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.

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A141070 Number of primes in rows of Pascal-like triangles with index of asymmetry y = 3 and index of obliquity z = 0 or z = 1.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 3, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 3, 4, 3, 3, 3, 3, 3, 3, 5, 4, 3, 3, 4, 3, 3, 3, 3, 3, 3, 3, 3, 3, 5, 3, 4, 3, 3, 3, 5, 3, 4, 3, 3, 3, 4, 3, 3, 3, 3, 3, 3, 4, 3, 4, 4, 3, 3, 3, 3, 5, 4, 4, 3, 3, 3, 3, 4, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 5, 3, 3, 3
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Jul 16 2008

Keywords

Comments

For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) + G(n+4, k) for n >= 0 and k = 1..(n+1). (This is array A140996.)
For the Pascal-like triangle with index of asymmetry y = 3 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, k) = G(n+1, k-3) + G(n+1, k-4) + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for n >= 0 and k = 4..(n+4). (This is array A140995.)
From Petros Hadjicostas, Jun 13 2019: (Start)
The two triangular arrays A140995 and A140996, which are described above, are mirror images of each other. Thus, we get the same sequence no matter which one we use.
Even though the numbering of the rows of both triangular arrays A140995 and A140996 starts at n = 0, the author of this sequence set up the offset at n = 1; that is, a(n) = number of primes in row n - 1 for A140995 (or for A140996) for n >= 1.
Finally, we mention that in the attached picture about Stepan's triangles, the letter s is used to describe the index of asymmetry and the letter e is used to describe the index of obliqueness (instead of the letters y and z, respectively).
(End)

Examples

			Pascal-like triangle with y = 3 and z = 0 (i.e, A140996) begins as follows:
  1, so a(1) = 0.
  1 1, so a(2) = 0.
  1 2 1, so prime 2 and a(3) = 1.
  1 4 2 1, so prime 2 and a(4) = 1.
  1 8 4 2 1, so prime 2 and a(5) = 1.
  1 16 8 4 2 1, so prime 2 and a(6) = 1.
  1 31 17 8 4 2 1, so primes 2, 17, 31 and a(7) = 3.
  1 60 35 17 8 4 2 1, so primes 2, 17 and a(8) = 2.
  1 116 72 35 17 8 4 2 1, so primes 2, 17 and a(9) = 2.
  1 224 148 72 35 17 8 4 2 1, so primes 2, 17 and a(10) = 2.
  1 432 303 149 72 35 17 8 4 2 1, so primes 2, 17, 149 and a(11) = 3.
  ...

Links

Crossrefs

Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141072, A141073.

Programs

  • Mathematica
    nlim = 100;
For[n = 0, n <= nlim, n++, G[n, 0] = 1];
For[n = 1, n <= nlim, n++, G[n, n] = 1];
For[n = 2, n <= nlim, n++, G[n, n-1] = 2];
For[n = 3, n <= nlim, n++, G[n, n-2] = 4];
For[n = 4, n <= nlim, n++, G[n, n-3] = 8];
For[n = 5, n <= nlim, n++, For[k = 1, k < n-3, k++,
   G[n, k] = G[n-4, k-1] + G[n-4, k] + G[n-3, k] + G[n-2, k] +
     G[n-1, k]]];
A141070 = {}; For[n = 0, n <= nlim, n++, c = 0;
 For[k = 0, k <= n, k++, If[PrimeQ[G[n, k]], c++]];
 AppendTo[A141070, c]];
A141070 (* Robert Price, Jul 03 2019 *)

Extensions

Partially edited by N. J. A. Sloane, Jul 18 2008
More terms and comments edited by Petros Hadjicostas, Jun 13 2019
a(52)-a(100) from Robert Price, Jul 03 2019

A141072 Sum of diagonal numbers in a Pascal-like triangle with index of asymmetry y = 3 and index of obliquity z = 0 (going upwards, left to right).

