A027941 -id:A027941 - cp's OEIS frontend

A210554 Triangle of coefficients of polynomials v(n,x) jointly generated with A208341; see the Formula section.

1, 2, 2, 3, 5, 4, 4, 9, 12, 8, 5, 14, 25, 28, 16, 6, 20, 44, 66, 64, 32, 7, 27, 70, 129, 168, 144, 64, 8, 35, 104, 225, 360, 416, 320, 128, 9, 44, 147, 363, 681, 968, 1008, 704, 256, 10, 54, 200, 553, 1182, 1970, 2528, 2400, 1536, 512

Offset: 1

Clark Kimberling, Mar 22 2012

For a discussion and guide to related arrays, see A208510.

Also the number of multisets of size k that fit within some normal multiset of size n. A multiset is normal if it spans an initial interval of positive integers. - Andrew Howroyd, Sep 18 2018

			Triangle begins:
  1;
  2,  2;
  3,  5,   4;
  4,  9,  12,   8;
  5, 14,  25,  28,  16;
  6, 20,  44,  66,  64,  32;
  7, 27,  70, 129, 168, 144, 64;
  ...
First three polynomials v(n,x): 1, 2 + 2x , 3 + 5x + 4x^2.
The T(3, 1) = 3 multisets: (1), (2), (3).
The T(3, 2) = 5 multisets: (11), (12), (13), (22), (23).
The T(3, 3) = 4 multisets: (111), (112), (122), (123).

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Row sums are A027941.

Cf. A160232, A208341, A208510, A303974.

T := (n,k) -> simplify((n + 1 - k)*hypergeom([1 - k, -k + n + 2], [2], -1)):
seq(seq(T(n,k), k=1..n), n=1..10); # Peter Luschny, Sep 18 2018

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + x*v[n - 1, x] + 1;
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]   (* A208341 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]   (* A210554 *)

T(n,k)={sum(i=1, k, binomial(k-1, i-1)*binomial(n-k+i, i))} \\ Andrew Howroyd, Sep 18 2018

u(n,x)=x*u(n-1,x)+x*v(n-1,x)+1,
v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
T(n,k) = Sum_{i=1..k} binomial(k-1, i-1)*binomial(n-k+i, i). - Andrew Howroyd, Sep 18 2018
T(n,k) = (n - k + 1)*hypergeom([1 - k, n - k + 2], [2], -1). - Peter Luschny, Sep 18 2018

Example corrected by Philippe Deléham, Mar 23 2012

A264236 Number of vertices at level n of the hyperbolic Pascal pyramid.

Original entry on oeis.org

1, 3, 6, 13, 36, 138, 736, 4908, 36351, 280228, 2190651, 17206203, 135357481, 1065387963, 8387050686, 66029196613, 519841755036, 4092692363058, 32221664474776, 253680537891828, 1997222414704551, 15724098193422028, 123795561597659331, 974640390569138163

Offset: 0

Michel Marcus, Nov 09 2015

Colin Barker, Table of n, a(n) for n = 0..1000
László Németh, Hyperbolic Pascal pyramid, arXiv:1511.02067 [math.CO], 2015 (6th line of Table 1).
Index entries for linear recurrences with constant coefficients, signature (12,-37,37,-12,1).

Cf. A076765, A027941, A055588, A264237.

LinearRecurrence[{12, -37, 37, -12, 1}, {1, 3, 6, 13, 36}, 30] (* Bruno Berselli, Nov 09 2015 *)

