A140996 -id:A140996 - cp's OEIS frontend

A107066 Expansion of 1/(1-2*x+x^5).

1, 2, 4, 8, 16, 31, 60, 116, 224, 432, 833, 1606, 3096, 5968, 11504, 22175, 42744, 82392, 158816, 306128, 590081, 1137418, 2192444, 4226072, 8146016, 15701951, 30266484, 58340524, 112454976, 216763936, 417825921, 805385358, 1552430192, 2992405408, 5768046880

Offset: 0

Paul Barry, May 10 2005

Row sums of number triangle A107065.
Same as A018922 plus first 3 additional terms. - Vladimir Joseph Stephan Orlovsky, Jul 08 2011
a(n) is the number of binary words of length n containing no subword 01011. - Alois P. Heinz, Mar 14 2012

			G.f. = 1 + 2*x + 4*x^2 + 8*x^3 + 16*x^4 + 31*x^5 + 60*x^6 + 116*x^7 + 224*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Otto Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see p. 356.
Jia Huang, A coin flip game and generalizations of Fibonacci numbers, arXiv:2501.07463 [math.CO], 2025. See p. 7.
Thomas Langley, Jeffrey Liese, and Jeffrey Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011), Article #11.4.2.
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,-1).

Cf. A018922, A119407 (partial sums), A000078 (first differences).
Cf. A209888. - Alois P. Heinz, Mar 14 2012
Column k = 1 of array A140996 (with a different offset) and second main diagonal of A140995.
Column k = 4 of A172119 (with a different offset).

a:=[1,2,4,8,16];; for n in [6..40] do a[n]:=2*a[n-1]-a[n-5]; od; a; # G. C. Greubel, Jun 12 2019

I:=[1,2,4,8,16]; [n le 5 select I[n] else 2*Self(n-1) - Self(n-5): n in [1..40]]; // G. C. Greubel, Jun 12 2019

CoefficientList[Series[1/(1 - 2*z + z^5), {z, 0, 40}], z] (* Vladimir Joseph Stephan Orlovsky, Jul 08 2011 *)
LinearRecurrence[{2,0,0,0,-1}, {1,2,4,8,16}, 40] (* G. C. Greubel, Jun 12 2019 *)

{a(n) = if( n<0, n = -n; polcoeff( -x^5 / (1 - 2*x^4 + x^5) + x * O(x^n), n), polcoeff( 1 / (1 - 2*x + x^5) + x * O(x^n), n))} /* Michael Somos, Dec 28 2012 */

(1/(1-2*x+x^5)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jun 12 2019

a(n) = 2*a(n-1) - a(n-5).
a(n) = Sum_{k=0..floor(n/5)} C(n-4*k, k) * 2^(n-2*k) *(-1)^k.
a(n) = A018922(n-3) for n >= 3. - R. J. Mathar, Mar 09 2007
First difference of A119407. - Michael Somos, Dec 28 2012
From Petros Hadjicostas, Jun 12 2019: (Start)
G.f.: 1/((1 - x)*(1 - x - x^2 - x^3 - x^4)).
Setting k = 1 in the double recurrence for array A140996, we get that a(n+5) = 1 + a(n+1) + a(n+2) + a(n+3) + a(n+4) for n >= 0, which of course we can prove using other methods as well. See also Dunkel (1925).
(End)
a(n) = Sum_{k=0..n+3} A000078(k). - Greg Dresden, Jan 01 2021

A141065 List of different composite numbers in Pascal-like triangles with index of asymmetry y = 1 and index of obliqueness z = 0 or z = 1.

