A002450 -id:A002450 - cp's OEIS frontend

A253563 Permutation of natural numbers: a(0) = 1, a(1) = 2; after which, a(2n) = A253560(a(n)), a(2n+1) = A253550(a(n)).

1, 2, 4, 3, 8, 6, 9, 5, 16, 12, 18, 10, 27, 15, 25, 7, 32, 24, 36, 20, 54, 30, 50, 14, 81, 45, 75, 21, 125, 35, 49, 11, 64, 48, 72, 40, 108, 60, 100, 28, 162, 90, 150, 42, 250, 70, 98, 22, 243, 135, 225, 63, 375, 105, 147, 33, 625, 175, 245, 55, 343, 77, 121, 13, 128, 96, 144, 80, 216, 120, 200, 56, 324, 180, 300, 84, 500, 140, 196, 44

Offset: 0

Antti Karttunen, Jan 03 2015

This sequence can be represented as a binary tree. Each child to the left is obtained by applying A253560 to the parent, and each child to the right is obtained by applying A253550 to the parent:
                                     1
                                     |
                  ...................2...................
                 4                                       3
       8......../ \........6                   9......../ \........5
      / \                 / \                 / \                 / \
     /   \               /   \               /   \               /   \
    /     \             /     \             /     \             /     \
  16       12         18       10         27       15         25       7
32  24   36  20     54  30   50  14     81  45   75  21    125  35   49 11
etc.
Sequence A253565 is the mirror image of the same tree. Also in binary trees A005940 and A163511 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) tells distance of n from 1 in all these trees. Of these four trees, this is the one where the left child is always larger than the right child.
Note that the indexing of sequence starts from 0, although its range starts from one.
a(n) (n>=1) can be obtained by the composition of a bijection between {1,2,3,4,...} and the set of integer partitions and a bijection between the set of integer partitions and {2,3,4,...}. Explanation on the example n=10. Write 2*n = 20 as a binary number: 10100. Consider a Ferrers board whose southeast border is obtained by replacing each 1 by an east step and each 0 by a north step. We obtain the Ferrers board of the partition p = (2,2,1). Finally, a(10) = 2'*2'*1', where m' = m-th prime. Thus, a(10)= 3*3*2 = 18. - Emeric Deutsch, Sep 17 2016

Antti Karttunen, Table of n, a(n) for n = 0..8191
Index entries for sequences that are permutations of the natural numbers

Inverse: A253564.
Cf. A252737 (row sums), A252738 (row products).
Cf. A002110, A002450, A005940, A122111, A252464, A253561, A253550, A253560, A253565, A054429, A252464.

a:= proc(n) local m; m:= n; [0]; while m>0 do `if`(1=
      irem(m, 2, 'm'), map(x-> x+1, %), [%[], 0]) od:
      `if`(n=0, 1, mul(ithprime(i), i=%))
    end:
seq(a(n), n=0..100);  # Alois P. Heinz, Aug 23 2017

p[n_] := p[n] = FactorInteger[n][[-1, 1]];
b[n_] := n p[n];
c[1] = 1; c[n_] := (n/p[n]) NextPrime[p[n]];
a[0] = 1; a[1] = 2; a[n_] := a[n] = If[EvenQ[n], b[a[n/2]], c[a[(n-1)/2]]];
a /@ Range[0, 100] (* Jean-François Alcover, Feb 15 2021 *)

a(0) = 1, a(1) = 2; after which, a(2n) = A253560(a(n)), a(2n+1) = A253550(a(n)).
As a composition of other permutations:
a(n) = A122111(A005940(n+1)).
a(n) = A253565(A054429(n)).
Other identities and observations. For all n >= 0:
A002110(n) =  a(A002450(n)). [Primorials occur at positions (4^n - 1)/3.]
For all n >= 1: a(2n) - a(2n+1) > 0. [See the comment above.]

A134941 Mountain numbers.

Original entry on oeis.org

1, 121, 131, 141, 151, 161, 171, 181, 191, 1231, 1241, 1251, 1261, 1271, 1281, 1291, 1321, 1341, 1351, 1361, 1371, 1381, 1391, 1421, 1431, 1451, 1461, 1471, 1481, 1491, 1521, 1531, 1541, 1561, 1571, 1581, 1591, 1621, 1631, 1641, 1651, 1671, 1681, 1691, 1721

Offset: 1

Omar E. Pol, Nov 22 2007

For n > 1 the structure of digits represents a mountain. The first digit is 1. The last digit is 1. The first digits are in increasing order. The last digits are in decreasing order. The numbers only have one largest digit. This sequence is finite. The last term is 12345678987654321.
The total number of terms is 21846. - Hans Havermann, Nov 25 2007
A002450(8) + 1 = 21846. - Reinhard Zumkeller, May 17 2010
From Reinhard Zumkeller, May 25 2010: (Start)
A178333 is the characteristic function of mountain numbers: A178333(a(n)) = 1;
A178334(n) is the number of mountain numbers <= n;
A178052 and A178053 give sums of digits and digital roots of mountain numbers;
A178051(n) is the peak value of the n-th mountain number. (End)

			The A-number of this sequence (A134941) is itself a mountain number:
  . . . 9 . .
  . . . . . .
  . . . . . .
  . . . . . .
  . . . . . .
  . . 4 . 4 .
  . 3 . . . .
  . . . . . .
  1 . . . . 1

Joshua Zucker, Table of n, a(n) for n = 1..21846 (shows all terms).

Cf. A134853, A135417, A134951, A182721.