Vec((1-9*x+7*x^2+15*x^3+3*x^4)/((1-x)*(1-3*x+x^2)*(1-8*x+x^2)) + O(x^50)) \\ Altug Alkan, Nov 09 2015

a(n) = 12*a(n-1) - 37*a(n-2) + 37*a(n-3) - 12*a(n-4) + a(n-5).
a(n) = (-3/2 + 9*sqrt(5)/10)*((3 + sqrt(5))/2)^n + (-3/2 - 9*sqrt(5)/10)*((3 - sqrt(5))/2)^n + (7/12 - 3*sqrt(15)/20)*(4 + sqrt(15))^n + (7/12 + 3*sqrt(15)/20)*(4 - sqrt(15))^n + 17/6. (See Németh paper, page 9.)
G.f.: (1 - 9*x + 7*x^2 + 15*x^3 + 3*x^4)/((1 - x)*(1 - 3*x + x^2)*(1 - 8*x + x^2)). [Bruno Berselli, Nov 09 2015]
a(n) = A076765(n-3) + 3*Fibonacci(2*(n-1)) + 3. - Ehren Metcalfe, Apr 18 2019

More terms from Bruno Berselli, Nov 09 2015

A214984 Array: T(m,n) = (F(m) + F(2m) + ... + F(nm))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).

Original entry on oeis.org

1, 2, 1, 4, 4, 1, 7, 12, 5, 1, 12, 33, 22, 8, 1, 20, 88, 94, 56, 12, 1, 33, 232, 399, 385, 134, 19, 1, 54, 609, 1691, 2640, 1487, 342, 30, 1, 88, 1596, 7164, 18096, 16492, 6138, 872, 48, 1, 143, 4180, 30348, 124033, 182900, 110143, 25319, 2256, 77, 1

Offset: 1

Clark Kimberling, Oct 28 2012

col 1: A001612 (except for initial term)
row 1: A000071
row 2: A027941
row 3: A049652
row 4: A092521
row 6: A049664
row 8: A156093 without minus signs

			Northwest corner:
1...2....4.....7......12......20
1...4....12....33.....88......232
1...5....22....94.....399.....1691
1...8....56....385....2640....18096
1...12...134...1487...16492...182900

Clark Kimberling, Antidiagonals n = 1..60, flattened

Cf. A214978, A214985, A214986.

F[n_] := Fibonacci[n]; L[n_] := LucasL[n];
t[m_, n_] := (1/F[m])*Sum[F[m*k], {k, 1, n}]
TableForm[Table[t[m, n], {m, 1, 10}, {n, 1, 10}]]
Flatten[Table[t[k, n + 1 - k], {n, 1, 12}, {k, 1, n}]]

For odd-numbered rows (m odd):
T(m,n) = (F(m*n+m) + F(m*n) - F(m))/(F(m)*L(m)).
For even-numbered rows (m even):
T(m,n) = (F(m*n+m) - F(m*n) - F(m))/(F(m)*(L(m)-2)).

A165278 Table read by antidiagonals: T(n, k) is the k-th number with n-1 even-indexed Fibonacci numbers in its Zeckendorf representation.

Original entry on oeis.org

2, 5, 1, 7, 3, 4, 13, 6, 9, 12, 15, 8, 11, 25, 33, 18, 10, 17, 30, 67, 88, 20, 14, 22, 32, 80, 177, 232, 34, 16, 24, 46, 85, 211, 465, 609, 36, 19, 27, 59, 87, 224, 554, 1219, 1596, 39, 21, 29, 64, 122, 229, 588, 1452, 3193, 4180, 41, 23, 31, 66, 156, 231, 601

Offset: 1

Clark Kimberling, Sep 13 2009

For n>=0, row n is the monotonic sequence of positive integers m such that the number of even-indexed Fibonacci numbers in the Zeckendorf representation of m is n.
We begin the indexing at 2; that is, 1=F(2), 2=F(3), 3=F(4), 5=F(5),...
Every positive integer occurs exactly once in the array, so that as a sequence it is a permutation of the positive integers.
For counts of odd-indexed Fibonacci numbers, see A165279.
Essentially, (row 0)=A062879, (column 1)=A027941, (column 2)=A069403.

			Northwest corner:
2....5....7...13...15...18...20...34...36...
1....3....6....8...10...14...16...19...20...
4....9...11...17...22...24...27...29...31...
12..25...30...32...46...59...64...66...72...
Examples:
20=13+5+2=F(7)+F(5)+F(3), zero evens, so 20 is in row 0.
19=13+5+1=F(7)+F(5)+F(2), one even, so 19 is in row 1.
22=21+1=F(8)+F(2), two evens, so 22 is in row 2.