Original entry on oeis.org

4, 12, 20, 28, 33, 46, 54, 63, 69, 88, 168, 70, 143, 161, 289, 232, 567, 594, 169, 376, 399, 817, 1194, 407, 609, 934, 1778, 1820, 2355, 408, 975, 986, 2150, 3789, 4570, 984, 1596, 2316, 4862, 5646, 7922, 8745, 985, 2367, 2583, 9849, 10801, 16281, 16532, 4180, 5667, 17091, 23585, 30923, 32948, 2378

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).
From Petros Hadjicostas, Jun 09 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.
As suggested by R. J. Mathar for sequence A141064, in each row of A140998, the composites not appearing in earlier rows are collected, sorted, and added to the sequence.
Obviously, 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.
Finally, we explain the meaning of the double recurrence in the attached photograph (about Stepan's triangles and Pascal's triangles).
The creator of the stone slab uses the notation G_n^k to denote either one of the two double arrays G(n, k) described above.
On the stone slab, the letter s is used to denote the "index of asymmetry" (denoted by y here) and the letter e is used to denote the 0-1 "index of obliqueness" (denoted by z here). Thus, as described above, there are two kinds of Stepan-Pascal triangles depending on whether e = z equals 0 or 1.
If e = 0, the value of k goes from 1 to n + 1, whereas if e = 1 the value of k goes from s + 1 = y + 1 (= 2 here) to n + s + 1 = n + y + 1.
The "index of asymmetry" s = y can take any (fixed) integer value from 0 to infinity. The fixed value of s = y determines the number of initial conditions: G(n + x + 1, n - e*n + e*x - e + 1) = 2^x for x = 0, 1, ..., s = y. In addition, there is one more initial condition: G(n, e*n) = 1.
The "index of asymmetry" s = y also determines the order of the recurrence (which is probably s + 2 = y + 2): 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).
Apparently, for convenience, the author of the current sequence has shifted the indices of the recurrences that appear on the stone slab (see at the beginning of the comments).
(End)

			Pascal-like triangle with y = 1 and z = 0 (i.e., A140998) begins as follows:
  1, so no composites.
  1 1, so no composites.
  1 2 1, so no composites.
  1 4 2 1, so a(1) = 4.
  1 7 5 2 1, so no composites.
  1 12 11 5 2 1, so a(2) = 12.
  1 20 23 12 5 2 1, so a(3) = 20.
  1 33 46 28 12 5 2 1, so a(4) = 28, a(5) = 33, and a(6) = 46.
  1 54 89 63 29 12 5 2 1, so a(7) = 54 and a(8) = 63.
  1 88 168 137 69 29 12 5 2 1, so a(9) = 69, a(10) = 88, and a(11) = 168.
  1 143 311 289 161 70 29 12 5 2 1, so a(12) = 70, a(13) = 143, a(14) = 161, and a(15) = 289.
  1 232 567 594 367 168 70 29 12 5 2 1, so a(16) = 232, a(17) = 567, and a(18) = 594.
  ... [example edited by _Petros Hadjicostas_, Jun 11 2019]

Petros Hadjicostas, Table of n, a(n) for n = 1..120
Juri-Stepan Gerasimov, Stepan's triangles and Pascal's triangle are connected by the recurrence relation ...

Cf. A140993 (mirror image of A140998 with y = 1 and z = 1), A140994 (triangle when y = 2 and z = 1), A140995 (triangle when y = 3 and z = 1), A140996 (triangle when y = 3 and z = 0), A140997 (triangle when y = 2 and z = 0), A140998 (has the above triangle with y = 1 and z = 0), A141020, A141021, A141064 (has primes for y = 1), A141066 (has composites when y = 2), A141067 (has primes when y = 2), A141068 (has primes when y = 3), A141069 (has composites when y = 3).

# This is a modification of R. J. Mathar's program for A141031 (for the case y = 4 and z = 0).
# Construction of 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;
# Construction of the current sequence:
A141065 := 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 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;
# Generation of terms of the current sequence:
A141065(24);
# If one wishes to sort composites, then one may replace RETURN(a) in the above Maple code 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

A141031 Nonprimes in the triangle A141020.