Cf. A115300, A175044. - Reinhard Zumkeller, May 25 2010

import Data.List (elemIndices)
a134941 n = a134941_list !! (n-1)
a134941_list = elemIndices 1 a178333_list
-- Reinhard Zumkeller, Oct 28 2001

mountainQ[n_] := MatchQ[ IntegerDigits[n], {1, a___, b_, c___, 1} /; OrderedQ[{1, a, b}, Less] && OrderedQ[ Reverse[{b, c, 1}], Less]]; mountainQ[1] = True; Select[Range[2000], mountainQ] (* Jean-François Alcover, Jun 13 2012 *)
Prepend[Union @@ ((FromDigits@#&/@Flatten[Table[Join[(k=Prepend[#,1]&/@
Subsets[Range[2,#-1]])[[i]], {#}, (Reverse@# & /@k)[[j]]],
{i, 2^(# - 2)}, {j, 2^(# - 2)}], 1])&/@Range[9]), 1] (* Hans Rudolf Widmer, Apr 30 2024 *)

from itertools import product
def ups():
    d = "23456789"
    for b in product([0, 1], repeat=8):
        yield "1" + "".join(d[i]*b[i] for i in range(8))
def downsfrom(apex):
    if apex < 3: yield "1"*int(apex==2); return
    d = "8765432"[-(apex-2):]
    for b in product([0, 1], repeat=len(d)):
        yield "".join(d[i]*b[i] for i in range(len(d))) + "1"
def A134941(): # return full sequence as a list
    mountain_strs = (u+d for u in ups() for d in downsfrom(int(u[-1])))
    return sorted(int(ms) for ms in mountain_strs)
print(A134941()[:45]) # Michael S. Branicky, Dec 31 2021

A007059 Number of balanced ordered trees with n nodes.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 14, 24, 43, 77, 140, 256, 472, 874, 1628, 3045, 5719, 10780, 20388, 38674, 73562, 140268, 268066, 513350, 984911, 1892875, 3643570, 7023562, 13557020, 26200182, 50691978, 98182666, 190353370, 369393466, 717457656

Offset: 0

N. J. A. Sloane, Mira Bernstein, R. Kemp

Essentially the same as A079500: a(0)=0 and a(n)=A079500(n-1) for n >= 1.
Diagonal sums of the "postage stamp" array: for rows n >= -1, column m >= 0 is given by F(n,m) = F(n-1,m) + F(n-2,m) + ... + F(n-m,m) with F(0,m)=1 (m >= 0), F(n,m)=0 (n < 0) and F(n,0)=0 (n > 0). (Rows indicate the required sum, columns indicate the integers available {0,...,m}, entries F(n,m) indicate number of ordered ways sum can be achieved (e.g., n=3, m=2: 3 = 1+1+1 = 1+2 = 2+1 so F(3,2)=3 ways)). - Richard L. Ollerton
Conjecture: for n > 0, a(n+1) is the number of "numbral" divisors of (4^n-1)/3 = A002450(n) (see A048888 for the definition of numbral arithmetic). This has been verified computationally up to n=15. - John W. Layman, Dec 18 2001 [This conjecture follows immediately from Proposition 2.3 of Frosini and Rinaldi. - N. J. A. Sloane, Apr 29 2011]
Also number of Dyck paths of semi-length n-1 with all peaks at the same height. (not mentioned in Frosini reference) - David Scambler, Nov 19 2010
For n >= 1, a(n) is the number of compositions of n where all parts are smaller than the first part, see example. For n >= 1, a(n-1) = A079500(n) is the number of compositions of n where no part exceeds the first part, see the example in A079500. - Joerg Arndt, Dec 29 2012

			F(-1,0)=0 so a(0)=0. F(0,0)=1, F(-1,1)=0 so a(1)=1. F(1,0)=0, F(0,1)=1, F(-1,2)=0 so a(2)=1. F(2,0)=0, F(1,1)=1, F(0,2)=1, F(-1,3)=0 so a(3)=2.
From _Joerg Arndt_, Dec 29 2012: (Start)
There are a(8)=24 compositions p(1) + p(2) + ... + p(m) = 8 such that p(k) < p(1):
[ 1]  [ 2 1 1 1 1 1 1 ]
[ 2]  [ 3 1 1 1 1 1 ]
[ 3]  [ 3 1 1 1 2 ]
[ 4]  [ 3 1 1 2 1 ]
[ 5]  [ 3 1 2 1 1 ]
[ 6]  [ 3 1 2 2 ]
[ 7]  [ 3 2 1 1 1 ]
[ 8]  [ 3 2 1 2 ]
[ 9]  [ 3 2 2 1 ]
[10]  [ 4 1 1 1 1 ]
[11]  [ 4 1 1 2 ]
[12]  [ 4 1 2 1 ]
[13]  [ 4 1 3 ]
[14]  [ 4 2 1 1 ]
[15]  [ 4 2 2 ]
[16]  [ 4 3 1 ]
[17]  [ 5 1 1 1 ]
[18]  [ 5 1 2 ]
[19]  [ 5 2 1 ]
[20]  [ 5 3 ]
[21]  [ 6 1 1 ]
[22]  [ 6 2 ]
[23]  [ 7 1 ]
[24]  [ 8 ]
(End)

Arnold Knopfmacher and Neville Robbins, Compositions with parts constrained by the leading summand, Ars Combin. 76 (2005), 287-295.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..1000 (first 401 terms from T. D. Noe)
D. Applegate, M. LeBrun, and N. J. A. Sloane, Dismal Arithmetic, J. Int. Seq. 14 (2011) # 11.9.8.
A. Frosini and S. Rinaldi, On the Sequence A079500 and Its Combinatorial Interpretations, J. Integer Seq., Vol. 9 (2006), Article 06.3.1.
R. Kemp, Balanced ordered trees, Random Structures Algorithms, 5 (1994), pp. 99-121.
Index entries for sequences related to trees

Cf. A048888, A092921.

