cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 21-29 of 29 results.

A022105 Fibonacci sequence beginning 1, 15.

Original entry on oeis.org

1, 15, 16, 31, 47, 78, 125, 203, 328, 531, 859, 1390, 2249, 3639, 5888, 9527, 15415, 24942, 40357, 65299, 105656, 170955, 276611, 447566, 724177, 1171743, 1895920, 3067663, 4963583, 8031246, 12994829
Offset: 0

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Author

Keywords

Comments

a(n-1)=sum(P(15;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=14. These are the SW-NE diagonals in P(15;n,k), the (15,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Crossrefs

a(n) = A109754(14, n+1).
a(k) = A118654(4, k).

Programs

  • Magma
    a0:=1; a1:=15; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
  • Mathematica
    a={};b=1;c=15;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,12,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 23 2008 *)
    LinearRecurrence[{1,1},{1,15},40] (* Harvey P. Dale, Oct 11 2015 *)

Formula

a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=15. a(-1):=14.
G.f.: (1+14*x)/(1-x-x^2).
a(n) = A101220(14,0,n+1). - Ross La Haye, May 02 2006

A022107 Fibonacci sequence beginning 1, 17.

Original entry on oeis.org

1, 17, 18, 35, 53, 88, 141, 229, 370, 599, 969, 1568, 2537, 4105, 6642, 10747, 17389, 28136, 45525, 73661, 119186, 192847, 312033, 504880, 816913, 1321793, 2138706, 3460499, 5599205, 9059704, 14658909
Offset: 0

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Author

Keywords

Comments

a(n-1)=sum(P(17;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=16. These are the SW-NE diagonals in P(17;n,k), the (17,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Crossrefs

a(n) = A109754(16, n+1) = A101220(16, 0, n+1).

Programs

  • Magma
    a0:=1; a1:=17; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
  • Mathematica
    a={};b=1;c=17;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,12,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 23 2008 *)
    LinearRecurrence[{1,1},{1,17},40] (* Harvey P. Dale, Aug 04 2017 *)

Formula

a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=17. a(-1):=16.
G.f.: (1+16*x)/(1-x-x^2).

A022108 Fibonacci sequence beginning 1, 18.

Original entry on oeis.org

1, 18, 19, 37, 56, 93, 149, 242, 391, 633, 1024, 1657, 2681, 4338, 7019, 11357, 18376, 29733, 48109, 77842, 125951, 203793, 329744, 533537, 863281, 1396818, 2260099, 3656917, 5917016, 9573933, 15490949
Offset: 0

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Author

Keywords

Comments

a(n-1)=sum(P(18;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=17. These are the SW-NE diagonals in P(18;n,k), the (18,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Crossrefs

a(n) = A109754(17, n+1) = A101220(17, 0, n+1).

Programs

  • Magma
    a0:=1; a1:=18; [GeneralizedFibonacciNumber(a0, a1, n): n in [0..30]]; // Bruno Berselli, Feb 12 2013
  • Mathematica
    a={};b=1;c=18;AppendTo[a,b];AppendTo[a,c];Do[b=b+c;AppendTo[a,b];c=b+c;AppendTo[a,c],{n,1,12,1}];a (* Vladimir Joseph Stephan Orlovsky, Jul 23 2008 *)
    LinearRecurrence[{1,1},{1,18},40] (* Harvey P. Dale, Apr 15 2018 *)

Formula

a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=18. a(-1):=17.
G.f.: (1+17*x)/(1-x-x^2).

A022391 Fibonacci sequence beginning 1, 21.

