A141066 -id:A141066 - cp's OEIS frontend

A141073 List of central integer pairs in Pascal-like triangles with index of asymmetry y = 3 and index of obliqueness z = 0 or z = 1.

1, 1, 4, 2, 8, 4, 17, 8, 35, 17, 72, 35, 149, 72, 308, 149, 636, 308, 1314, 636, 2715, 1314, 5609, 2715, 11588, 5609, 23941, 11588, 49462, 23941, 102188, 49462, 211120, 102188, 436173, 211120, 901131, 436173, 1861732, 901131, 3846329, 1861732, 7946496, 3846329

Offset: 1

Juri-Stepan Gerasimov, Jul 16 2008

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 k = 4..(n+4). (This is array A140995.)
Arrays A140995 and A140996 are mirror images of each other. For discussion about their properties and their connection to Stepan's triangles, see their documentation. See also the documentation of the sequences in the CROSSREFS. - Petros Hadjicostas, Jun 13 2019

			Pascal-like triangle with y = 3 and z = 0 (i.e., A140996) begins as follows:
  1, so no central pair.
  1 1, so a(1) = 1 and a(2) = 1.
  1 2 1, so no central pair.
  1 4 2 1, so a(3) = 4 and a(4) = 2.
  1 8 4 2 1, so no central pair.
  1 16 8 4 2 1, so a(5) = 8 and a(6) = 4.
  1 31 17 8 4 2 1, so no central pair.
  1 60 35 17 8 4 2 1, so a(7) = 17 and a(8) = 8.
  1 116 72 35 17 8 4 2 1, so no central pair.
  1 224 148 72 35 17 8 4 2 1, so a(9) = 35 and a(10) = 17.
  1 432 303 149 72 35 17 8 4 2 1, so no central pair.
  1 833 618 308 149 72 35 17 8 4 2 1, so a(11) = 72 and a(12) = 35.
... [edited by _Petros Hadjicostas_, Jun 13 2019]

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A129847, A140993, A140994, A140995, A140996, A140997, A140998, A141065, A141066, A141067, A141068, A141069, A141070, A141072.

Rest[CoefficientList[Series[x*(x^8 + 3*x^6 + x^5 + 3*x^4 + x^3 + 3*x^2 + x + 1)/(1 - x^2 - x^4 - 2*x^6 -x^8),{x,0,44}],x]] (* James C. McMahon, Jul 16 2025 *)

From Petros Hadjicostas, Jun 13 2019: (Start)
a(2*n - 1) = A140996(2*n - 1, n - 1) = A140995(2*n - 1, n) and a(2*n) = A140996(2*n - 1, n) = A140995(2*n - 1, n - 1) for n >= 1.
a(2*n) = a(2*n - 3) for n >= 3.
a(n) = 2*a(n-2) + A129847(floor(n/2) - (4 + (-1)^n)) for n >= 9.
G.f.: x*(x^8 + 3*x^6 + x^5 + 3*x^4 + x^3 + 3*x^2 + x + 1)/(1 - x^2 - x^4 - 2*x^6 -x^8). (End)

Partially edited by N. J. A. Sloane, Jul 18 2008

More terms from Petros Hadjicostas, Jun 13 2019

A141064 List of different primes in Pascal-like triangles with index of asymmetry y = 1 and index of obliquity z = 0 or z = 1.

Original entry on oeis.org

2, 5, 7, 11, 23, 29, 89, 137, 311, 367, 1021, 3217, 5441, 2377, 12619, 65761, 5741, 144593, 13859, 78511, 1462397, 33461, 469957, 2552939, 11096497, 5930669, 6343133, 26512597, 470831, 127626137, 372222703, 15955507, 538270693, 531077333, 11401285549, 38613943, 15433507333, 92554537183, 113828092793