b:= proc(n, m) option remember; `if`(n=0, 1,
      `if`(m=0, add(b(n-j, j), j=1..n),
      add(b(n-j, min(n-j, m)), j=1..min(n, m))))
    end:
a:= n-> b(n-1, 0):
seq(a(n), n=0..40);  # Alois P. Heinz, May 01 2014

f[ n_, m_ ] := f[ n, m ]=Which[ n>0, Sum[ f[ n-i, m ], {i, 1, m} ], n<0, 0, n==0, 1 ] Table[ Sum[ f[ i, n-i ], {i, 0, n} ], {n, -1, 40} ]

from functools import cache
@cache
def F(k, n):
    return sum(F(k, n-j) for j in range(1, min(k, n))) if n > 1 else n
def A007059(n): return sum(F(k, n+1-k) for k in range(1, n+1))
print([A007059(n) for n in range(36)])  # Peter Luschny, Jan 05 2024

Define generalized Fibonacci numbers by Sum_{h>=0} F(p, h)z^n = z^(p-1)(1-z)/(1-2z+z^p+1). Then a(n) = 1 + Sum_{h=2..n} F(h-1, n-2).
G.f.: Sum_{k>0} x^k*(1 - 2*x + x^2 + (1-x)*x^(k+1))/(1 - 2*x + x^(k+1)). - Vladeta Jovovic, Feb 25 2003
G.f.: -(1 + x^2 + 1/(x-1))*(1 + x*(x-1)^3*(1-x+x^3)/(Q(0)- x*(x-1)^3*(1-x+x^3))), where Q(k) = (x+1)*(2*x-1)*(1-x)^2 + x^(k+2)*(x + x^2 + x^3 - 2*x^4 - 1 - x^(k+3) + x^(k+5)) - x*(-1 + 2*x - x^(k+3))*(1 - 2*x + x^2 + x^(k+4) - x^(k+5))*(-1 + 4*x - 5*x^2 + 2*x^3 - x^(k+2) - x^(k+5) + 2*x^(k+3) - x^(2*k+5) + x^(2*k+6))/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 14 2013
G.f.: Sum_{n>=1} q^n/(1-q*(1-q^n)/(1-q)) = Sum_{n>=1} q^n/(1 - Sum_{k=1..n} q^k ). - Joerg Arndt, Jan 03 2024

More terms from Vladeta Jovovic, Apr 08 2000

A053985 Replace 2^k with (-2)^k in binary expansion of n.

Original entry on oeis.org

0, 1, -2, -1, 4, 5, 2, 3, -8, -7, -10, -9, -4, -3, -6, -5, 16, 17, 14, 15, 20, 21, 18, 19, 8, 9, 6, 7, 12, 13, 10, 11, -32, -31, -34, -33, -28, -27, -30, -29, -40, -39, -42, -41, -36, -35, -38, -37, -16, -15, -18, -17, -12, -11, -14, -13, -24, -23, -26, -25, -20, -19

Offset: 0

Henry Bottomley, Apr 03 2000

Base 2 representation for n (in lexicographic order) converted from base -2 to base 10.
Maps natural numbers uniquely onto integers; within each group of positive values, maximum is in A002450; a(n)=n iff n can be written only with 1's and 0's in base 4 (A000695).
a(n) = A004514(n) - n. - Reinhard Zumkeller, Dec 27 2003
Schroeppel gives formula n = (a(n) + b) XOR b where b = binary ...101010, and notes this formula is reversible.  The reverse a(n) = (n XOR b) - b is a bit twiddle to transform 1 bits to -1.  Odd position 0 or 1 in n is flipped by "XOR b" to 1 or 0, then "- b" gives 0 or -1.  Only odd position 1's are changed, so b can be any length sure to cover those. - Kevin Ryde, Jun 26 2020

			a(9)=-7 because 9 is written 1001 base 2 and (-2)^3 + (-2)^0 = -8 + 1 = -7.
Or by Schroeppel's formula, b = binary 1010 then a(9) = (1001 XOR 1010) - 1010 = decimal -7. - _Kevin Ryde_, Jun 26 2020

Rémy Sigrist, Table of n, a(n) for n = 0..8191
Michael Beeler, R. William Gosper, and Richard Schroeppel, HAKMEM, MIT Artificial Intelligence Laboratory report AIM-239, February 1972, item 128 by Schroeppel, page 62. Also HTML transcription. Figure 7 path drawn 0, 1, -2, -1, 4, ... is the present sequence.
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, pp. 45, 49.
Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625 [math.CO], 2020-2021.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions

Inverse of A005351. Cf. A039724, A007088, A065369, A073791, A073792, A073793, A073794, A073795, A073796 and A073835.

f[n_Integer, b_Integer] := Block[{l = IntegerDigits[n]}, Sum[l[[ -i]]*(-b)^(i - 1), {i, 1, Length[l]}]]; a = Table[ FromDigits[ IntegerDigits[n, 2]], {n, 0, 80}]; b = {}; Do[b = Append[b, f[a[[n]], 2]], {n, 1, 80}]; b
(* Second program: *)
Array[FromDigits[IntegerDigits[#, 2], -2] &, 62, 0] (* Michael De Vlieger, Jun 27 2020 *)

a(n) = fromdigits(binary(n), -2) \\ Rémy Sigrist, Sep 01 2018

def A053985(n): return  -(b:=int('10'*(n.bit_length()+1>>1),2)) + (n^b) if n else 0 # Chai Wah Wu, Nov 18 2022

From Ralf Stephan, Jun 13 2003: (Start)
G.f.: (1/(1-x)) * Sum_{k>=0} (-2)^k*x^2^k/(1+x^2^k).
a(0) = 0, a(2*n) = -2*a(n), a(2*n+1) = -2*a(n)+1. (End)
a(n) = Sum_{k>=0} A030308(n,k)*A122803(k). - Philippe Deléham, Oct 15 2011
a(n) = (n XOR b) - b where b = binary ..101010 [Schroeppel].  Any b of this form (A020988) with bitlength(b) >= bitlength(n) suits. - Kevin Ryde, Jun 26 2020

A162795 Total number of toothpicks in the toothpick structure A139250 that are parallel to the initial toothpick, after n odd rounds.

