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.

Showing 1-10 of 18 results. Next

A163267 Partial sums of A118977.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 10, 13, 14, 16, 19, 22, 25, 30, 36, 40, 41, 43, 46, 49, 52, 57, 63, 67, 70, 75, 81, 87, 95, 106, 116, 121, 122, 124, 127, 130, 133, 138, 144, 148, 151, 156, 162, 168, 176, 187, 197, 202, 205, 210, 216, 222, 230, 241, 251, 258, 266, 277, 289, 303, 322, 343, 358
Offset: 0

Views

Author

Gary W. Adamson, Jul 24 2009

Keywords

Comments

Can be considered toothpick sequence for N=1.
Based on a consistent set of rules for generating toothpick sequences.
Cf. A139250 where (1, 1, 3, 1, 3, 5, 7, ...) is convolved with (1, 2, 2, 2, ...) and A162958 where A162956 is convolved with (1, 3, 3, 3, ...); the present sequence can be considered toothpick N=1 since A118977 is convolved with (1, 1, 1, ...).
Arranged in array fashion, the first three toothpick sequences would be:
N=1: A163267: (1, 2, 4, 5, 7, 10, 13, 14, ...)
N=2: A139250: (1, 3, 7, 11, 15, 23, 35, 43, ...)
N=3: A162958: (1, 4, 10, 19, 25, 40, 67, 94, ...)
...
Is there an illustration of this sequence using toothpicks? - Omar E. Pol, Dec 13 2016

Crossrefs

Programs

  • Mathematica
    a[0] = 0; a[1] = 1; a[n_] := a[n] = (j = n - 2^Floor[Log[2, n]]; a[j] + a[j + 1]); Table[Sum[a[n], {n, 0, k}], {k, 0, 20}] (* G. C. Greubel, Dec 12 2016 *)

Extensions

Edited and extended by N. J. A. Sloane, Jan 07 2010

A160552 a(0)=0, a(1)=1; a(2^i+j) = 2*a(j) + a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 5, 11, 17, 15, 1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31, 1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31, 5, 11, 17, 19, 21, 39, 49, 35, 21, 39, 53, 59, 81, 127, 129, 63, 1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31
Offset: 0

Views

Author

David Applegate, May 18 2009

Keywords

Comments

This recurrence is patterned after the one for A152980, but without the special cases.
Sequence viewed as triangle:
0,
1,
1, 3,
1, 3, 5, 7,
1, 3, 5, 7, 5, 11, 17, 15,
1, 3, 5, 7, 5, 11, 17, 15, 5, 11, 17, 19, 21, 39, 49, 31.
The rows converge to A151548.
Also the sum of the terms in the k-th row (k >= 1) is 4^(k-1). Proof by induction. - N. J. A. Sloane, Jan 21 2010
If this sequence [1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 5, 11, 17, 15, ...] is convolved with [1, 2, 2, 2, 2, 2, 2, ...] we obtain A139250, the toothpick sequence. Example: A139250(5) = 15 = (1, 2, 2, 2, 2) * (3, 1, 3, 1, 1). - Gary W. Adamson, May 19 2009
Starting with 1 and convolved with [1, 2, 0, 0, 0, ...] = A151548. - Gary W. Adamson, Jun 04 2009
Refer to A162956 for the analogous triangle using N=3. - Gary W. Adamson, Jul 20 2009
It appears that the sums of two successive terms give the positive terms of A139251. - Omar E. Pol, Feb 18 2015

Examples

			a(2) = a(2^1+0) = 2*a(0) + a(1) = 1, a(3) = a(2^1+1) = 2*a(1) + a(2) = 3*a(2^i) = 2*a(0) + a(1) = 1.
		

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.

Programs

  • Maple
    S:=proc(n) option remember; local i,j; if n <= 1 then RETURN(n); fi; i:=floor(log(n)/log(2)); j:=n-2^i; 2*S(j)+S(j+1); end; # N. J. A. Sloane, May 18 2009
    H := x*(1+2*x)/(1+x) + (4*x^2/(1+2*x))*(mul(1+x^(2^k-1)+2*x^(2^k),k=1..20)-1); series(H,x,120); # N. J. A. Sloane, May 23 2009
  • Mathematica
    Nest[Join[#, 2 # + Append[Rest@#, 1]] &, {0}, 7] (* Ivan Neretin, Feb 09 2017 *)

Formula

G.f.: x*(1+2*x)/(1+x) + (4*x^2/(1+2*x))*(-1 + Product_{k>=1} (1 + x^(2^k-1) + 2*x^(2^k))). - N. J. A. Sloane, May 23 2009, based on Gary W. Adamson's comment above and the known g.f. for A139250.
It appears that a(n) = A169708(n)/4, n >= 1. - Omar E. Pol, Feb 15 2015
It appears that a(n) = A139251(n) - a(n-1), n >= 1. - Omar E. Pol, Feb 18 2015