Cf. A165276, A165277, A165279.

f[n_] := Module[{i = Ceiling[Log[GoldenRatio, Sqrt[5]*n]], v = {}, m = n}, While[i > 1, If[Fibonacci[i] <= m, AppendTo[v, 1]; m -= Fibonacci[i], If[v != {}, AppendTo[v, 0]]]; i--]; Total[Reverse[v][[1 ;; -1 ;; 2]]]]; T = GatherBy[SortBy[ Range[10^4], f], f]; Table[Table[T[[n - k + 1, k]], {k, n, 1, -1}], {n, 1, Length[T]}] // Flatten (* Amiram Eldar, Feb 04 2020 *)

More terms from Amiram Eldar, Feb 04 2020

A303974 Regular triangle where T(n,k) is the number of aperiodic multisets of size k that fit within some normal multiset of size n.

Original entry on oeis.org

1, 2, 1, 3, 3, 3, 4, 6, 10, 6, 5, 10, 22, 23, 15, 6, 15, 40, 57, 62, 27, 7, 21, 65, 115, 165, 129, 63, 8, 28, 98, 205, 356, 385, 318, 120, 9, 36, 140, 336, 676, 914, 1005, 676, 252, 10, 45, 192, 518, 1176, 1885, 2524, 2334, 1524, 495, 11, 55, 255, 762, 1918, 3528, 5495, 6319, 5607, 3261, 1023

Offset: 1

Gus Wiseman, May 03 2018

A multiset is normal if it spans an initial interval of positive integers. It is aperiodic if its multiplicities are relatively prime.

			Triangle begins:
1
2    1
3    3    3
4    6   10    6
5   10   22   23   15
6   15   40   57   62   27
7   21   65  115  165  129   63
8   28   98  205  356  385  318  120
9   36  140  336  676  914 1005  676  252
The a(4,3) = 10 multisets: (112), (113), (122), (123), (124), (133), (134), (223), (233), (234).
The a(5,4) = 23 multisets:
(1112), (1222),
(1113), (1123), (1223), (1233), (1333), (2223), (2333),
(1124), (1134), (1224), (1234), (1244), (1334), (1344), (2234), (2334), (2344),
(1235), (1245), (1345), (2345).

Andrew Howroyd, Table of n, a(n) for n = 1..1275

Row sums are A303976.

Cf. A000740, A000837, A001597, A007716, A007916, A027941, A178472, A210554, A301700, A303431, A303546, A303551, A303945.

allnorm[n_Integer]:=Function[s,Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1];
Table[Length/@GatherBy[Select[Union@@Rest/@Subsets/@allnorm[n],GCD@@Length/@Split[#]===1&],Length],{n,10}]

T(n,k)={sumdiv(k, d, moebius(k/d)*sum(i=1, d, binomial(d-1, i-1)*binomial(n-k+i, i)))} \\ Andrew Howroyd, Sep 18 2018

T(n,k) = Sum_{d|k} mu(k/d) * Sum_{i=1..d} binomial(d-1, i-1)*binomial(n-k+i, i). - Andrew Howroyd, Sep 18 2018

Terms a(56) and beyond from Andrew Howroyd, Sep 18 2018

A188378 Partial sums of A005248.

Original entry on oeis.org

2, 5, 12, 30, 77, 200, 522, 1365, 3572, 9350, 24477, 64080, 167762, 439205, 1149852, 3010350, 7881197, 20633240, 54018522, 141422325, 370248452, 969323030, 2537720637, 6643838880, 17393796002, 45537549125, 119218851372, 312119004990, 817138163597, 2139295485800

Offset: 0

Gabriele Fici, Mar 29 2011

Different from A024851.

Luo proves that these integers cannot be uniquely decomposed as the sum of distinct and nonconsecutive terms of the Lucas number sequence. - Michel Marcus, Apr 20 2020

Robert Israel, Table of n, a(n) for n = 0..1000
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
David C. Luo, Nonuniqueness Properties of Zeckendorf Related Decompositions, arXiv:2004.08316 [math.NT], 2020.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A002878, A027941, A001654.