Original entry on oeis.org

1, 5, 9, 21, 25, 37, 53, 85, 89, 101, 117, 149, 165, 201, 261, 341, 345, 357, 373, 405, 421, 457, 517, 597, 613, 649, 709, 793, 853, 965, 1173, 1365, 1369, 1381, 1397, 1429, 1445, 1481, 1541, 1621, 1637, 1673, 1733, 1817, 1877, 1989, 2197, 2389, 2405, 2441, 2501

Offset: 1

Omar E. Pol, Jul 14 2009

nonn

Partial sums of A162793.
Also, total number of ON cells at stage n of the two-dimensional cellular automaton defined as follows: replace every "vertical" toothpick of length 2 with a centered unit square "ON" cell, so we have a cellular automaton which is similar to both A147562 and A169707 (this is the "one-step bishop" version). For the "one-step rook" version we use toothpicks of length sqrt(2), then rotate the structure 45 degrees and then replace every toothpick with a unit square "ON" cell. For the illustration of the sequence as a cellular automaton we now have three versions: the original version with toothpicks, the one-step rook version and one-step bishop version. Note that the last two versions refer to the standard ON cells in the same way as the two versions of A147562 and the two versions of A169707. It appears that the graph of this sequence lies between the graphs of A147562 and A169707. Also, it appears that this sequence shares infinitely many terms with both A147562 and A169707, see Formula section and Example section. - Omar E. Pol, Feb 20 2015
It appears that this is also a bisection (the odd terms) of A255747.

			From _Omar E. Pol_, Feb 18 2015: (Start)
Written as an irregular triangle T(j,k), k>=1, in which the row lengths are the terms of A011782:
    1;
    5;
    9, 21;
   25, 37, 53, 85;
   89,101,117,149,165,201,261,341;
  345,357,373,405,421,457,517,597,613,649,709,793,853,965,1173,1365;
  ...
The right border gives the positive terms of A002450.
(End)
It appears that T(j,k) = A147562(j,k) = A169707(j,k), if k is a power of 2, for example: it appears that the three mentioned triangles only share the elements of the columns 1, 2, 4, 8, 16, ... - _Omar E. Pol_, Feb 20 2015

Michel Marcus, Table of n, a(n) for n = 1..10000
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS

Cf. A002450, A048645, A139250, A139251, A147562, A153000, A159791, A159792, A160164, A160552, A162793, A162794, A162796, A162797, A169707, A255263, A255264, A255747.

It appears that a(n) = A147562(n) = A169707(n), if n is a term of A048645, otherwise A147562(n) < a(n) < A169707(n). - Omar E. Pol, Feb 20 2015
It appears that a(n) = (A169707(2n) - 1)/4 = A255747(2n-1). - Omar E. Pol, Mar 07 2015
a(n) = 1 + 4*A255737(n-1). - Omar E. Pol, Mar 08 2015

More terms from N. J. A. Sloane, Dec 28 2009

A157154 Triangle T(n, k, m) = (m(n-k) + 1)T(n-1, k-1, m) + (mk + 1)T(n-1, k, m) - mk(n-k)*T(n-2, k-1, m) with T(n, 0, m) = T(n, n, m) = 1 and m = 3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 5, 1, 1, 21, 21, 1, 1, 85, 234, 85, 1, 1, 341, 2110, 2110, 341, 1, 1, 1365, 17163, 35882, 17163, 1365, 1, 1, 5461, 131751, 505979, 505979, 131751, 5461, 1, 1, 21845, 976876, 6395471, 11433118, 6395471, 976876, 21845, 1, 1, 87381, 7089360, 75400800, 220599330, 220599330, 75400800, 7089360, 87381, 1

Offset: 0

Roger L. Bagula, Feb 24 2009

			Triangle begins as:
  1;
  1,     1;
  1,     5,       1;
  1,    21,      21,        1;
  1,    85,     234,       85,         1;
  1,   341,    2110,     2110,       341,         1;
  1,  1365,   17163,    35882,     17163,      1365,        1;
  1,  5461,  131751,   505979,    505979,    131751,     5461,       1;
  1, 21845,  976876,  6395471,  11433118,   6395471,   976876,   21845,     1;
  1, 87381, 7089360, 75400800, 220599330, 220599330, 75400800, 7089360, 87381, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A007318 (m=0), A157152 (m=1), A157153 (m=2), this sequence (m=3), A157155 (m=4), A157156 (m=5).
Cf. A157147, A157148, A157149, A157150, A157151, A157207, A157208, A157209, A157210, A157211, A157212, A157268, A157272, A157273, A157274, A157275.
Cf. A002450.

T[n_, k_, m_]:= T[n, k, m]= If[k==0 || k==n, 1, (m*(n-k)+1)*T[n-1,k-1,m] + (m*k+1)*T[n-1,k,m] - m*k*(n-k)*T[n-2,k-1,m]];
Table[T[n,k,3], {n,0,10}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Jan 10 2022 *)

@CachedFunction
def T(n,k,m):  # A157154
    if (k==0 or k==n): return 1
    else: return (m*(n-k) +1)*T(n-1,k-1,m) + (m*k+1)*T(n-1,k,m) - m*k*(n-k)*T(n-2,k-1,m)
flatten([[T(n,k,3) for k in (0..n)] for n in (0..20)]) # G. C. Greubel, Jan 10 2022

T(n, k, m) = (m*(n-k) + 1)*T(n-1, k-1, m) + (m*k + 1)*T(n-1, k, m) - m*k*(n-k)*T(n-2, k-1, m) with T(n, 0, m) = T(n, n, m) = 1 and m = 1.
T(n, n-k, m) = T(n, k, m).
T(n, 1, 3) = A002450(n). - G. C. Greubel, Jan 10 2022

Edited by G. C. Greubel, Jan 10 2022

A016127 Expansion of g.f. 1/((1-2x)(1-5*x)).

