A051340 -id:A051340 - cp's OEIS frontend

A130296 Triangle read by rows: T[i,1]=i, T[i,j]=1 for 1 < j <= i = 1,2,3,...

1, 2, 1, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Gary W. Adamson, May 20 2007

This sequence was formerly named "reversals of A051340", but it is actually the truncation of A051340 to its lower left triangular part, re-indexed to start rows and columns with 1. - M. F. Hasler, Aug 15 2015

			First few rows of the triangle are:
1;
2, 1;
3, 1, 1;
4, 1, 1, 1;
5, 1, 1, 1, 1;
...

Cf. A051340, A130297, A005408 (row sums).

A130296(i,j)=if(j==1,i,j<=i) \\ The sequence should not be defined for j>i but it is used in several places as infinite square matrix with upper right part equal to zero. - M. F. Hasler, Aug 15 2015

from math import isqrt
def A130296(n): return comb((m:=isqrt(k:=n<<1))+(k>m*(m+1)),2)-comb((m2:=isqrt(k-2))+(k-2>m2*(m2+1)),2)+1 # Chai Wah Wu, Nov 09 2024

Truncation of A051340 to its lower left. By rows, "n" followed by (n-1) 1's. (1,2,3...) in the left border, all 1's in other columns.
A130296^2 = A130297.
a(n) = A004201(n) - A004201(n-1) for n>1. - Reinhard Zumkeller, Jul 16 2008

A141419 Triangle read by rows: T(n, k) = A000217(n) - A000217(n - k) with 1 <= k <= n.

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 4, 7, 9, 10, 5, 9, 12, 14, 15, 6, 11, 15, 18, 20, 21, 7, 13, 18, 22, 25, 27, 28, 8, 15, 21, 26, 30, 33, 35, 36, 9, 17, 24, 30, 35, 39, 42, 44, 45, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55

Offset: 1

Roger L. Bagula, Aug 05 2008

As a rectangle, the accumulation array of A051340.
From Clark Kimberling, Feb 05 2011: (Start)
Here all the weights are divided by two where they aren't in Cahn.
As a rectangle, A141419 is in the accumulation chain
... < A051340 < A141419 < A185874 < A185875 < A185876 < ...
(See A144112 for the definition of accumulation array.)
row 1: A000027
col 1: A000217
diag (1,5,...): A000326 (pentagonal numbers)
diag (2,7,...): A005449 (second pentagonal numbers)
diag (3,9,...): A045943 (triangular matchstick numbers)
diag (4,11,...): A115067
diag (5,13,...): A140090
diag (6,15,...): A140091
diag (7,17,...): A059845
diag (8,19,...): A140672
(End)
Let N=2*n+1 and k=1,2,...,n. Let A_{N,n-1} = [0,...,0,1; 0,...,0,1,1; ...; 0,1,...,1; 1,...,1], an n X n unit-primitive matrix (see [Jeffery]). Let M_n=[A_{N,n-1}]^4. Then t(n,k)=[M_n](1,k), that is, the n-th row of the triangle is given by the first row of M_n. - _L. Edson Jeffery, Nov 20 2011
Conjecture. Let N=2*n+1 and k=1,...,n. Let A_{N,0}, A_{N,1}, ..., A_{N,n-1} be the n X n unit-primitive matrices (again see [Jeffery]) associated with N, and define the Chebyshev polynomials of the second kind by the recurrence U_0(x) = 1, U_1(x) = 2*x and U_r(x) = 2*x*U_(r-1)(x) - U_(r-2)(x)  (r>1). Define the column vectors V_(k-1) = (U_(k-1)(cos(Pi/N)), U_(k-1)(cos(3*Pi/N)), ..., U_(k-1)(cos((2*n-1)*Pi/N)))^T, where T denotes matrix transpose. Let S_N = [V_0, V_1, ..., V_(n-1)] be the n X n matrix formed by taking V_(k-1) as column k-1. Let X_N = [S_N]^T*S_N, and let [X_N](i,j) denote the entry in row i and column j of X_N, i,j in {0,...,n-1}. Then t(n,k) = [X_N](k-1,k-1), and row n of the triangle is given by the main diagonal entries of X_N. Remarks: Hence t(n,k) is the sum of squares t(n,k) = sum[m=1,...,n (U_(k-1)(cos((2*m-1)*Pi/N)))^2]. Finally, this sequence is related to A057059, since X_N = [sum_{m=1,...,n} A057059(n,m)*A_{N,m-1}] is also an integral linear combination of unit-primitive matrices from the N-th set. - L. Edson Jeffery, Jan 20 2012
Row sums: n*(n+1)*(2*n+1)/6. - L. Edson Jeffery, Jan 25 2013
n-th row = partial sums of n-th row of A004736. - Reinhard Zumkeller, Aug 04 2014
T(n,k) is the number of distinct sums made by at most k elements in {1, 2, ... n}, for 1 <= k <= n, e.g., T(6,2) = the number of distinct sums made by at most 2 elements in {1,2,3,4,5,6}. The sums range from 1, to 5+6=11. So there are 11 distinct sums. - Derek Orr, Nov 26 2014
A number n occurs in this sequence A001227(n) times, the number of odd divisors of n, see A209260. - Hartmut F. W. Hoft, Apr 14 2016
Conjecture: 2*n + 1 is composite if and only if gcd(t(n,m),m) != 1, for some m. - L. Edson Jeffery, Jan 30 2018
From Peter Munn, Aug 21 2019 in respect of the sequence read as a triangle: (Start)
A number m can be found in column k if and only if A286013(m, k) is nonzero, in which case m occurs in column k on row A286013(m, k).
The first occurrence of m is in row A212652(m) column A109814(m), which is the rightmost column in which m occurs. This occurrence determines where m appears in A209260. The last occurrence of m is in row m column 1.
Viewed as a sequence of rows, consider the subsequences (of rows) that contain every positive integer. The lexicographically latest of these subsequences consists of the rows with row numbers in A270877; this is the only one that contains its own row numbers only once.
(End)

