A141020 -id:A141020 - cp's OEIS frontend

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.

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

A001949 Solutions of a fifth-order probability difference equation.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 4, 8, 16, 32, 63, 124, 244, 480, 944, 1856, 3649, 7174, 14104, 27728, 54512, 107168, 210687, 414200, 814296, 1600864, 3147216, 6187264, 12163841, 23913482, 47012668, 92424472, 181701728, 357216192, 702268543, 1380623604, 2714234540

Offset: 0

N. J. A. Sloane

This sequence is the case r = 5 in the solution to an r-th order probability difference equation that can be found in Eqs. (4) and (3) on p. 356 of Dunkel (1925). (Equation (3) follows equation (4) in the paper!) For r = 2, we get a shifted version of A000071. For r = 3, we get a shifted version of A008937. For r = 4, we get a shifted version of A107066. For r = 6, we get a shifted version of A172316. See also the table in A172119. - Petros Hadjicostas, Jun 15 2019

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
O. Dunkel, Solutions of a probability difference equation, Amer. Math. Monthly, 32 (1925), 354-370; see pp. 356 and 369.
T. Langley, J. Liese, and J. Remmel, Generating Functions for Wilf Equivalence Under Generalized Factor Order, J. Int. Seq. 14 (2011), Article #11.4.2.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (2,0,0,0,0,-1)

Column k = 1 of A141020 (with a different offset) and second main diagonal of A141021 (with no zeros).
Column k = 5 of A172119.
Partial sums of A001591.

A001949:=1/(z-1)/(z**5+z**4+z**3+z**2+z-1); # Simon Plouffe in his 1992 dissertation

t={0,0,0,0,0};Do[AppendTo[t,t[[-5]]+t[[-4]]+t[[-3]]+t[[-2]]+t[[-1]]+1],{n,40}];t (* Vladimir Joseph Stephan Orlovsky, Jan 21 2012 *)
LinearRecurrence[{2,0,0,0,0,-1},{0,0,0,0,0,1},40] (* Harvey P. Dale, Jan 17 2015 *)

a(n):=sum(sum((-1)^j*binomial(n-5*j-5,k-1)*binomial(n-k-5*j-4,j),j,0,(n-k-4)/5),k,1,n-4); /* Vladimir Kruchinin, Oct 19 2011 */

x='x+O('x^99); concat(vector(5), Vec(x^5/((x-1)*(x^5+x^4+x^3+x^2+x-1)))) \\ Altug Alkan, Oct 04 2017

For n >= 6, a(n+1) = 2*a(n) - a(n-5).
G.f.: x^5 / ( (x-1)*(x^5 + x^4 + x^3 + x^2 + x - 1) ).
a(n) = Sum_{k=1..n-4} Sum_{j=0..floor((n-k-4)/5)} (-1)^j*binomial(n-5*j-5, k-1)*binomial(n-k-5*j-4, j). - Vladimir Kruchinin, Oct 19 2011
4*a(n) = A000322(n+1) - 1. - R. J. Mathar, Aug 16 2017
From Petros Hadjicostas, Jun 15 2019: (Start)
a(n) = 1 + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) for n >= 5. (See Eq. (4) and the Theorem with r = 5 on p. 356 of Dunkel (1925).)
a(n) = T(n - 5, 5) for n >= 5, where T(n, k) = Sum_{j = 0..floor(n/(k+1))} (-1)^j * binomial(n - k*j, n - (k+1)*j) * 2^(n - (k+1)*j) for 0 <= k <= n. This is Richard Choulet's formula in A172119.
(End)

Name edited by Petros Hadjicostas, Jun 15 2019

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

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

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).

A141019 a(n) is the largest number in the n-th row of triangle A140996.

Original entry on oeis.org

1, 1, 2, 4, 8, 16, 31, 60, 116, 224, 432, 833, 1606, 3096, 5968, 11504, 22175, 42744, 84752, 169880, 340013, 679604, 1356641, 2704954, 5387340, 10718620, 21304973, 42308331, 83945336, 166423276, 329683867, 652627294, 1291020297, 2552209710, 5042305104

Offset: 0

Juri-Stepan Gerasimov, Jul 11 2008

nonn

Also the largest number in the n-th row of A140995.

			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 16 8 4 2 1 is a(5) = 16.
The largest number of 1 31 17 8 4 2 1 is a(6) = 31.

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.

A140996 := 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 ; else procname(n-1,k)+procname(n-2,k) +procname(n-3,k)+procname(n-4,k)+procname(n-4,k-1) ; fi; end:
A141019 := proc(n) max(seq(A140996(n,k),k=0..n)) ; end: for n from 0 to 50 do printf("%d,",A141019(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, k == n-3, 8, True, T[n-1, k] + T[n-2, k] + T[n-3, k] + T[n-4, k] + T[n-4, k-1]];
a[n_] := Table[T[n, k], {k, 0, n}] // Max;
Table[a[n], {n, 0, 34}] (* Jean-François Alcover, Jan 28 2024, after R. J. Mathar *)

a(n) = max_{k=0..n} A140996(n,k).

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

Simplified definition and extended by R. J. Mathar, Sep 19 2008

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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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A001949 Solutions of a fifth-order probability difference equation.

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

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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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A141019 a(n) is the largest number in the n-th row of triangle A140996.

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