A162741 -id:A162741 - cp's OEIS frontend

A114197 A Pascal-Fibonacci triangle.

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 13, 13, 5, 1, 1, 6, 21, 31, 21, 6, 1, 1, 7, 31, 61, 61, 31, 7, 1, 1, 8, 43, 106, 142, 106, 43, 8, 1, 1, 9, 57, 169, 286, 286, 169, 57, 9, 1, 1, 10, 73, 253, 520, 659, 520, 253, 73, 10, 1

Offset: 0

Paul Barry, Nov 16 2005

T(2n,n) is A114198. Row sums are A114199. Row sums of inverse are 0^n.

			Triangle begins
  1;
  1,   1;
  1,   2,   1;
  1,   3,   3,   1;
  1,   4,   7,   4,   1;
  1,   5,  13,  13,   5,   1;
  1,   6,  21,  31,  21,   6,   1;
  1,   7,  31,  61,  61,  31,   7,   1;
  1,   8,  43, 106, 142, 106,  43,   8,   1;

Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A114197, A162741, A228074.

As a number triangle, T(n,k) = Sum_{j=0..n-k} C(n-k, j)C(k, j)F(j);
As a number triangle, T(n,k) = Sum_{j=0..n} C(n-k, n-j)C(k, j-k)F(j-k);
As a number triangle, T(n,k) = Sum_{j=0..n} C(k, j)C(n-k, n-j)F(k-j) if k <= n, 0 otherwise.
As a square array, T(n,k) = Sum_{j=0..n} C(n, j)C(k, j)F(j);
As a square array, T(n,k) = Sum_{j=0..n+k} C(n, n+k-j)C(k, j-k)F(j-k);
Column k has g.f.: (Sum_{j=0..k} C(k, j)F(j+1)(x/(1-x))^j)*x^k/(1-x);
G.f.: -((x^2-x)*y-x+1)/((x^4+x^3-x^2)*y^2+(x^3-3*x^2+2*x)*y-x^2+2*x-1). - Vladimir Kruchinin, Jan 15 2018

A109906 A triangle based on A000045 and Pascal's triangle: T(n,m) = Fibonacci(n-m+1) * Fibonacci(m+1) * binomial(n,m).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 6, 6, 3, 5, 12, 24, 12, 5, 8, 25, 60, 60, 25, 8, 13, 48, 150, 180, 150, 48, 13, 21, 91, 336, 525, 525, 336, 91, 21, 34, 168, 728, 1344, 1750, 1344, 728, 168, 34, 55, 306, 1512, 3276, 5040, 5040, 3276, 1512, 306, 55, 89, 550, 3060, 7560, 13650, 16128, 13650, 7560, 3060, 550, 89

Offset: 0

Roger L. Bagula and Gary W. Adamson, Aug 24 2008

Row sums give A081057.

			Triangle T(n,k) begins:
   1;
   1,   1;
   2,   2,    2;
   3,   6,    6,    3;
   5,  12,   24,   12,     5;
   8,  25,   60,   60,    25,     8;
  13,  48,  150,  180,   150,    48,    13;
  21,  91,  336,  525,   525,   336,    91,   21;
  34, 168,  728, 1344,  1750,  1344,   728,  168,   34;
  55, 306, 1512, 3276,  5040,  5040,  3276, 1512,  306,  55;
  89, 550, 3060, 7560, 13650, 16128, 13650, 7560, 3060, 550, 89;
  ...

Reinhard Zumkeller, Rows n = 0..120 of table, flattened
Peter McCalla, Asamoah Nkwanta, Catalan and Motzkin Integral Representations, arXiv:1901.07092 [math.NT], 2019.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A141611, A141617, A000045, A081057.