			As a triangle:
   1,
   2,  3,
   3,  5,  6,
   4,  7,  9, 10,
   5,  9, 12, 14, 15,
   6, 11, 15, 18, 20, 21,
   7, 13, 18, 22, 25, 27, 28,
   8, 15, 21, 26, 30, 33, 35, 36,
   9, 17, 24, 30, 35, 39, 42, 44, 45,
  10, 19, 27, 34, 40, 45, 49, 52, 54, 55;
As a rectangle:
   1   2   3   4   5   6   7   8   9  10
   3   5   7   9  11  13  15  17  19  21
   6   9  12  15  18  21  24  27  30  33
  10  14  18  22  26  30  34  38  42  46
  15  20  25  30  35  40  45  50  55  60
  21  27  33  39  45  51  57  63  69  75
  28  35  42  49  56  63  70  77  84  91
  36  44  52  60  68  76  84  92 100 108
  45  54  63  72  81  90  99 108 117 126
  55  65  75  85  95 105 115 125 135 145
Since the odd divisors of 15 are 1, 3, 5 and 15, number 15 appears four times in the triangle at t(3+(5-1)/2, 5) in column 5 since 5+1 <= 2*3, t(5+(3-1)/2, 3), t(1+(15-1)/2, 2*1) in column 2 since 15+1 > 2*1, and t(15+(1-1)/2, 1). - _Hartmut F. W. Hoft_, Apr 14 2016

R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Johann Cigler, Some elementary observations on Narayana polynomials and related topics, arXiv:1611.05252 [math.CO], 2016. See p. 24.
Carlton Gamer, David W. Roeder, and John J. Watkins, Trapezoidal Numbers, Mathematics Magazine 58:2 (1985), pp. 108-110.
L. E. Jeffery, Unit-primitive matrices
M. A. Nyblom, On the representation of the integers as a difference of nonconsecutive triangular numbers, Fibonacci Quarterly 39:3 (2001), pp. 256-263.

Cf. A000330 (row sums), A004736, A057059, A070543.
A144112, A051340, A141419, A185874, A185875, A185876 are accumulation chain related.
A141418 is a variant.
Cf. A001227, A209260. - Hartmut F. W. Hoft, Apr 14 2016
A109814, A212652, A270877, A286013 relate to where each natural number appears in this sequence.
A000027, A000217, A000326, A005449, A045943, A059845, A115067, A140090, A140091, A140672 are rows, columns or diagonals - refer to comments.

a141419 n k =  k * (2 * n - k + 1) `div` 2
a141419_row n = a141419_tabl !! (n-1)
a141419_tabl = map (scanl1 (+)) a004736_tabl
-- Reinhard Zumkeller, Aug 04 2014

a:=(n,k)->k*n-binomial(k,2): seq(seq(a(n,k),k=1..n),n=1..12); # Muniru A Asiru, Oct 14 2018