Offset: 1

Juri-Stepan Gerasimov, Jul 14 2008

nonn

For the Pascal-like triangle G(n, k) with index of asymmetry y = 1  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, k) = G(n+1, k-1) + G(n+1, k) + G(n+2, k) for n >= 0 and k = 1..(n+1).
For the Pascal-like triangle G(n, k) with index of asymmetry y = 1 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, k) = G(n+1, k-1) + G(n+1, k-2) + G(n+2, k-1) for n >= 0 and k = 2..(n+2).
In each row of A140998, the primes not appearing in earlier rows are collected, sorted, and added to the sequence. [R. J. Mathar, Apr 28 2010]
From Petros Hadjicostas, Jun 10 2019: (Start)
For the triangle with index of asymmetry y = 1 and index of obliqueness z = 0, read by rows, we have G(n, k) = A140998(n, k) for 0 <= k <= n.
For the triangle with index of asymmetry y = 1 and index of obliqueness z = 1, read by rows, we have  G(n, k) = A140993(n+1, k+1) for 0 <= k <= n.
Thus, except for the (unfortunate) shifting of the indices by 1, triangular arrays A140998 and A140993 are mirror images of each other.
Hence, instead of working with A140998, we may work with A140993: in each row of A140993, the primes not appearing in earlier rows may be collected, sorted, and added to the sequence (paraphrasing R. J. Mathar above!).
(End)

			Pascal-like triangle with y = 1 and z = 0 (i.e, A140998) begins as follows:
  1, so no prime.
  1 1, so no primes.
  1 2 1, so a(1) = 2.
  1 4 2 1, so no new primes.
  1 7 5 2 1, so a(2) = 5 and a(3) = 7.
  1 12 11 5 2 1, so a(4) = 11.
  1 20 23 12 5 2 1, so a(5) = 23.
  1 33 46 28 12 5 2 1, so no new primes.
  1 54 89 63 29 12 5 2 1, so a(6) = 29 and a(7) = 89.
  1 88 168 137 69 29 12 5 2 1, so a(8) = 137.
  1 143 311 289 161 70 29 12 5 2 1, so a(9) = 311.
  1 232 567 594 367 168 70 29 12 5 2 1, so a(10) = 367.
...
[edited by _Petros Hadjicostas_, Jun 11 2019]

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A140993, A140994, A140995, A140996, A140997, A140998, A141021, A141022, A141031, A141066, A141067.

# This is a modification R. J. Mathar's program from A141031 (for the case y = 4 and z = 0).
# Construct array A140998 (y = 1 and z = 0):
A140998 := proc(n, k) option remember; if k < 0 or n < k then 0; elif k = 0 or k = n then 1; elif k = n - 1 then 2; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 2, k - 1); end if; end proc;
# Construct the current sequence:
A141064 := proc(nmax) local a, b, n, k, new; a := []; for n from 0 to nmax do b := []; for k from 0 to n do new := A140998(n, k); if not (new = 1 or not isprime(new) or new in a or new in b) then b := [op(b), new]; end if; end do; a := [op(a), op(sort(b))]; end do; RETURN(a); end proc;
# Generate terms of the current sequence:
A141064(38);
# If one wants to get the primes sorted, then replace RETURN(a) in the Maple code above with RETURN(sort(a)). In such a case, however, the output sequence is not uniquely defined because it depends on the maximum n. - Petros Hadjicostas, Jun 15 2019

Partially edited by N. J. A. Sloane, Jul 18 2008
More terms from R. J. Mathar, Apr 28 2010
More terms from Petros Hadjicostas, Jun 11 2019

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

Juri-Stepan Gerasimov, Jul 16 2008

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)

			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.
  ...

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

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

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 *)

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

Juri-Stepan Gerasimov, Jul 16 2008

nonn

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)

			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]

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

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

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)

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

Petros Hadjicostas, Jun 25 2019

nonn

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.

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

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

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

Petros Hadjicostas, Aug 03 2019

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.

Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

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

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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A141073 List of central integer pairs in Pascal-like triangles with index of asymmetry y = 3 and index of obliqueness z = 0 or z = 1.

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A141064 List of different primes in Pascal-like triangles with index of asymmetry y = 1 and index of obliquity z = 0 or z = 1.

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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.

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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).

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

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

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