entry on oeis.org
2, 1, 1, 3, 2, 3, 4, 5, 5, 4, 7, 9, 10, 9, 7, 11, 16, 19, 19, 16, 11, 18, 27, 35, 38, 35, 27, 18, 29, 45, 62, 73, 73, 62, 45, 29, 47, 74, 107, 135, 146, 135, 107, 74, 47, 76, 121, 181, 242, 281, 281, 242, 181, 121, 76, 123, 197, 302, 423, 523, 562, 523, 423, 302, 197, 123
Offset: 0
The first two Lucas numbers (for n=0 and n=1) are 2 and 1, so the first two rows (again, for n=0 and n=1) of the triangle are 2 and 1, 1 respectively.
Triangle begins:
2;
1, 1;
3, 2, 3;
4, 5, 5, 4;
7, 9, 10, 9, 7;
11, 16, 19, 19, 16, 11;
18, 27, 35, 38, 35, 27, 18;
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T[n_, 0] := LucasL[n]; T[n_, n_] := LucasL[n];
T[n_, k_] := T[n - 1, k - 1] + T[n - 1, k];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten
A379837
Triangle read by rows formed using Pascal's rule except that n-th row begins and ends with Fibonacci(n+3).
Original entry on oeis.org
2, 3, 3, 5, 6, 5, 8, 11, 11, 8, 13, 19, 22, 19, 13, 21, 32, 41, 41, 32, 21, 34, 53, 73, 82, 73, 53, 34, 55, 87, 126, 155, 155, 126, 87, 55, 89, 142, 213, 281, 310, 281, 213, 142, 89, 144, 231, 355, 494, 591, 591, 494, 355, 231, 144
Offset: 0
Triangle begins:
k= 0 1 2 3 4 5 6
n=0: 2;
n=1: 3, 3;
n=2: 5, 6, 5;
n=3: 8, 11, 11, 8;
n=4: 13, 19, 22, 19, 13;
n=5: 21, 32, 41, 41, 32, 21;
n=6: 34, 53, 73, 82, 73, 53, 34;
...
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// As triangle // t={};Do[r={};Do[If[k==0||k==n,m=Fibonacci[n+3],m=t[[n,k]]+t[[n,k+1]]];r=AppendTo[r,m],{k,0,n}];AppendTo[t,r],{n,0,11}];t
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