Original entry on oeis.org

1, 7, 39, 203, 1031, 5187, 25999, 130123, 650871, 3254867, 16275359, 81378843, 406898311, 2034499747, 10172515119, 50862608363, 254313107351, 1271565667827, 6357828601279, 31789143530683, 158945718701991, 794728595607107, 3973642982229839, 19868214919537803, 99341074614466231

Offset: 0

N. J. A. Sloane

With leading zero, binomial transform of A002450. - Paul Barry, Apr 11 2003
The sequence of fractions a(n)/(n+1) is the 3rd binomial transform of the sequence of fractions J(n+1)/(n+1) where J(n) is A001045(n). - Paul Barry, Aug 05 2005
Equals term (1,2) in M^n, M = the 3 X 3 matrix [1, 1, 3; 1, 3, 1; 3, 1, 1]. a(n)/a(n-1) tends to 5, a root to the charpoly x^3 - 5*x^2 - 4*x + 20. - Gary W. Adamson, Mar 12 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Kalika Prasad, Munesh Kumari, Rabiranjan Mohanta, and Hrishikesh Mahato, The sequence of higher order Mersenne numbers and associated binomial transforms, arXiv:2307.08073 [math.NT], 2023.
Index entries for linear recurrences with constant coefficients, signature (7,-10).

Cf. A001045, A002450, A005057, A020989.

[(5^(n+1)-2^(n+1))/3: n in [0..30]]; // Vincenzo Librandi, Jun 08 2011

Join[{a=1,b=7},Table[c=7*b-10*a;a=b;b=c,{n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
CoefficientList[Series[1/((1 - 2 x) (1 - 5 x)), {x, 0, 19}], x] (* Michael De Vlieger, Jan 31 2018 *)
LinearRecurrence[{7,-10},{1,7},30] (* Harvey P. Dale, Aug 18 2020 *)

Vec(1/((1-2*x)*(1-5*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

[lucas_number1(n,7,10) for n in range(1, 21)] # Zerinvary Lajos, Apr 23 2009

[(5^n - 2^n)/3 for n in range(1,21)] # Zerinvary Lajos, Jun 05 2009

a(n) = (5^(n+1) - 2^(n+1))/3 = Sum_{i=0..n} 5^i*2^(n-1) = 5*a(n-1) + 2^n = 2*a(n-1) + 5^n. - Henry Bottomley, Apr 07 2003
Binomial transform of A020989. - Paul Barry, May 18 2003
From Paul Barry, Aug 05 2005: (Start)
a(n) = Sum_{k=0..n} Sum_{j=0..n} 5^(n-j)*binomial(j,k).
a(n) = Sum_{k=0..n} 2^k*5^(n-k) = Sum_{k=0..n} 5^k*2^(n-k). (End)
For n > 2, a(n) = 9*a(n-1) - 24*a(n-2) + 20*a(n-3). - Gary W. Adamson, Dec 26 2007
From Elmo R. Oliveira, Mar 31 2025: (Start)
E.g.f.: exp(2*x)*(5*exp(3*x) - 2)/3.
a(n) = A005057(n+1)/3.
a(n) = 7*a(n-1) - 10*a(n-2). (End)

A187220 Gullwing sequence (see Comments lines for precise definition).

Original entry on oeis.org

0, 1, 3, 7, 15, 23, 31, 47, 71, 87, 95, 111, 135, 159, 191, 247, 311, 343, 351, 367, 391, 415, 447, 503, 567, 607, 639, 695, 767, 847, 967, 1143, 1303, 1367, 1375, 1391, 1415, 1439, 1471, 1527, 1591, 1631, 1663, 1719, 1791, 1871, 1991, 2167, 2327, 2399, 2431

Offset: 0

Omar E. Pol, Mar 07 2011

The Gullwing (or G-toothpick) sequence is a special type of toothpick sequence. It appears that this is a superstructure of A139250.
We define a "G-toothpick" to consist of two arcs of length Pi/2 forming a "gullwing" whose total length is equal to Pi = 3.141592...
A gullwing-shaped toothpick or G-toothpick or simply "gull" is formed by two Q-toothpicks (see A187210).
A G-toothpick has a midpoint and two endpoints. An endpoint is said to be "exposed" if it is not the midpoint or endpoint of any other G-toothpick.
The sequence gives the number of G-toothpicks in the structure after n stages. A187221 (the first differences) gives the number of G-toothpicks added at n-th stage.
a(n) is also the diameter of a circle whose circumference equals the total length of all gulls in the gullwing structure after n stages.
It appears that the gullwing pattern has a recursive, fractal-like structure. The animation shows the fractal-like behavior.
Note that the structure contains many different types of geometrical figures, for example: circles, hearts, etc. All figures are formed by arcs.
It appears that there are infinitely many types of circular shapes, which are related to the rectangles of the toothpick structure of A139250.
It also appears that the structure contains a nice pattern formed by distinct modular substructures: one central cross surrounded by several asymmetrical crosses (or "hidden crosses") of distinct sizes, and also several "nuclei" of crosses. This pattern is essentially similar to the crosses of A139250 but here the structure is harder to see. For example, consider the nucleus of a cross; in the toothpick structure a nucleus is formed by two squares and two rectangles but here a nucleus is formed by two circles and two hearts.
It appears furthermore that this structure has connections with the square-cross fractal and with the T-square fractal, just as in the case of the toothpick structure of A139250.
For more information see A139250 and A187210.
It appears this is also the connection between A147562 (the Ulam-Warburton cellular automaton) and the toothpick sequence A139250. The behavior of the function is similar to A147562 but here the structure is more complex. (see Plot 2 button: A147562 vs A187220). - Omar E. Pol, Mar 11 2011, Mar 13 2011
From Omar E. Pol, Mar 25 2011: (Start)
If we remove the first gull of the structure so we can see that there is a correspondence between the gullwing structure and the I-toothpick structure of A139250, for example: a pair of opposite gulls in horizontal position in the gullwing structure is equivalent to a vertical I-toothpick with length 4 in the I-toothpick structure, such that the midpoint of each horizontal gull coincides with the midpoint of each vertical toothpick of the I-toothpick. See A160164.
Also, B-toothpick sequence. We define a "B-toothpick" to consist of four arcs of length Pi/2 forming a "bell" similar to the Gauss function. A Bell-shaped toothpick or B-toothpick or simply "bell" is formed by four Q-toothpicks (see A187210). A B-toothpick has length 2*Pi.  The sequence gives the number of B-toothpicks in the structure after n stages.
Also, if we remove the first bell of the structure, we can find a correspondence between this structure and the I-toothpick structure of A139250. In this case, for example, a pair of opposite bells in horizontal position is equivalent to a vertical I-toothpick with length 8 in the I-toothpick structure, such that the midpoint of each horizontal bell coincides with the midpoint of each vertical toothpick of the I-toothpick. See A160164.
Also, for this sequence there is a third structure formed by isosceles right triangles since gulls or bells can be replaced by these triangles.
Note that the size of the gulls, bells and triangles can be adjusted such that two or three of these structures can be overlaid.
(End)
Also, it appears that if we let k=floor(log_2(n)), then for n >= 1, a(2^k) = (4^(k+1) + 5)/3 - 2^(k+1).  Otherwise, a(n)=(4^(k+1) + 5)/3 + 8*A153006(n-1-2^k). - Christopher Hohl, Dec 19 2018