Original entry on oeis.org

1, 21, 22, 43, 65, 108, 173, 281, 454, 735, 1189, 1924, 3113, 5037, 8150, 13187, 21337, 34524, 55861, 90385, 146246, 236631, 382877, 619508, 1002385, 1621893, 2624278, 4246171, 6870449, 11116620, 17987069, 29103689, 47090758, 76194447, 123285205, 199479652, 322764857, 522244509
Offset: 0

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Author

Keywords

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(21;n-1-k,k), n>=1, with a(-1)=20. These are the SW-NE diagonals in P(21;n,k), the (21,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Programs

  • Magma
    [Fibonacci(n+2) + 19*Fibonacci(n): n in [0..50]]; // G. C. Greubel, Mar 02 2018
  • Mathematica
    LinearRecurrence[{1, 1}, {1, 21}, 30] (* Jean-François Alcover, Feb 25 2018 *)
    Table[Fibonacci[n + 2] + 19*Fibonacci[n], {n, 0, 50}] (* G. C. Greubel, Mar 02 2018 *)
  • PARI
    for(n=0, 50, print1(fibonacci(n+2) + 19*fibonacci(n), ", ")) \\ G. C. Greubel, Mar 02 2018
    

Formula

a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=21. a(-1):=20.
G.f.: (1+20*x)/(1-x-x^2).

Extensions

Terms a(30) onward added by G. C. Greubel, Mar 02 2018

A056125 a(n) = (5*n + 4)*binomial(n+7,7)/4.

Original entry on oeis.org

1, 18, 126, 570, 1980, 5742, 14586, 33462, 70785, 140140, 262548, 469404, 806208, 1337220, 2151180, 3368244, 5148297, 7700814, 11296450, 16280550, 23088780, 32265090, 44482230, 60565050, 81516825, 108548856, 143113608, 186941656
Offset: 0

Views

Author

Barry E. Williams, Jul 07 2000

Keywords

Crossrefs

Cf. A052254.
Cf. A093645 ((10, 1) Pascal, column m=8).
Partial sums of A052254.

Programs

  • GAP
    List([0..30], n-> (5*n+4)*Binomial(n+7,7)/4 ); # G. C. Greubel, Jan 19 2020
  • Magma
    [(5*n+4)*Binomial(n+7,7)/4: n in [0..30]]; // G. C. Greubel, Jan 19 2020
    
  • Maple
    seq( (5*n+4)*binomial(n+7,7)/4, n=0..30); # G. C. Greubel, Jan 19 2020
  • Mathematica
    Table[((5n+4)Binomial[n+7,7])/4,{n,0,30}] (* or *) LinearRecurrence[{9,-36,84, -126,126,-84,36,-9,1},{1,18,126,570,1980,5742,14586,33462,70785},30] (* Harvey P. Dale, Jan 18 2013 *)
  • PARI
    vector(31, n, (5*n-1)*binomial(n+6,7)/4 ) \\ G. C. Greubel, Jan 19 2020
    
  • Sage
    [(5*n+4)*binomial(n+7,7)/4 for n in (0..30)] # G. C. Greubel, Jan 19 2020
    

Formula

G.f.: (1+9*x)/(1-x)^9.
a(0)=1, a(1)=18, a(2)=126, a(3)=570, a(4)=1980, a(5)=5742, a(6)=14586, a(7)=33462, a(8)=70785, a(n) = 9*a(n-1) -36*a(n-2) +84*a(n-3) -126*a(n-4) + 126*a(n-5) -84*a(n-6) +36*a(n-7) -9*a(n-8) +a(n-9). - Harvey P. Dale, Jan 18 2013
From G. C. Greubel, Jan 19 2020: (Start)
a(n) = 10*binomial(n+8,8) - 9*binomial(n+7,7).
E.g.f.: (20160 + 342720*x + 917280*x^2 + 823200*x^3 + 323400*x^4 + 62328*x^5 + 6076*x^6 + 284*x^7 + 5*x^8)*exp(x)/20160. (End)

A022392 Fibonacci sequence beginning 1, 22.

Original entry on oeis.org

1, 22, 23, 45, 68, 113, 181, 294, 475, 769, 1244, 2013, 3257, 5270, 8527, 13797, 22324, 36121, 58445, 94566, 153011, 247577, 400588, 648165, 1048753, 1696918, 2745671, 4442589, 7188260, 11630849, 18819109, 30449958, 49269067, 79719025, 128988092, 208707117, 337695209, 546402326
Offset: 0

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Author

Keywords

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(22;n-1-k,k), n>=1, with a(-1)=21. These are the SW-NE diagonals in P(22;n,k), the (22,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Programs