Original entry on oeis.org

1, 4, 8, 16, 32, 33, 63, 124, 136, 244, 276, 480, 560, 561, 944, 1135, 1140, 1856, 2298, 2316, 3649, 4705, 7174, 9398, 9558, 9559, 14104, 18984, 19415, 27728, 38320, 39432, 39457, 54512, 77298, 80075, 80163, 107168, 155823, 162583, 162863, 162864, 210687, 313927, 330878, 414200, 632080, 669872, 814296, 1271960, 1600864

Offset: 1

Juri-Stepan Gerasimov, Jul 12 2008

nonn

From Petros Hadjicostas, Jun 22 2019: (Start)
This is a dynamically defined sequence. Since the nonprimes from each row are mixed with the nonprimes of previous rows and then sorted, the value of a(n) may change each time we add a new row.
For a modification of R. J. Mathar's program below so that nonprimes are sorted only within each row (so as to get a uniquely defined sequence) see the documentation of sequences A141064, A141065, A141066, A141067, A141068, and A141069.
(End)

			Scanning rows of A141020 or A141021 and sorting new nonprimes into the list we get:
  1 yields a(1) = 1.
  1 1 yields no new member.
  1 2 1 yields no new member.
  1 4 2 1 yields a(2) = 4.
  1 8 4 2 1 yields a(3) = 8.
  1 16 8 4 2 1 yields a(4) = 16.
  1 32 16 8 4 2 1 yields a(5) = 32.
  1 63 33 16 8 4 2 1 yields a(6) = 33 and a(7) = 63.
  1 124 67 33 16 8 4 2 1 yields a(8) = 124.
  1 244 136 67 33 16 8 4 2 1 yields a(9) = 136 and a(10) = 244.
  1 480 276 136 67 33 16 8 4 2 1 yields a(11) = 276 and a(12) = 480.
  1 944 560 276 136 67 33 16 8 4 2 1 yields a(13) = 560 and a(14) = 944.
  ...
From _Petros Hadjicostas_, Jun 22 2019: (Start)
In the above example, we only sort the nonprimes up to row 11; we get the same output from _R. J. Mathar_'s program below if we say A141031(11). If, however, we include more rows in the program, the indexing of the nonprimes changes.
For example, the nonprimes in the data above come from the nonprimes of 22 rows. If we include more rows, then the indexing again changes and the value of each a(n) may not stay the same.
(End)

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, A141064, A141065, A141066, A141067, A141068, A141069.

A141020 := proc(n,k) option remember ; if k<0 or k>n then 0 ; elif k=0 or k=n then 1 ; elif k=n-1 then 2 ; elif k=n-2 then 4 ; elif k=n-3 then 8 ; elif k=n-4 then 16 ; else procname(n-1,k)+procname(n-2,k)+procname(n-3,k)+procname(n-4,k) +procname(n-5,k)+procname(n-5,k-1) ; fi; end:
A141031 := proc(nmax) local a,n,k ; a := [] ; for n from 0 to nmax do for k from 0 to n do a141020 := A141020(n,k) ; if not isprime(a141020) and not a141020 in a then a := [op(a),a141020] ; fi; od: od: RETURN(sort(a)) ; end: A141031(30) ; # R. J. Mathar, Sep 19 2008

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

Simplified definition, corrected values by R. J. Mathar, Sep 19 2008

A141069 List of different composites in Pascal-like triangles with index of asymmetry y = 3 and index of obliqueness z = 0 or z = 1.