Cf. A016945, A267797.

[5*Fibonacci(n)*Fibonacci(n+1)+1+(-1)^n: n in [0..40]]; // Vincenzo Librandi, Jan 24 2016

f:= gfun:-rectoproc({a(n+3)-4*a(n+2)+4*a(n+1)-a(n), a(0) = 2, a(1) = 5, a(2) = 12}, a(n), remember):
map(f, [$0..60]); # Robert Israel, Feb 02 2016

LinearRecurrence[{4,-4,1},{2,5,12},30] (* Harvey P. Dale, Oct 05 2015 *)
Accumulate@ LucasL@ Range[0, 58, 2] (* Michael De Vlieger, Jan 24 2016 *)

a(n) = 5*fibonacci(n)*fibonacci(n+1) + 1 + (-1)^n; \\ Michel Marcus, Aug 26 2013

Vec((-2+3*x)/((x-1)*(x^2-3*x+1)) + O(x^100)) \\ Altug Alkan, Jan 24 2016

a(n) = A000032(2n+1)+1 = A002878(n)+1 = 2*A027941(n+1)-3*A027941(n).
G.f.: ( -2+3*x ) / ( (x-1)*(x^2-3*x+1) ). - R. J. Mathar, Mar 30 2011
a(n) = 5*A001654(n) + 1 + (-1)^n, n>=0. [Wolfdieter Lang, Jul 23 2012]
(a(n)^3 + (a(n)-2)^3) / 2 = A000032(A016945(n)) = Lucas(6*n+3) = A267797(n), for n>0. - Altug Alkan, Jan 31 2016
a(n) = 2^(-1-n)*(2^(1+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n). - Colin Barker, Nov 02 2016

A217472 Coefficient table for polynomials used for the formula of partial sums of odd powers of even-indexed Fibonacci numbers.

Original entry on oeis.org

1, -3, 1, 25, -15, 4, -553, 455, -224, 44, 32220, -32664, 22500, -8316, 1276, -4934996, 5825600, -5028452, 2640220, -771980, 96976, 1985306180, -2636260484, 2688531560, -1791505144, 751934040, -181539072, 19298224, -2096543510160, 3060180107600, -3555908800752, 2830338574800, -1521052125120, 530958146400, -109131456720, 10054374704

Offset: 0

Wolfdieter Lang, Oct 12 2012

The following formula is due to Ozeki (see the reference, Theorem 2, p. 109) and also to Prodinger (see the reference, p. 207). Here the version of Prodinger is given which coincides with the one of Ozeki (up to a misprint P instead of 1 in the latter).
  sum(F(2*k)^(2*m+1),k=0..n) =  sum(lambda(m,l)*F(2*n+1)^(2*l+1),l=0..m) + C(m), m>=0, n>= 0, with F=A000045 (Fibonacci), L=A000032 (Lucas),
  lambda(m,l) = (-5)^(l-m)* sum(binomial(2*m+1,j)*binomial(m-j+l,m-j-l)*
(2*(m-j)+1)/L(2*(m-j)+1) ,j=0..m-l)/(2*l+1) and
  C(m) = (1/5^m)*sum((-1)^(j-1)* binomial(2*m+1,j)*F(2*(m-j)+1)/L(2*(m-j)+1),j=0..m).
In order to have an integer triangle T(m,l) instead of the rational lambda(m,l) one uses the sequence pL(m) = product(L(2*i+1),i=0..m), m >= 0, given in A217473, with T(m,l) = pL(m)*lambda(m,l). Similarly, c(m) = pL(m)*C(m) gives the integer sequence A217474 = [-1, 2, -14, 278, -15016, 2172632, -835765304, 851104689248, ...].
Thus, pL(m)*sum(F(2*k)^(2*m+1),k=0..n) =  sum(T(m,l)*F(2*n+1)^(2*l+1),l=0..m) + c(m), m >= 0, n >= 0.
For Melham's conjecture on  pL(m)*sum(F(2*k)^(2*m+1),k=0..n) see A217475 where also the reference is given.