			On the infinite square grid we start at stage 0 with no G-toothpicks, so a(0) = 0.
At stage 1 we place a G-toothpick:
Midpoint : (0,-1)
Endpoints: (-1,0) and (1,0)
So a(1) = 1. There are two exposed endpoints.
At stage 2 we place two G-toothpicks:
Midpoint of the left G-toothpick : (-1,0)
Endpoints of the left G-toothpick: (-2,1) and (-2,-1)
Midpoint of the right G-toothpick : (1,0)
Endpoints of the right G-toothpick: (2,1) and (2,-1)
So a(2) = 1+2 = 3. There are four exposed endpoints.
And so on...

Iain Fox, Table of n, a(n) for n = 0..10000
David Applegate, The movie version
David Applegate, Illustration for a(16) = 311 [Created from the movie version, see previous link]
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
Brady Haran and N. J. A. Sloane, Terrific Toothpick Patterns, Numberphile video (2018).
Omar E. Pol, Illustration of initial terms
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS

Cf. A002450, A139250, A147562, A160164, A187210, A187221, A187222.

Join[{0, 1}, Rest[CoefficientList[Series[(2 x / ((1 - x) (1 + 2 x))) (1+2 x Product[1 + x^(2^k - 1) + 2 x^(2^k), {k, 0, 20}]), {x, 0, 53}], x] + 1 ]] (* Vincenzo Librandi, Jul 02 2017 *)

A139250(n) = my(msb(m) = 2^(#binary(m)-1), k = (2*msb(n)^2 + 1) / 3); if(n==msb(n), k , k + 2*A139250(n-msb(n)) + A139250(n - msb(n) + 1) - 1)
a(n) = if(n<2, n, 1 + 2*A139250(n-1)) \\ Iain Fox, Dec 10 2018

def msb(n):
    t=0
    while n>>t>0: t+=1
    return 2**(t - 1)
def a139250(n):
    k=(2*msb(n)**2 + 1)//3
    return 0 if n==0 else k if n==msb(n) else k + 2*a139250(n - msb(n)) + a139250(n - msb(n) + 1) - 1
def a(n): return 0 if n==0 else 1 + 2*a139250(n - 1)
print([a(n) for n in range(101)]) # Indranil Ghosh, Jul 01 2017

a(n) = 1 + 2*A139250(n-1), for n >= 1.
a(n) = 1 + A160164(n-1), for n >= 1. - [Suggested by Omar E. Pol, Mar 13 2011, proved by Nathaniel Johnston, Mar 22 2011]
The formula involving A160164 can be seen by identifying a Gullwing in the n-th generation (n >= 2) with midpoint at (x,y) and endpoints at (x-1,y+1) and (x+1,y+1) with a toothpick in the (n-1)st generation with endpoints at (x,y-1) and (x,y+1) -- this toothpick from (x,y-1) to (x,y+1) should be considered as having length ONE (i.e., it is HALF of an I-toothpick). The formula involving A139250 follows as a result of the relationship between A139250 and A160164.
a(n) = A147614(n-1) + A160124(n-1), n >= 2. - Omar E. Pol, Feb 15 2013
a(n) = 7 + 8*A153000(n-3), n >= 3. - Omar E. Pol, Feb 16 2013

A282629 Sheffer triangle (exp(x), exp(3*x) - 1). Named S2[3,1].

Original entry on oeis.org

1, 1, 3, 1, 15, 9, 1, 63, 108, 27, 1, 255, 945, 594, 81, 1, 1023, 7380, 8775, 2835, 243, 1, 4095, 54729, 109890, 63180, 12393, 729, 1, 16383, 395388, 1263087, 1151010, 387828, 51030, 2187, 1, 65535, 2816865, 13817034, 18752391, 9658278, 2133054, 201204, 6561, 1, 262143, 19914660, 146620935, 285232185, 210789621, 69502860, 10825650, 767637, 19683