Some other Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A111006, A114197, A162741, A228074.

a109906 n k = a109906_tabl !! n !! k
a109906_row n = a109906_tabl !! n
a109906_tabl = zipWith (zipWith (*)) a058071_tabl a007318_tabl
-- Reinhard Zumkeller, Aug 15 2013

f:= n-> combinat[fibonacci](n+1):
T:= (n, k)-> binomial(n, k)*f(k)*f(n-k):
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 26 2023

Clear[t, n, m] t[n_, m_] := Fibonacci[(n - m + 1)]*Fibonacci[(m + 1)]*Binomial[n, m]; Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}]; Flatten[%]

T(n,m) = Fibonacci(n-m+1)*Fibonacci(m+1)*binomial(n,m).

T(n,k) = A058071(n,k) * A007318(n,k). - Reinhard Zumkeller, Aug 15 2013

Offset changed by Reinhard Zumkeller, Aug 15 2013

A162849 Pairs of numbers that add up to the 'backward decimal expansion' of fraction 1/109 and whose difference is the 'backward decimal expansion' of fraction 1/89.

Original entry on oeis.org

0, 1, 10, 101, 2010, 10201, 303010, 1040201, 40703010, 107050201, 5140803010, 11112050201, 625200803010, 1162613050201, 74146210803010, 122513313050201, 8639754210803010, 12992793413050201, 993903355210803010

Offset: 1

Mark Dols, Jul 14 2009

Sum of pairs also (consecutive) cumulative sum of 110^n (or numerators of 1/110^1 + 1/110^2 + ... + 1/110^n, representing fraction 1/109).

Difference of pairs also cumulative sum of 90^n (or numerators of 1/90^1 + 1/90^2 + ... + 1/90^n, representing fraction 1/89).

			In pairs:
           0,           1;
          10,         101;
        2010,       10201;
      303010,     1040201;
    40703010,   107050201;
  5140803010, 11112050201;

Index entries for linear recurrences with constant coefficients, signature (0,201,0,-10100,0,9900).

Cf. A162741, A161999, A007318, A000045.

For n even: a(n) = 100*a(n-2)+10*a(n-1), for n odd: a(n) = 100*a(n-2)+10*a(n-3)+1; with a(0)=0, a(1)=1.
From R. J. Mathar, Feb 11 2010: (Start)
a(n) = 201*a(n-2) - 10100*a(n-4) + 9900*a(n-6).
G.f.: x^2*(-1-10*x+100*x^2)/((x-1)*(1+x)*(90*x^2-1)*(110*x^2-1)). (End)

More terms from R. J. Mathar, Feb 11 2010

A165155 a(n) = 100*a(n-1) + 11^(n-1) for n>0, a(0)=0.

Original entry on oeis.org

0, 1, 111, 11221, 1123431, 112357741, 11235935151, 1123595286661, 112359548153271, 11235955029685981, 1123595505326545791, 112359550558592003701, 11235955056144512040711, 1123595505617589632447821, 112359550561793485956926031, 11235955056179728345526186341

Offset: 0

Mark Dols, Sep 05 2009

Generalization of A000225. - Mark Dols, Jan 28 2010

			From _Mark Dols_, Jan 28 2010: (Start)
As triangle:
  ........... 1
  ......... 1 1 1
  ....... 1 1 2 2 1
  ..... 1 1 2 3 4 3 1
  ... 1 1 2 3 5 7 7 4 1
  . 1 1 2 3 5 9 3 5 1 5 1
  1 1 2 3 5 9 5 2 8 6 6 6 1
(Mirrored version of A162741) (End)

G. C. Greubel, Table of n, a(n) for n = 0..495
Index entries for linear recurrences with constant coefficients, signature (111,-1100).

Cf. A000225, A021093, A094704, A162741, A164913.

[(1/89)*(100^n-11^n): n in [0..40]] // Vincenzo Librandi, Dec 05 2010

RecurrenceTable[{a[0]==0,a[n]==100a[n-1]+11^(n-1)},a,{n,40}] (* Harvey P. Dale, Feb 20 2016 *)

[(10^(2*n) - 11^n)/89 for n in range(41)] # G. C. Greubel, Feb 09 2023

G.f.: x/((1-11*x)*(1-100*x)). - R. J. Mathar, Nov 02 2016

E.g.f.: (1/89)*(exp(100*x) - exp(11*x)). - G. C. Greubel, Feb 09 2023

a(0) prepended by Bruno Berselli, Oct 02 2015

A172162 a(n) = ( A165154(n) + A165155(n) )/2.