T[n_, m_] = m*(2*n - m + 1)/2; a = Table[Table[T[n, m], {m, 1, n}], {n, 1, 10}]; Flatten[a]

t(n,m) = m*(2*n - m + 1)/2.
t(n,m) = A000217(n) - A000217(n-m). - L. Edson Jeffery, Jan 16 2013
Let v = d*h with h odd be an integer factorization, then v = t(d+(h-1)/2, h) if h+1 <= 2*d, and v = t(d+(h-1)/2, 2*d) if h+1 > 2*d; see A209260. - Hartmut F. W. Hoft, Apr 14 2016
G.f.: y*(-x + y)/((-1 + x)^2*(-1 + y)^3). - Stefano Spezia, Oct 14 2018
T(n, 2) = A060747(n) for n > 1. T(n, 3) = A008585(n - 1) for n > 2. T(n, 4) = A016825(n - 2) for n > 3. T(n, 5) = A008587(n - 2) for n > 4. T(n, 6) = A016945(n - 3) for n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7.r n > 5. T(n, 7) = A008589(n - 3) for n > 6. T(n, 8) = A017113(n - 4) for n > 7. T(n, 9) = A008591(n - 4) for n > 8. T(n, 10) = A017329(n - 5) for n > 9. T(n, 11) = A008593(n - 5) for n > 10.  T(n, 12) = A017593(n - 6) for n > 11. T(n, 13) = A008595(n - 6) for n > 12. T(n, 14) = A147587(n - 7) for n > 13. T(n, 15) = A008597(n - 7) for n > 14. T(n, 16) = A051062(n - 8) for n > 15. T(n, 17) = A008599(n - 8) for n > 16. - Stefano Spezia, Oct 14 2018
T(2*n-k, k) = A070543(n, k). - Peter Munn, Aug 21 2019

Simpler name by Stefano Spezia, Oct 14 2018

A131821 Triangle read by rows: row n consists of n followed by (n-2) ones then n.

Original entry on oeis.org

1, 2, 2, 3, 1, 3, 4, 1, 1, 4, 5, 1, 1, 1, 5, 6, 1, 1, 1, 1, 6, 7, 1, 1, 1, 1, 1, 7, 8, 1, 1, 1, 1, 1, 1, 8, 9, 1, 1, 1, 1, 1, 1, 1, 9, 10, 1, 1, 1, 1, 1, 1, 1, 1, 10, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 14, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 14

Offset: 1

Gary W. Adamson, Jul 19 2007

			First few rows of the triangle are:
  1;
  2, 2;
  3, 1, 3;
  4, 1, 1, 4;
  5, 1, 1, 1, 5;
  6, 1, 1, 1, 1, 6;
  7, 1, 1, 1, 1, 1, 7;
  ...

Row sums: A016777.

Cf. A130296, A051340, A016777, A131830.

A131821 := proc(n,c) if c=1 or c=n then n ; else 1 ; fi ; end: for n from 1 to 16 do for c from 1 to n do printf("%d,",A131821(n,c)) ; od: od: # R. J. Mathar, May 30 2008

T[n_, k_] := If[k == 1 || k == n, n, 1];
Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 05 2023 *)

T(n,k) := if k = 1 or k = n then n else 1$
create_list(T(n, k), n, 1, 12, k, 1, n); /* Franck Maminirina Ramaharo, Dec 19 2018 */

from math import isqrt
def A131821(n): return m+(k>r) if 0<=(k:=n<<1)-(r:=(m:=isqrt(k))*(m+1))<=2 or n<=2 else 1 # Chai Wah Wu, Nov 07 2024

T(n,k) = A130296(n,k) + A051340(k-1,n-k) - 1.
T(n,1) = T(n,n) = n for n >= 1, and T(n,k) = 1 for 2 <= k <= n - 1, n >= 3.
From Franck Maminirina Ramaharo, Dec 19 2018: (Start)
T(n,k) = A130296(n,k) + A130296(n,n-k+1) - 1.
G.f.: y*(x - 3*x^2*y^2 + (x^2 + x^3)*y^3)/((1 - y)^2*(1 - x*y)^2).
E.g.f.: (1 - x^2 +(x^2 + (x - x^2)*y)*exp(y) - (1 - (x - x^2)*y)*exp(x*y))/(1 - x). (End)

More terms from R. J. Mathar, May 30 2008

A063656 Numbers k such that the truncated square root of k is equal to the rounded square root of k.