A151702 a(0)=1, a(1)=0; a(2^i + j) = a(j) + a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 3, 4, 3, 1, 1, 2, 2, 2, 3, 4, 3, 2, 3, 4, 4, 5, 7, 7, 4, 1, 1, 2, 2, 2, 3, 4, 3, 2, 3, 4, 4, 5, 7, 7, 4, 2, 3, 4, 4, 5, 7, 7, 5, 5, 7, 8, 9, 12, 14, 11, 5, 1, 1, 2, 2, 2, 3, 4, 3, 2, 3, 4, 4, 5, 7, 7, 4, 2, 3, 4, 4, 5, 7, 7, 5, 5, 7, 8, 9, 12, 14, 11, 5, 2, 3, 4, 4, 5, 7, 7
Offset: 0

Views

Author

N. J. A. Sloane, Jun 06 2009

Keywords

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.
If first two terms are dropped, same as A151552.

Programs

  • Maple
    f:=proc(r,s,a,b) local s1,n,i,j;
    s1:=array(0..120);
    s1[0]:=r; s1[1]:=s;
    for n from 2 to 120 do i:=floor(log(n)/log(2));
    j:=n-2^i; s1[n]:=a*s1[j]+b*s1[j+1]; od:
    [seq(s1[n],n=0..120)];
    end;
    f(1,0,1,1);
  • Mathematica
    a = {1, 0}; Do[AppendTo[a, a[[j]] + a[[j + 1]]], {i, 6}, {j, 2^i}]; a (* Ivan Neretin, Jun 28 2017 *)

A151703 a(0)=1, a(1)=0; a(2^i+j) = a(j) + 2*a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

1, 0, 1, 2, 1, 2, 5, 4, 1, 2, 5, 4, 5, 12, 13, 6, 1, 2, 5, 4, 5, 12, 13, 6, 5, 12, 13, 14, 29, 38, 25, 8, 1, 2, 5, 4, 5, 12, 13, 6, 5, 12, 13, 14, 29, 38, 25, 8, 5, 12, 13, 14, 29, 38, 25, 16, 29, 38, 41, 72, 105, 88, 41, 10, 1, 2, 5, 4, 5, 12, 13, 6, 5, 12, 13, 14, 29, 38, 25, 8, 5, 12, 13, 14
Offset: 0

Views

Author

N. J. A. Sloane, Jun 06 2009

Keywords

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.
If first two terms are dropped, same as A151691.

Programs

  • Maple
    See A151702 for Maple code.
  • Mathematica
    a = {1, 0}; Do[AppendTo[a, a[[j]] + 2 a[[j + 1]]], {i, 6}, {j, 2^i}]; a (* Ivan Neretin, Jul 04 2017 *)

A151568 a(0)=1, a(1)=1; a(2^i+j)=2*a(j)+a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

1, 1, 3, 5, 3, 5, 11, 13, 3, 5, 11, 13, 11, 21, 35, 29, 3, 5, 11, 13, 11, 21, 35, 29, 11, 21, 35, 37, 43, 77, 99, 61, 3, 5, 11, 13, 11, 21, 35, 29, 11, 21, 35, 37, 43, 77, 99, 61, 11, 21, 35, 37, 43, 77, 99, 69, 43, 77, 107, 117, 163, 253, 259, 125, 3, 5, 11, 13, 11, 21, 35, 29, 11
Offset: 0

Views

Author

N. J. A. Sloane, May 25 2009

Keywords

Comments

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.

Programs

  • Mathematica
    a = {1, 1}; Do[AppendTo[a, 2 a[[j]] + a[[j + 1]]], {i, 6}, {j, 2^i}]; a (* Ivan Neretin, Jul 04 2017 *)

A151569 a(0)=1, a(1)=2; a(2^i+j)=2*a(j)+a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

1, 2, 4, 8, 4, 8, 16, 20, 4, 8, 16, 20, 16, 32, 52, 44, 4, 8, 16, 20, 16, 32, 52, 44, 16, 32, 52, 56, 64, 116, 148, 92, 4, 8, 16, 20, 16, 32, 52, 44, 16, 32, 52, 56, 64, 116, 148, 92, 16, 32, 52, 56, 64, 116, 148, 104, 64, 116, 160, 176, 244, 380, 388, 188, 4, 8, 16, 20, 16, 32, 52
Offset: 0

Views

Author

N. J. A. Sloane, May 25 2009

Keywords

Comments

Equals 2*A160552+A151704.

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.

Programs

  • Mathematica
    a = {1, 2}; Do[AppendTo[a, 2 a[[j]] + a[[j + 1]]], {i, 6}, {j, 2^i}]; a (* Ivan Neretin, Jul 04 2017 *)

A151570 a(0)=1, a(1)=1; a(2^i + j) = a(j) + a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

1, 1, 2, 3, 2, 3, 5, 5, 2, 3, 5, 5, 5, 8, 10, 7, 2, 3, 5, 5, 5, 8, 10, 7, 5, 8, 10, 10, 13, 18, 17, 9, 2, 3, 5, 5, 5, 8, 10, 7, 5, 8, 10, 10, 13, 18, 17, 9, 5, 8, 10, 10, 13, 18, 17, 12, 13, 18, 20, 23, 31, 35, 26, 11, 2, 3, 5, 5, 5, 8, 10, 7, 5, 8, 10, 10, 13, 18, 17, 9, 5, 8, 10, 10, 13, 18, 17, 12
Offset: 0

Views

Author

N. J. A. Sloane, May 25 2009

Keywords

Comments

Equals A118977 + A151702.