			The triangle T(m,l) begins:
m\l        0        1         2        3        4      5  ...
0:         1
1:        -3        1
2:        25      -15         4
3:      -553      455      -224       44
4:     32220   -32664     22500    -8316     1276
5:  -4934996  5825600  -5028452  2640220  -771980  96976
...
row 6:  1985306180   -2636260484   2688531560   -1791505144   751934040   -181539072    19298224.
row 7: -2096543510160  3060180107600 -3555908800752 2830338574800  -1521052125120  530958146400  -109131456720 10054374704.
m=0: 1*sum(F(2*k)^1,k=0..n) = 1*F(2*n+1)^1  - 1, the last term comes from c(0) = A217474 = -1. See A027941.
m=1: 1*4*sum(F(2*k)^3,k=0..n) = -3*F(2*n+1)^1 +1*F(2*n+1)^3  +  2. See 4*A163198.
m=2: 1*4*11*sum(F(2*k)^5,k=0..n) = 25*F(2*n+1)^1 - 15*F(2*n+1)^3 + 4*F(2*n+1)^5 - 14. See 44*A217471.

K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
H. Prodinger, On a sum of Melham and its variants, The Fibonacci Quart. 46/47 (2008/2009), no. 3, 207-215.

Cf. A217474, A217475.

T(m,l) = pL(m)*lambda(m,l), m >= 0, l = 0..m, with pL(m) = A217473(m) and lambda(m,l) given in a comment above.

A303976 Number of different aperiodic multisets that fit within some normal multiset of size n.

Original entry on oeis.org

1, 3, 9, 26, 75, 207, 565, 1518, 4044, 10703, 28234, 74277, 195103, 511902, 1342147, 3517239, 9214412, 24134528, 63204417, 165505811, 433361425, 1134664831, 2970787794, 7777975396, 20363634815, 53313819160, 139579420528, 365427311171, 956707667616, 2504704955181

Offset: 1

Gus Wiseman, May 03 2018

nonn

A multiset is normal if it spans an initial interval of positive integers. It is aperiodic if its multiplicities are relatively prime.

			The a(4) = 26 aperiodic multisets:
(1), (2), (3), (4),
(12), (13), (14), (23), (24), (34),
(112), (113), (122), (123), (124), (133), (134), (223), (233), (234),
(1112), (1123), (1222), (1223), (1233), (1234).

Andrew Howroyd, Table of n, a(n) for n = 1..500

Row sums of A303974.

Cf. A000740, A000837, A007916, A027941, A178472, A210554, A301700, A303431, A303546, A303551, A303945.

allnorm[n_Integer]:=Function[s,Array[Count[s,y_/;y<=#]+1&,n]]/@Subsets[Range[n-1]+1];
Table[Length[Select[Union@@Rest/@Subsets/@allnorm[n],GCD@@Length/@Split[#]===1&]],{n,10}]

seq(n)={Vec(sum(d=1, n, moebius(d)*x^d/(1 - x - x^d*(2-x)) + O(x*x^n))/(1-x))} \\ Andrew Howroyd, Feb 04 2021

a(n) = Sum_{k=1..n} Sum_{d|k} mu(k/d) * Sum_{i=1..d} binomial(d-1, i-1)*binomial(n-k+i, i). - Andrew Howroyd, Sep 18 2018

G.f.: Sum_{d>=1} mu(d)*x^d/((1 - x - x^d*(2-x))*(1-x)). - Andrew Howroyd, Feb 04 2021

Terms a(13) and beyond from Andrew Howroyd, Sep 18 2018

A155110 a(n) = 8*Fibonacci(2n+1).

Original entry on oeis.org

8, 16, 40, 104, 272, 712, 1864, 4880, 12776, 33448, 87568, 229256, 600200, 1571344, 4113832, 10770152, 28196624, 73819720, 193262536, 505967888, 1324641128, 3467955496, 9079225360, 23769720584, 62229936392, 162920088592, 426530329384, 1116670899560

Offset: 0

Paul Curtz, Jan 20 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (3,-1).