Original entry on oeis.org

4, 8, 16, 35, 60, 72, 116, 148, 224, 303, 432, 308, 618, 833, 636, 1257, 1606, 1313, 2550, 3096, 1314, 2709, 5160, 5968, 2715, 5584, 10418, 11504, 5609, 11499, 20991, 22175, 23655, 42215, 42744, 11588, 23934, 48607, 82392, 84752, 23941, 99763, 158816, 169880

Offset: 1

Juri-Stepan Gerasimov, Jul 16 2008

nonn

For the Pascal-like triangle 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 arrays A140995 and A140996, which are described above, are mirror images of one another.
To make the current sequence uniquely defined, we follow the suggestion of R. J. Mathar for sequence A141064. For each row of array A140996, the composites not appearing in earlier rows are collected, sorted, and added to the sequence. We get exactly the same sequence by working with array A140995 instead.
Finally, we explain the meaning of the double recurrence in the attached photograph. It concerns the connection between Stepan's triangles and Pascal's triangles. The creator of the stone slab uses the notation G_n^k to denote the double array G(n, k), where 0 <= k <= n.
On the stone slab, the letter s is used to denote the "index of asymmetry" (denoted by y here) and the letter e is used to denote the 0-1 "index of obliqueness" (denoted by z here). Thus, as described above, there are two kinds of Stepan-Pascal triangles depending on whether e is equal to 0 or 1. (The case s = 0 corresponds to Pascal's triangle A007318.)
If e = 0, the value of k goes from 1 to n + 1, whereas if e = 1 the value of k goes from s + 1 to n + s + 1.
The "index of asymmetry" s = y can take any (fixed) integer value from 0 to infinity. The fixed value of s = y determines the number of initial conditions: G(n + x + 1, n - e*n + e*x - e + 1) = 2^x for x = 0, 1, ..., s. In addition, there is one more initial condition: G(n, e*n) = 1.
The "index of asymmetry" s = y also determines the order of the recurrence (which is probably s + 2 = y + 2): 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).
(End)

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

Petros Hadjicostas, Table of n, a(n) for n = 1..110
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, A141064, A141065, A141066, A141067, A141069, A141070, A141072, A141073.

# This is a modification of R. J. Mathar's program from sequence A141031 (for the case y = 4 and z = 0).
# Definition of sequence A140996 (y = 3 and z = 0):
A140996 := 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; elif k = n - 2 then 4; elif k = n - 3 then 8; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 3, k) + procname(n - 4, k) + procname(n - 4, k - 1); end if; end proc;
# Definition of current sequence:
A141069 := 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 := A140996(n, k); if not (new = 1 or 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;
# Generation of current sequence until row n = 30:
A141069(30);
# If one wishes the composites to be sorted, then replace RETURN(a) with RETURN(sort(a)) in the above Maple code. In such a case, however, the output may not necessarily be uniquely defined (because it changes with the value of n). - Petros Hadjicostas, Jun 15 2019

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

More terms from Petros Hadjicostas, Jun 13 2019

A141068 List of different primes in Pascal-like triangles with index of asymmetry y = 3 and index of obliquity z = 0 or z = 1.

Original entry on oeis.org

2, 17, 31, 149, 11587, 49429, 15701951, 21304973, 3846277, 251375273, 5449276159, 296410704409, 750391353973, 205109154121, 875366796349, 72210869205443, 139884035510017, 79014319582741129, 94461530406533783, 2562508045902551

Offset: 1

Juri-Stepan Gerasimov, Jul 16 2008

nonn

For the Pascal-like triangle 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 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 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.
To make the current sequence uniquely defined, we follow the suggestion of R. J. Mathar for sequence A141064. For each row of array A140996, the primes not appearing in earlier rows are collected, sorted, and added to the sequence. We get exactly the same sequence by working with array A140995 instead.
Finally, we mention that in the attached picture about the connection between Stepan's triangles and the Pascal triangle, 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). The Pascal triangle A007318 has index of asymmetry s = y = 0.
(End)

			Pascal-like triangle with y = 3 and z = 0 (i.e., A140996) begins as follows:
  1, so no primes.
  1 1, so no primes
  1 2 1, so a(1) = 2.
  1 4 2 1, so no new primes.
  1 8 4 2 1, so no new primes.
  1 16 8 4 2 1, so new primes.
  1 31 17 8 4 2 1, so a(2) = 17 and a(3) = 31.
  1 60 35 17 8 4 2 1, so no new primes.
  1 116 72 35 17 8 4 2 1, so no new primes.
  1 224 148 72 35 17 8 4 2 1, so new primes.
  1 432 303 149 72 35 17 8 4 2 1, so a(4) = 149.
  ...