Offset: 0

Wolfdieter Lang, Apr 03 2017

For Sheffer triangles (infinite lower triangular exponential convolution matrices) see the W. Lang link under A006232, with references).
The e.g.f. for the sequence of column m is (Sheffer property) exp(x)*(exp(3*x) - 1)^m/m!.
This is a generalization of the Sheffer triangle Stirling2(n, m) = A048993(n, m) denoted by (exp(x), exp(x)-1), which could be named S2[1,0].
The a-sequence for this Sheffer triangle has e.g.f. 3*x/log(1+x) and is 3*A006232(n)/ A006233(n) (Cauchy numbers of the first kind).
The z-sequence has e.g.f. (3/(log(1+x)))*(1 - 1/(1+x)^(1/3)) and is A284857(n) / A284858(n).
The main diagonal gives A000244.
The row sums give A284859. The alternating row sums give A284860.
The triangle appears in the o.g.f. G(n, x) of the sequence {(1 + 3*m)^n}{m>=0}, as G(n, x) = Sum{m=0..n} T(n, m)*m!*x^m/(1-x)^(m+1), n >= 0. Hence the corresponding e.g.f. is, by the linear inverse Laplace transform, E(n, t) = Sum_{m >=0}(1 + 3*m)^n t^m/m! = exp(t)*Sum_{m=0..n} T(n, m)*t^m.
The corresponding Euler triangle with reversed rows is rEu(n, k) = Sum_{m=0..k} (-1)^(k-m)*binomial(n-m, k-m)*T(n, k)*k!, 0 <= k <= n. This is A225117 with row reversion.
The first column k sequences divided by 3^k are A000012, A002450 (with a leading 0), A016223, A021874. For the e.g.f.s and o.g.f.s see below. - Wolfdieter Lang, Apr 09 2017
From Wolfdieter Lang, Aug 09 2017: (Start)
The general row polynomials R(d,a;n,x) = Sum_{k=0..n} T(d,a;n,m)*x^m of the Sheffer triangle S2[d,a] satisfy, as special polynomials of the Boas-Buck class, the identity (see the reference, and we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function)
  (E_x - n*1)*R(d,a;n,x) = - n*a*R(d,a;n-1,x) - Sum_{k=0..n-1} binomial(n, k+1)*(-d)^(k+1)*Bernoulli(k+1)*E_x*R(d,a;n-1-k,x), with E_x = x*d/dx (Euler operator).
This entails a recurrence for the sequence of column m, for n > m:
  T(d,a;n,m) = (1/(n - m))*[(n/2)*(2*a + d*m)*T(d,a;n-1,m) + m*Sum_{p=m..n-2} binomial(n,p)(-d)^(n-p)*Bernoulli(n-p)*T(d,a;p,m)], with input T(d,a;n,n) = d^n. For the present [d,a] = [3,1] case see the formula and example sections below. - Wolfdieter Lang, Aug 09 2017 (End)
The inverse of this triangular Sheffer matrix S2[3,1] is S1[3,1] with rational elements S1[3,1](n, k) = (-1)^(n-k)*A286718(n, k)/3^k. - Wolfdieter Lang, Nov 15 2018
Named after the American mathematician Isador Mitchell Sheffer (1901-1992). - Amiram Eldar, Jun 19 2021

			The triangle T(n, m) begins:
  n\m 0      1        2         3         4         5        6        7      8     9
  0:  1
  1:  1      3
  2:  1     15        9
  3:  1     63      108        27
  4:  1    255      945       594        81
  5:  1   1023     7380      8775      2835       243
  6:  1   4095    54729    109890     63180     12393      729
  7:  1  16383   395388   1263087   1151010    387828    51030     2187
  8:  1  65535  2816865  13817034  18752391   9658278  2133054   201204   6561
  9:  1 262143 19914660 146620935 285232185 210789621 69502860 10825650 767637 19683
  ...
------------------------------------------------------------------------------------
Nontrivial recurrence for m=0 column from z-sequence: T(4,0) = 4*(1*1 + 63*(-1/6) + 108*(11/54) + 27*(-49/108)) = 1.
Recurrence for m=2 column from a-sequence: T(4, 2) = (4/2)*(1*63*3 + 2*108*(3/2) + 3*27*(-3/6)) = 945.
Recurrence for row polynomial R(3, x) (Meixner type): ((3*x + 1) + 3*x*d_x)*(1 + 15*x + 9*x^2) = 1 + 63*x + 108*x^2 + 27*x^3.
E.g.f. and o.g.f. of n = 1 powers {(1 + 3*m)^1}_{m>=0} A016777: E(1, x) = exp(x) * (T(1, 0) + T(1, 1)*x) = exp(x)*(1+3*x). O.g.f.: G(1, x) = T(1, 0)*0!/(1-x) + T(1, 1)*1!*x/(1-x)^2 = (1+2*x)/(1-x)^2.
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (1/2)*(2*(2 + 3*2)*T(3, 2) + 2*6*(-3)^2*bernoulli(2)*T(2, 2)) = (1/2)*(16*108 + 12*9*(1/6)*9) = 945. - _Wolfdieter Lang_, Aug 09 2017

Ralph P. Boas, Jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Michael De Vlieger, Table of n, a(n) for n = 0..11475, rows n = 0..150, flattened.
Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 9.
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.
Wolfdieter Lang, On Generating functions of Diagonals Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.

Cf. A000012, A000244, A006232/A006233, A016777, A024036, A111577, A225117, A225466, A284857, A284858, A284859, A284860, A284861, A286718.

Table[Sum[Binomial[m, k] (-1)^(k - m) (1 + 3 k)^n/m!, {k, 0, m}], {n, 0, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Apr 08 2017 *)

T(n, m) = sum(k=0, m, binomial(m, k) * (-1)^(k - m) * (1 + 3*k)^n/m!);
for(n=0, 9, for(m=0, n, print1(T(n, m),", ");); print();) \\ Indranil Ghosh, Apr 08 2017