Cf. A001519, A027941, A099496, A122367, A153873, A154811, A156551.

[8*Fibonacci(2*n+1): n in [0..30]]; // Vincenzo Librandi, Aug 07 2011

A155110 := proc(n) 8*combinat[fibonacci](2*n+1) ; end: seq(A155110(n),n=0..50) ; # R. J. Mathar, Oct 06 2009

8*Fibonacci[2*Range[0,30]+1] (* G. C. Greubel, Apr 21 2021 *)

[8*fibonacci(2*n+1) for n in (0..30)] # G. C. Greubel, Apr 21 2021

a(n) = 8*A001519(n+1) = 8*A122367(n) = 8 *|A099496(n)|.
a(n) == A154811(n+6) (mod 9).
a(n) == A156551(n) (mod 10).
a(n) = A153873(n) - A027941(n).
G.f.: 8*(1 - x)/(1 - 3*x + x^2). - G. C. Greubel, Apr 21 2021

Comments converted to formulas by R. J. Mathar, Oct 06 2009

A203169 Sum of the fourth powers of the first n even-indexed Fibonacci numbers.

Original entry on oeis.org

0, 1, 82, 4178, 198659, 9349284, 439330980, 20639983621, 969645224182, 45552722051318, 2140008541351943, 100534850436141384, 4722997973709689160, 221880369994471370761, 10423654392318557192602, 489689876072761951752602

Offset: 0

Stuart Clary, Dec 30 2011

Natural bilateral extension (brackets mark index 0): ..., -9349284, -198659, -4178, -82, -1, 0, [0], 1, 82, 4178, 198659, 9349284, ... That is, a(-n) = -a(n-1).

Index entries for linear recurrences with constant coefficients, signature (56,-440,770,-440,56,-1).

Cf. A203170, A203171, A203172.

Cf. A027941, A103434, A163198.

a[n_Integer] := (1/75)(Fibonacci[8n+4] - 12*Fibonacci[4n+2] + 9*(2*n+1)); Table[a[n], {n, 0, 20}]

Let F(n) be the Fibonacci number A000045(n).
a(n) = sum_{k=1..n} F(2k)^4.
Closed form: a(n) = (1/75)(F(8n+4) - 12 F(4n+2) + 9(2 n + 1)).
Recurrence: a(n) - 56 a(n-1) + 440 a(n-2) - 770 a(n-3) + 440 a(n-4) - 56 a(n-5) + a(n-6) = 0.
G.f.: A(x) = (x + 26 x^2 + 26 x^3 + x^4)/(1 - 56 x + 440 x^2 - 770 x^3 + 440 x^4 - 56 x^5 + x^6) = x(1 + x)(1 + 25 x + x^2)/((1 - x)^2 (1 - 7 x + x^2)(1 - 47 x + x^2)).

cp's OEIS Frontend

A210554 Triangle of coefficients of polynomials v(n,x) jointly generated with A208341; see the Formula section.

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A264236 Number of vertices at level n of the hyperbolic Pascal pyramid.

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A214984 Array: T(m,n) = (F(m) + F(2*m) + ... + F(n*m))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).

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A165278 Table read by antidiagonals: T(n, k) is the k-th number with n-1 even-indexed Fibonacci numbers in its Zeckendorf representation.

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Mathematica

Extensions

A303974 Regular triangle where T(n,k) is the number of aperiodic multisets of size k that fit within some normal multiset of size n.

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Mathematica

PARI

Formula

Extensions

A188378 Partial sums of A005248.

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Magma

Maple

Mathematica

PARI

PARI

Formula

A217472 Coefficient table for polynomials used for the formula of partial sums of odd powers of even-indexed Fibonacci numbers.

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A214984 Array: T(m,n) = (F(m) + F(2m) + ... + F(nm))/F(m), by antidiagonals, where F = A000045 (Fibonacci numbers).