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.

Programs

  • Mathematica
    a = {1, 1}; Do[AppendTo[a, a[[j]] + a[[j + 1]]], {i, 6}, {j, 2^i}]; a (* Ivan Neretin, Jun 28 2017 *)

A151571 a(0)=1, a(1)=2; a(2^i + j) = a(j) + a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

1, 2, 3, 5, 3, 5, 8, 8, 3, 5, 8, 8, 8, 13, 16, 11, 3, 5, 8, 8, 8, 13, 16, 11, 8, 13, 16, 16, 21, 29, 27, 14, 3, 5, 8, 8, 8, 13, 16, 11, 8, 13, 16, 16, 21, 29, 27, 14, 8, 13, 16, 16, 21, 29, 27, 19, 21, 29, 32, 37, 50, 56, 41, 17, 3, 5, 8, 8, 8, 13, 16, 11, 8, 13, 16, 16, 21, 29, 27, 14, 8, 13
Offset: 0

Views

Author

N. J. A. Sloane, May 25 2009

Keywords

Comments

Equals 2*A118977 + A151702.

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.

Programs

  • Mathematica
    a = {1, 2}; Do[AppendTo[a, a[[j]] + a[[j + 1]]], {i, 6}, {j, 2^i}]; a (* Ivan Neretin, Jun 28 2017 *)

A151572 a(0)=0, a(1)=1; a(2^i + j) = a(j) + 2*a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

0, 1, 2, 5, 2, 5, 12, 9, 2, 5, 12, 9, 12, 29, 30, 13, 2, 5, 12, 9, 12, 29, 30, 13, 12, 29, 30, 33, 70, 89, 56, 17, 2, 5, 12, 9, 12, 29, 30, 13, 12, 29, 30, 33, 70, 89, 56, 17, 12, 29, 30, 33, 70, 89, 56, 37, 70, 89, 96, 173, 248, 201, 90, 21, 2, 5, 12, 9, 12, 29, 30, 13, 12, 29, 30, 33, 70
Offset: 0

Views

Author

N. J. A. Sloane, May 25 2009

Keywords

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.

Programs

  • Mathematica
    a = {0, 1}; Do[AppendTo[a, a[[j]] + 2 a[[j + 1]]], {i, 6}, {j, 2^i}]; a (* Ivan Neretin, Jun 28 2017 *)

A151573 a(0)=1, a(1)=1; a(2^i + j) = a(j) + 2*a(j+1) for 0 <= j < 2^i.

Original entry on oeis.org

1, 1, 3, 7, 3, 7, 17, 13, 3, 7, 17, 13, 17, 41, 43, 19, 3, 7, 17, 13, 17, 41, 43, 19, 17, 41, 43, 47, 99, 127, 81, 25, 3, 7, 17, 13, 17, 41, 43, 19, 17, 41, 43, 47, 99, 127, 81, 25, 17, 41, 43, 47, 99, 127, 81, 53, 99, 127, 137, 245, 353, 289, 131, 31, 3, 7, 17, 13, 17, 41, 43, 19, 17
Offset: 0

Views

Author

N. J. A. Sloane, May 25 2009

Keywords

Comments

Equals A151572 + A151703.

Crossrefs

For the recurrence a(2^i+j) = C*a(j) + D*a(j+1), a(0) = A, a(1) = B for following values of (A B C D) see: (0 1 1 1) A118977, (1 0 1 1) A151702, (1 1 1 1) A151570, (1 2 1 1) A151571, (0 1 1 2) A151572, (1 0 1 2) A151703, (1 1 1 2) A151573, (1 2 1 2) A151574, (0 1 2 1) A160552, (1 0 2 1) A151704, (1 1 2 1) A151568, (1 2 2 1) A151569, (0 1 2 2) A151705, (1 0 2 2) A151706, (1 1 2 2) A151707, (1 2 2 2) A151708.

Programs

  • Maple
    N:= 10: # to get a(0) to a(2^(N+1)-1)
    a[0]:= 1:
    a[1]:= 1:
    for i from 1 to N do
      for j from 0 to 2^i-1 do
        a[2^i+j]:= a[j]+2*a[j+1]
      od
    od:
    seq(a[i],i=0..2^(N+1)-1); # Robert Israel, May 28 2014
  • Mathematica
    a = {1, 1}; Do[AppendTo[a, a[[j]] + 2 a[[j + 1]]], {i, 6}, {j, 2^i}]; a (* Ivan Neretin, Jun 28 2017 *)
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