Original entry on oeis.org

0, 1, 101, 10201, 1020401, 102050701, 10205121101, 1020513261601, 102051333512201, 10205133479922901, 1020513348977553701, 102051334912467474601, 10205133491373712765601, 1020513349139081705516701, 102051334913924160974827901, 10205133491392617410795809201

Offset: 0

Mark Dols, Jan 27 2010

Colin Barker, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (102,-101,-9900).

Cf. A162741, A162849, A165154, A165155, A172163.

Table[99 100^n/9701 - 11^n/178 - (-9)^n/218, {n, 0, 20}] (* Bruno Berselli, Oct 02 2015 *)

concat(0, Vec(-x*(x-1)/((9*x+1)*(11*x-1)*(100*x-1)) + O(x^30))) \\ Colin Barker, Oct 02 2015

[(-89*(-9)^n - 109*11^n + 198*10^(2*n))/19402 for n in (0..50)] # G. C. Greubel, Apr 24 2022

a(n) = 99*100^n/9701 - 11^n/178 - (-9)^n/218. [Bruno Berselli, Oct 02 2015]
From Colin Barker, Oct 02 2015: (Start)
a(n) = 102*a(n-1) - 101*a(n-2) - 9900*a(n-3) for n>3.
G.f.: x*(1-x) / ((1+9*x)*(1-11*x)*(1-100*x)).
(End)

a(0) and more terms added by Bruno Berselli, Oct 02 2015

A172163 a(n) = ( A165155(n) - A165154(n) )/2.

Original entry on oeis.org

0, 0, 10, 1020, 103030, 10307040, 1030814050, 103082025060, 10308214641070, 1030821549763080, 103082156348992090, 10308215646124529100, 1030821564770799275110, 103082156478507926931120, 10308215647869324982098130, 1030821564787110934730377140

Offset: 0

Mark Dols, Jan 27 2010

Colin Barker, Table of n, a(n) for n = 0..501
Index entries for linear recurrences with constant coefficients, signature (102,-101,-9900).

Cf. A162741, A162849, A165154, A165155, A172162.

Table[10^(2 n + 1)/9701 - 11^n/178 + (-9)^n/218, {n, 0, 20}] (* Bruno Berselli, Oct 02 2015 *)
LinearRecurrence[{102,-101,-9900},{0,0,10},20] (* Harvey P. Dale, Aug 17 2021 *)

concat([0,0], Vec(10*x^2/((9*x+1)*(11*x-1)*(100*x-1)) + O(x^30))) \\ Colin Barker, Oct 02 2015

[(89*(-9)^n + 2*10^(2*n+1) - 109*11^n)/19402 for n in (0..50)] # G. C. Greubel, Apr 24 2022

a(n) = 10^(2*n+1)/9701 - 11^n/178 + (-9)^n/218. [Bruno Berselli, Oct 02 2015]
From Colin Barker, Oct 02 2015: (Start)
a(n) = 102*a(n-1) - 101*a(n-2) - 9900*a(n-3) for n>2.
G.f.: 10*x^2 / ((1+9*x)*(1-11*x)*(1-100*x)).
(End)

a(0)=0 and more terms added by Bruno Berselli, Oct 02 2015

A208245 Triangle read by rows: a(n,k) = a(n-2,k) + a(n-2,k-1).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 2, 1, 1, 1, 3, 4, 3, 2, 1, 1, 1, 4, 6, 5, 3, 2, 1, 1, 1, 4, 7, 7, 5, 3, 2, 1, 1, 1, 5, 10, 11, 8, 5, 3, 2, 1, 1, 1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1, 1, 6, 15, 21, 19, 13, 8, 5, 3, 2, 1, 1, 1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1

Offset: 1

Richard R. Forberg, Apr 22 2013

Sum of terms in each row are given by sequence A052955.
Columns (at constant k) converge toward Fibonacci starting first from high value of k).
First seven rows are same as A008242. The odd numbered rows of this sequence equal the rows of A123736.  Also it has some similarities to A162741.
Columns (constant k), prior to convergence to Fibonacci, appear as various other sequences (e.g. k = 4, is sequence A055803, with other columns in same referenced family).