A nontrivial recurrence for the column m=0 entries T(n, 0) = 1 from the z-sequence given above: T(n,0) = n*Sum_{j=0..n-1} z(j)*T(n-1,j), n >= 1, T(0, 0) = 1.
Recurrence for column m >= 1 entries from the a-sequence given above: T(n, m) = (n/m)*Sum_{j=0..n-m} binomial(m-1+j, m-1)*a(j)*T(n-1, m-1+j), m >= 1.
Recurrence for row polynomials R(n, x) (Meixner type): R(n, x) = ((3*x+1) + 3*x*d_x)*R(n-1, x), with differentiation d_x, for n >= 1, with input R(0, x) = 1.
T(n, m) = Sum_{k=0..m} binomial(m,k)*(-1)^(k-m)*(1 + 3*k)^n/m!, 0 <= m <= n.
E.g.f. of triangle: exp(z)*exp(x*(exp(3*z)-1)) (Sheffer type).
E.g.f. for sequence of column m is exp(x)*((exp(3*x) - 1)^m)/m! (Sheffer property).
From Wolfdieter Lang, Apr 09 2017: (Start)
Standard three-term recurrence: T(n, m) = 0 if n < m, T(n,-1) = 0, T(0, 0) = 1, T(n, m) = 3*T(n-1, m-1) + (1+3*m)*T(n-1, m) for n >= 1. From the T(n, m) formula. Compare with the recurrence of S2[3,2] given in A225466.
The o.g.f. for sequence of column m is 3^m*x^m/Product_{j=0..m} (1 - (1+3*j)*x). (End)
In terms of Stirling2 = A048993: T(n, m) = Sum_{k=0..n} binomial(n, k)* 3^k*Stirling2(k, m), 0 <= m <= n. - Wolfdieter Lang, Apr 13 2017
Boas-Buck recurrence for column sequence m: T(n, m) = (1/(n - m))*((n/2)*(2 + 3*m)*T(n-1, m) + m*Sum_{p=m..n-2} binomial(n,p)*(-3)^(n-p)*Bernoulli(n-p)*T(p, m)), for n > m >= 0, with input T(m, m) = 3^m. See a comment above. - Wolfdieter Lang, Aug 09 2017

A014825 a(n) = 4*a(n-1) + n with n > 1, a(1)=1.

Original entry on oeis.org

1, 6, 27, 112, 453, 1818, 7279, 29124, 116505, 466030, 1864131, 7456536, 29826157, 119304642, 477218583, 1908874348, 7635497409, 30541989654, 122167958635, 488671834560, 1954687338261, 7818749353066

Offset: 1

N. J. A. Sloane

			G.f. = x + 6*x^2 + 27*x^3 + 112*x^4 + 453*x^5 + 1818*x^6 + 7279*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See pp. 9, 18.
Hacène Belbachir and El-Mehdi Mehiri, Enumerating moves in the optimal solution of the Tower of Hanoi, arXiv:2210.08657 [math.CO], 2022.
László Tóth, On Schizophrenic Patterns in b-ary Expansions of Some Irrational Numbers, arXiv:2002.06584 [math.NT], 2020. See also Proc. Amer. Math. Soc. 148 (2020), 461-469.
Index entries for linear recurrences with constant coefficients, signature (6,-9,4).

Cf. A002450 (first differences), A052161 (partial sums).
Cf. A001045, A006314, A014916, A053142, A091919.
Cf. A171654 (mod 10).

[(4^(n+1)-3*n-4)/9: n in [1..30]]; // Vincenzo Librandi, Aug 23 2011

RecurrenceTable[{a[1]==1,a[n]==4a[n-1]+n},a[n],{n,30}] (* Harvey P. Dale, Oct 12 2011 *)
a[ n_]:= SeriesCoefficient[x/((1-4x)(1-x)^2), {x, 0, n}] (* Michael Somos, Jun 20 2012 *)

{a(n) = polcoeff( x / ((1 - x)^2 * (1 - 4*x)) + x * O(x^n), n)} /* Michael Somos, Jun 20 2012 */

def A014825(n): return (((1<<(n+1<<1))-4)//3-n)//3 # Chai Wah Wu, Nov 12 2024

[(4^(n+1) -3*n -4)/9 for n in (1..30)] # G. C. Greubel, Feb 18 2020

a(n) = (4^(n+1) - 3*n - 4)/9.
G.f.: x/((1-4*x)*(1-x)^2).
a(n) = Sum_{k=0..n} (n-k)*4^k = Sum_{k=0..n} k*4^(n-k). - Paul Barry, Jul 30 2004
a(n) = Sum_{k=0..n} binomial(n+2, k+2)*3^k [Offset 0]. - Paul Barry, Jul 30 2004
a(n) = Sum_{k=0..n} Sum_{j=0..2k} (-1)^(j+1)*J(j)*J(2k-j), J(n) = A001045(n). - Paul Barry, Oct 23 2009
Convolution square of A006314. - Michael Somos, Jun 20 2012
E.g.f.: (4*exp(4*x) - (4+3*x)*exp(x))/9. - G. C. Greubel, Feb 18 2020
a(n) = A014916(-n-1)*4^(n+1) = A091919(2*n-2) for all n in Z. - Michael Somos, Oct 02 2020
a(n) = Sum_{k=0..n} A002450(k). - Joseph Brown, May 11 2021
Last digits give A171654. - Paul Curtz, Oct 10 2021

cp's OEIS Frontend

A253563 Permutation of natural numbers: a(0) = 1, a(1) = 2; after which, a(2n) = A253560(a(n)), a(2n+1) = A253550(a(n)).

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A134941 Mountain numbers.

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A007059 Number of balanced ordered trees with n nodes.

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A053985 Replace 2^k with (-2)^k in binary expansion of n.

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A162795 Total number of toothpicks in the toothpick structure A139250 that are parallel to the initial toothpick, after n odd rounds.

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A157154 Triangle T(n, k, m) = (m*(n-k) + 1)*T(n-1, k-1, m) + (m*k + 1)*T(n-1, k, m) - m*k*(n-k)*T(n-2, k-1, m) with T(n, 0, m) = T(n, n, m) = 1 and m = 3, read by rows.

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A016127 Expansion of g.f. 1/((1-2*x)*(1-5*x)).

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A157154 Triangle T(n, k, m) = (m(n-k) + 1)T(n-1, k-1, m) + (mk + 1)T(n-1, k, m) - mk(n-k)*T(n-2, k-1, m) with T(n, 0, m) = T(n, n, m) = 1 and m = 3, read by rows.

A016127 Expansion of g.f. 1/((1-2x)(1-5*x)).