  • Magma
    [Fibonacci(n+2) + 20*Fibonacci(n): n in [0..50]]; // G. C. Greubel, Mar 02 2018
  • Mathematica
    Table[Fibonacci[n + 2] + 20*Fibonacci[n], {n, 0, 50}] (* or *) LinearRecurrence[{1,1}, {1,22}, 50] (* G. C. Greubel, Mar 02 2018 *)
  • PARI
    for(n=0, 50, print1(fibonacci(n+2) + 20*fibonacci(n), ", ")) \\ G. C. Greubel, Mar 02 2018
    

Formula

a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=22. a(-1):=21.
G.f.: (1+21*x)/(1-x-x^2).

Extensions

Terms a(30) onward added by G. C. Greubel, Mar 02 2018

A022393 Fibonacci sequence beginning 1, 23.

Original entry on oeis.org

1, 23, 24, 47, 71, 118, 189, 307, 496, 803, 1299, 2102, 3401, 5503, 8904, 14407, 23311, 37718, 61029, 98747, 159776, 258523, 418299, 676822, 1095121, 1771943, 2867064, 4639007, 7506071, 12145078, 19651149, 31796227, 51447376, 83243603, 134690979, 217934582, 352625561, 570560143
Offset: 0

Views

Author

Keywords

Comments

a(n-1) = Sum_{k=0..ceiling((n-1)/2)} P(23;n-1-k,k), n>=1, with a(-1)=22. These are the SW-NE diagonals in P(23;n,k), the (23,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.

Programs

  • GAP
    List([0..40],n->Fibonacci(n+2)+21*Fibonacci(n)); # Muniru A Asiru, Mar 03 2018
  • Magma
    [Fibonacci(n+2) + 21*Fibonacci(n): n in [0..50]]; // G. C. Greubel, Mar 02 2018
    
  • Mathematica
    a[1]=1; a[2]=23; a[n_]:=a[n]=a[n - 1]+a[n - 2] (*  José María Grau Ribas, Feb 15 2010 *)
    LinearRecurrence[{1,1},{1,23},30] (* Harvey P. Dale, Sep 30 2011 *)
    Table[Fibonacci[n + 2] + 21*Fibonacci[n], {n, 0, 50}] (* G. C. Greubel, Mar 02 2018 *)
  • PARI
    for(n=0, 50, print1(fibonacci(n+2) + 21*fibonacci(n), ", ")) \\ G. C. Greubel, Mar 02 2018
    

Formula

a(n) = a(n-1) + a(n-2), n>=2, a(0)=1, a(1)=23. a(-1):=22.
G.f.: (1+22*x)/(1-x-x^2).

Extensions

Terms a(30) onward added by G. C. Greubel, Mar 02 2018

A093646 Higher dimensional figurate numbers based on 12-gonal numbers A051624.

Original entry on oeis.org

1, 19, 145, 715, 2695, 8437, 23023, 56485, 127270, 267410, 529958, 999362, 1805570, 3142790, 5293970, 8662214, 13810511, 21511325, 32807775, 49088325, 72177105, 104442195, 148924425, 209489475, 291006300, 399555156, 542668764, 729610420
Offset: 0

Views

Author

Wolfdieter Lang, Apr 22 2004

Keywords

Crossrefs

Cf. A093645 ((10, 1) Pascal, column m=9).

Formula

a(n)= (10*n+9)*binomial(n+8, 8)/9. G.f.: (1+9*x)/(1-x)^10.

A180053 a(1)=1, a(2)=101, a(n) = 1001*a(n-1) for n > 2.

Original entry on oeis.org

1, 101, 101101, 101202101, 101303303101, 101404606404101, 101506011010505101, 101607517021515606101, 101709124538537121707101, 101810833663075658828808101, 101912644496738734487636909101, 102014557141235473222124546010101
Offset: 1

Views

Author

Mark Dols, Aug 08 2010

Keywords

Examples

			In a column:
                1
              101
           101101
        101202101
     101303303101
  101404606404101
		

Crossrefs

Cf. A093645.

Programs

Formula

G.f.: x(1-900*x)/(1-1001*x). - Philippe Deléham, Oct 08 2011
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