Original entry on oeis.org

1, 1, 2, 3, 6, 11, 22, 42, 83, 162, 319, 626, 1231, 2419, 4756, 9349, 18380, 36133, 71036, 139652, 274549, 539748, 1061117, 2086100, 4101165, 8062677, 15850806, 31161863, 61262610, 120439119, 236777074, 465491470, 915132135, 1799102406, 3536942203, 6953445286
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Jul 16 2008

Keywords

Comments

For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 0, which is read by rows, we have G(n, 0) = G(n+1, n+1) = 1, G(n+2, n+1) = 2, G(n+3, n+1) = 4, G(n+4, n+1) = 8, and G(n+5, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) + G(n+3, k) + G(n+4, k) for n >= 0 and k = 1..(n+1). (This is array A140996.)
For the Pascal-like triangle G(n, k) with index of asymmetry y = 3 and index of obliqueness z = 1, which is read by rows, we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, k) = G(n+1, k-3) + G(n+1, k-4) + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for n >= 0 and for k = 4..(n+4). (This is array A140995.)
From Petros Hadjicostas, Jun 13 2019: (Start)
In the example below the author uses array A140996 to create this sequence. If we use array A140995, which is the mirror image of A140996, and follow the same process, we get a different sequence: 1, 1, 2, 3, 4, 7, 8, 15, 16, 31, 33, 63, 68, 127, 140, 255, 288, 512, 592, ...
Even though array A140996 starts at row n = 0, the offset of the current sequence was set at n = 1. In other words, a(n) = Sum_{ceiling((n-1)/2) <= i <= n-1} G(i, n-1-i) = G(n-1, 0) + G(n-2, 1) + ... + G(ceiling((n-1)/2), floor((n-1)/2)) for n >= 1, where G(n, k) = A140996(n, k).
To get the g.f. of this sequence, we take the bivariate g.f. of sequence A140996, and set x = y. We multiply the result by x because the offset here was set at n = 1.
Finally, we mention that in the attached photograph about Stepan's triangle, the index of asymmetry is denoted by s (rather than y) and the index of obliqueness is denoted by e (rather than z). For the Pascal triangle, s = y = 0.
(End)

Examples

			Pascal-like triangle with y = 3 and z = 0 (i.e, A140996) begins as follows:
  1, so a(1) = 1.
  1   1, so a(2) = 1.
  1   2   1, so a(3) = 1 + 1 = 2.
  1   4   2   1, so a(4) = 1 + 2 = 3.
  1   8   4   2  1, so a(5) = 1 + 4 + 1 = 6.
  1  16   8   4  2  1, so a(6) = 1 + 8 + 2 = 11.
  1  31  17   8  4  2  1, so a(7) = 1 + 16 + 4 + 1 = 22.
  1  60  35  17  8  4  2 1, so a(8) = 1 + 31 + 8 + 2 = 42.
  1 116  72  35 17  8  4 2 1, so a(9) = 1 + 60 + 17 + 4 + 1 = 83.
  1 224 148  72 35 17  8 4 2 1, so a(10) = 1 +  116 + 35 + 8 + 2 = 162.
  1 432 303 149 72 35 17 8 4 2 1, so a(11) = 1 + 224 + 72 + 17 + 4 + 1 = 319.
... [edited by _Petros Hadjicostas_, Jun 13 2019]

Links

Crossrefs

Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141073.

Formula

From Petros Hadjicostas, Jun 13 2019: (Start)
a(n) = Sum_{ceiling((n-1)/2) <= i <= n-1} G(i, n-1-i) for n >= 1, where G(n, k) = A140996(n, k) for 0 <= k <= n.
G.f.: x*(1 - x^2 - x^3 - x^4 - x^5)/((1 - x)*(1 + x)*(1 - x - x^2 - x^3 - x^4 - x^5)) = x*(1 - x^2 - x^3 - x^4 - x^5)/(1 - x - 2*x^2 + x^6 + x^7).
Recurrence: a(n) = -(3 + (-1)^n)/2 + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n >= 6 with a(1) = a(2) = 1, a(3) = 2, a(4) = 3, and a(5) = 6.
(End)

Extensions

Partially edited by N. J. A. Sloane, Jul 18 2008
Name edited by and more terms from Petros Hadjicostas, Jun 13 2019

A308808 Limiting row sequence of Pascal-like triangle A141021 (with index of asymmetry s = 4).

Original entry on oeis.org

1, 2, 4, 8, 16, 33, 67, 136, 276, 561, 1140, 2316, 4705, 9559, 19421, 39457, 80163, 162864, 330885, 672247, 1365779, 2774802, 5637462, 11453422, 23269491, 47275758, 96048397, 195137952, 396454511, 805461867, 1636426882, 3324667561, 6754603284, 13723075972, 27880662448, 56644103708
Offset: 0