			The first 13 rows are (as above) where n is the row index:
1
1, 1
1, 1, 1
1, 2, 1, 1
1, 2, 2, 1, 1
1, 3, 3, 2, 1, 1
1, 3, 4, 3, 2, 1, 1
1, 4, 6, 5, 3, 2, 1, 1
1, 4, 7, 7, 5, 3, 2, 1, 1
1, 5, 10, 11, 8, 5, 3, 2, 1, 1
1, 5, 11, 14, 12, 8, 5, 3, 2, 1, 1
1, 6, 15, 21, 19, 13, 8, 5, 3, 2, 1, 1
1, 6, 16, 25, 26, 20, 13, 8, 5, 3, 2, 1, 1,

Reinhard Zumkeller, Rows n = 1..100 of triangle, flattened

Cf. A000045 (central terms).

a208245 n k = a208245_tabl !! (n-1) !! (k-1)
a208245_row n = a208245_tabl !! (n-1)
a208245_tabl = map fst $ iterate f ([1], [1, 1]) where
   f (us, vs) = (vs, zipWith (+) ([0] ++ us ++ [0]) (us ++ [0, 1]))
-- Reinhard Zumkeller, Jul 28 2013

a(n,k) = a(n-2,k) + a(n-2,k-1); if n=k or k=1 then a(n,k)=1; if n

A261507 Fibonacci-numbered rows of Pascal's triangle. Triangle read by rows: T(n,k)= binomial(Fibonacci(n), k).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 10, 10, 5, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1, 1, 21, 210, 1330, 5985, 20349, 54264, 116280, 203490, 293930, 352716, 352716, 293930, 203490, 116280, 54264, 20349, 5985, 1330, 210, 21, 1

Offset: 0

Maghraoui Abdelkader, Aug 22 2015

Subsequence of A007318.

			1,
1,  1,
1,  1,
1,  2,  1,
1,  3,  3,   1,
1,  5, 10,  10,   5,    1,
1,  8, 28,  56,  70,   56,   28,    8,    1,
1, 13, 78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13, 1

Jean-François Alcover, Table of n, a(n) for n = 0..388 (corrected by _N. J. A. Sloane_, Jan 18 2019)

Cf. A000045, A007318.

Cf. A036355, A037027, A228074, A162741, A034871, A034870, A191797.

Table[Binomial[Fibonacci[n], k], {n, 0, 8}, {k, 0, Fibonacci[n]}]//Flatten (* Jean-François Alcover, Nov 12 2015*)

v = vector(101,j,fibonacci(j)); i=0; n=0; while(n<100, for(k=0, n, print1(binomial(n, k), ", ","")); print(); i=i+1; n=v[i] ;)

T(n, k) = binomial(fibonacci(n), k).
T(n, 1) = fibonacci(n) = A000045(n).
T(n, 2) = A191797(n) for n>3.

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cp's OEIS Frontend

A114197 A Pascal-Fibonacci triangle.

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A109906 A triangle based on A000045 and Pascal's triangle: T(n,m) = Fibonacci(n-m+1) * Fibonacci(m+1) * binomial(n,m).

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A162849 Pairs of numbers that add up to the 'backward decimal expansion' of fraction 1/109 and whose difference is the 'backward decimal expansion' of fraction 1/89.

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A165155 a(n) = 100*a(n-1) + 11^(n-1) for n>0, a(0)=0.

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A172162 a(n) = ( A165154(n) + A165155(n) )/2.

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A172163 a(n) = ( A165155(n) - A165154(n) )/2.

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A208245 Triangle read by rows: a(n,k) = a(n-2,k) + a(n-2,k-1).

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