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

Cf. A007318, A140993, A140994, A140995, A140996, A140997, A140998, A141020, A141021, A141031, A141064, A141065, A141066, A141067, A141069, A141070, A141072, A141073.

# This is a modification of R. J. Mathar's program for A141031 (for the case y = 4 and z = 0).
# Definition of sequence A140996 (y = 3 and z = 0):
A140996 := 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; elif k = n - 2 then 4; elif k = n - 3 then 8; else procname(n - 1, k) + procname(n - 2, k) + procname(n - 3, k) + procname(n - 4, k) + procname(n - 4, k - 1); end if; end proc;
# Definition of the current sequence:
A141068 := 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 := A140996(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;
# Generation of the current sequence:
A141068(80);
# If one wishes to get the primes sorted (as R. J. Mathar does in A141031), then replace RETURN(a) in the 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 Petros Hadjicostas, Jun 13 2019

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.

Original entry on oeis.org

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

A141018 a(n) is the largest number in the n-th row of triangle A140997.

Original entry on oeis.org

1, 1, 2, 4, 8, 15, 28, 52, 96, 177, 345, 694, 1386, 2751, 5431, 10672, 20885, 40724, 79153, 153402, 296528, 571845, 1129293, 2264749, 4527029, 9021498, 17926740, 35527082, 70230422, 138504765, 272545323, 535184340, 1048842743, 2051669285, 4006253136, 7954830148

Offset: 0

Juri-Stepan Gerasimov, Jul 11 2008

nonn

Also the largest number of the n-th row of A140994.

			The largest number of 1 is a(0) = 1.
The largest number of 1 1 is a(1) = 1.
The largest number of 1 2 1 is a(2) = 2.
The largest number of 1 4 2 1 is a(3) = 4.
The largest number of 1 8 4 2 1 is a(4) = 8.
The largest number of 1 15 9 4 2 1 is a(5) = 15.
The largest number of 1 28 19 9 4 2 1 is a(6) = 28.
The largest number of 1 52 40 19 9 4 2 1 is a(7) = 52.

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.

A140997 := proc(n,k) option remember ; if k<0 or k>n then 0 ; elif k=0 or k=n then 1 ; elif k=n-1 then 2 ; elif k=n-2 then 4 ; else procname(n-1,k)+procname(n-2,k)+procname(n-3,k)+procname(n-3,k-1) ; fi; end:
A141018 := proc(n) max(seq(A140997(n,k),k=0..n)) ; end: for n from 0 to 60 do printf("%d,",A141018(n)) ; od: # R. J. Mathar, Sep 19 2008

T[n_, k_] := T[n, k] = Which[k < 0 || k > n, 0, k == 0 || k == n, 1, k == n-1, 2, k == n-2, 4, True, T[n-1, k]+T[n-2, k]+T[n-3, k]+T[n-3, k-1]];
a[n_] := Max[Table[T[n, k], {k, 0, n}]];
Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Oct 18 2023, after R. J. Mathar *)

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

Simplified definition, corrected from a(12) on and extended by R. J. Mathar, Sep 19 2008

cp's OEIS Frontend

A107066 Expansion of 1/(1-2*x+x^5).

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A141065 List of different composite numbers in Pascal-like triangles with index of asymmetry y = 1 and index of obliqueness z = 0 or z = 1.

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A141031 Nonprimes in the triangle A141020.

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A141069 List of different composites in Pascal-like triangles with index of asymmetry y = 3 and index of obliqueness z = 0 or z = 1.

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A141068 List of different primes in Pascal-like triangles with index of asymmetry y = 3 and index of obliquity z = 0 or z = 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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