Views

Author

Petros Hadjicostas, Jun 25 2019

Keywords

Comments

In the attached photograph, we see that the index of asymmetry is denoted by s and the index of obliqueness by e. The general recurrence is G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{1 <= m <= s+1} G(n+m, k-e*s+m*e-2*e) for n >= 0 with k = 1..(n+1) when e = 0 and k = (s+1)..(n+s+1) when e = 1. The initial conditions are G(n+x+1, n-e*n+e*x-e+1) = 2^x for x=0..s and n >= 0. There is one more initial condition, namely, G(n, e*n) = 1 for n >= 0.
For s = 0, we get Pascal's triangle A007318. For s = 1, we get A140998 (e = 0) and A140993 (e = 1). For s = 2, we get A140997 (e = 0) and A140994 (e = 1). For s = 3, we get A140996 (e = 0) and A140995 (e = 1). For s = 4, we have arrays A141020 (with e = 0) and A141021 (with e = 1). In some of these arrays, the indices n and k are shifted.
For the triangular array G(n, k) = A141021(n, k), we have G(n+6, k) = G(n+1, k-4) + G(n+1, k-5) + G(n+2, k-4) + G(n+3, k-3) + G(n+4, k-2) + G(n+5, k-1) for n >= 0 and k = 5..(n+5) with G(n+x+1, x) = 2^x for x = 0..4 and n >= 0.
With G(n, k) = A141021(n, k), the current sequence (a(k): k >= 0) is defined by a(k) = lim_{n -> infinity} G(n, k) for k >= 0. Then a(k) = a(k-5) + 2*a(k-4) + a(k-3) + a(k-2) + a(k-1) for k >= 5 with a(x) = 2^x for x = 0..4.

Links

Crossrefs

Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141015, A141018, A141020, A141021, A141031, A141065, A141066, A141067.

Formula

a(k) = a(k-5) + 2*a(k-4) + a(k-3) + a(k-2) + a(k-1) for k >= 5 with a(k) = 2^k for k = 0..4.
G.f.: -(x + 1)*(x^2 + 1)/(x^5 + 2*x^4 + x^3 + x^2 + x - 1).

A309462 Limiting row sequence for Pascal-like triangle A140995 (Stepan's triangle with index of asymmetry s = 3).

Original entry on oeis.org

1, 2, 4, 8, 17, 35, 72, 149, 308, 636, 1314, 2715, 5609, 11588, 23941, 49462, 102188, 211120, 436173, 901131, 1861732, 3846329, 7946496, 16417420, 33918306, 70075047, 144774689, 299103768, 617946857, 1276675050, 2637604132, 5449276664, 11258177753, 23259337731
Offset: 0

Views

Author

Petros Hadjicostas, Aug 03 2019

Keywords

Comments

In the attached photograph, we see that the index of asymmetry is denoted by s and the index of obliqueness by e. The general recurrence is G(n+s+2, k) = G(n+1, k-e*s+e-1) + Sum_{1 <= m <= s+1} G(n+m, k-e*s+m*e-2*e) for n >= 0 with k = 1..(n+1) when e = 0 and k = (s+1)..(n+s+1) when e = 1. The initial conditions are G(n+x+1, n-e*n+e*x-e+1) = 2^x for x=0..s and n >= 0. There is one more initial condition, namely, G(n, e*n) = 1 for n >= 0.
For s = 0, we get Pascal's triangle A007318. For s = 1, we get A140998 (e = 0) and A140993 (e = 1). For s = 2, we get A140997 (e = 0) and A140994 (e = 1). For s = 3, we get A140996 (e = 0) and A140995 (e = 1). For s = 4, we have arrays A141020 (with e = 0) and A141021 (with e = 1). In some of these arrays, the indices n and k are shifted.
For the triangular array G(n, k) = A140995(n, k), we have G(n, n) = G(n+1, 0) = 1, G(n+2, 1) = 2, G(n+3, 2) = 4, G(n+4, 3) = 8, and G(n+5, k) = G(n+1, k-3) + G(n+1, k-4) + G(n+2, k-3) + G(n+3, k-2) + G(n+4, k-1) for n >= 0 and k = 4..(n+4).
With G(n, k) = A140995(n, k), the current sequence (a(k): k >= 0) is defined by a(k) = lim_{n -> infinity} G(n, k) for k >= 0. Then a(k) = a(k-4) + 2*a(k-3) + a(k-2) + a(k-1) for k >= 4 with a(x) = 2^x for x = 0..3.

Links

Crossrefs

Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141021, A141031, A141065, A141066, A141067, A141068, A141069, A141070, A141072, A141073, A308808.

Formula

a(0) = 1, a(1) = 2, a(2) = 4, a(3) = 8, and a(k) = a(k-1) + a(k-2) + 2*a(k-3) + a(k-4) for k >= 4.
G.f.: (x^2 + x + 1)/(-x^4 - 2*x^3 - x^2 - x + 1).
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