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 9, 10, 11, 12, 16, 17, 18, 19, 20, 25, 26, 27, 28, 29, 30, 36, 37, 38, 39, 40, 41, 42, 49, 50, 51, 52, 53, 54, 55, 56, 64, 65, 66, 67, 68, 69, 70, 71, 72, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 121

Offset: 0

Floor van Lamoen, Jul 24 2001

Also: take 1, skip 0, take 2, skip 1, take 3, skip 2, ...
The union of sets of numbers in closed intervals [k^2,k^2+k], k >= 0, intervals 0 to 1, 1 to 2, 4 to 6, 9 to 12 etc. - J. M. Bergot, Jun 27 2013
Conjecture: the following definition produces a(n) for n >= 1: a(1) = 1; for n > 1, smallest number > a(n-1) satisfying the condition that a(n) is a square if and only if n is a triangular number. - J. Lowell, May 13 2014
Thus a(2) = 2, because 2 is not a triangular number and not a square; a(3) != 3, because 3 is not a square but is a triangular number; a(3) = 4 is OK because 4 is a square and 3 is a triangular number; etc. [Examples supplied by N. J. A. Sloane, May 13 2014]

			The triangle begins as:
   0;
   1,  2;
   4,  5,  6;
   9, 10, 11, 12;
  16, 17, 18, 19, 20;
  25, 26, 27, 28, 29, 30;
  36, 37, 38, 39, 40, 41, 42;
  49, 50, 51, 52, 53, 54, 55, 56;
  ... - _Stefano Spezia_, Oct 19 2024

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Harry J. Smith)

Cf. A063657, A004201-A004202.
Cf. A128217, A217575.
Essentially partial sums of A051340.

a063656 n = a063656_list !! n
a063656_list = f 1 [0..] where
   f k xs = us ++ f (k + 1) (drop (k - 1) vs) where
                    (us, vs) = splitAt k xs
-- Reinhard Zumkeller, Jun 20 2015

Select[Range[121],Floor[Sqrt[#]]==Round[Sqrt[#]] &] (* Stefano Spezia, Oct 19 2024 *)

{ n=-1; for (m=0, 10^9, if (sqrt(m)%1 < .5, write("b063656.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 27 2009

As a triangle from Stefano Spezia, Oct 19 2024: (Start)
T(n,k) = n^2 + k with 0 <= k <= n.
G.f.: x*(1 + x + 2*y - 4*x*y + 3*x^3*y^2 - x^2*y*(2 + y))/((1 - x)^3*(1 - x*y)^3). (End)

A126615 Denominators in a harmonic triangle.

Original entry on oeis.org

1, 2, 2, 2, 6, 3, 2, 6, 12, 4, 2, 6, 12, 20, 5, 2, 6, 12, 20, 30, 6, 2, 6, 12, 20, 30, 42, 7, 2, 6, 12, 20, 30, 42, 56, 8, 2, 6, 12, 20, 30, 42, 56, 72, 9, 2, 6, 12, 20, 30, 42, 56, 72, 90, 10, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 11, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 12, 2, 6

Offset: 1

Gary W. Adamson, Feb 09 2007

The harmonic triangle uses the terms of this sequence as denominators, with numerators = 1: (1/1; 1/2, 1/2; 1/2, 1/6, 1/3; 1/2, 1/6, 1/12, 1/4; 1/2, 1/6, 1/12, 1/10, 1/5; ...). Row sums of the harmonic triangle = 1.

			Triangle T(n,k) begins:
  1;
  2,  2;
  2,  6,  3;
  2,  6, 12,  4;
  2,  6, 12, 20,  5;
  2,  6, 12, 20, 30,  6;
  2,  6, 12, 20, 30, 42,  7;
  ...
1/1 = 1,
1/2 + 1/2 = 1,
1/2 + 1/6 + 1/3 = 1,
1/2 + 1/6 + 1/12 + 1/4 = 1, etc.

Row sums are A006527.

Cf. A000012, A002378, A051340, A127949, A003506.

A126615 := (n,k) -> `if`(n=k, n, 1/Beta(k,2));
seq(print(seq(A126615(n,k), k=1..n)), n=1..9); # Peter Luschny, Jul 27 2014

Flatten@Table[{Array[2PolygonalNumber@#&,n],n+1},{n,0,10}] (* Giorgos Kalogeropoulos, Mar 31 2021 *)

Denominators of the inverse of A127949; numerators = 1. Triangle read by rows, first (n-1) terms of 1*2, 2*3, 3*4, ...; followed by "n".

T(n,k) = k*(k+1) = A002378(k) for k < n; T(n,n) = n. - Andrés Ventas, Mar 26 2021

Gary W. Adamson submitted two different triangles numbered A127899 based on the harmonic numbers. This is the second of them, which I am renumbering as A126615. Unfortunately there were several other entries defined in terms of "A127899" and I may not have guessed which version of A127899 was being referred to. - N. J. A. Sloane, Jan 09 2007

More terms from Philippe Deléham, Dec 17 2008

A131229 Numbers congruent to {1,7} mod 10.

Original entry on oeis.org

1, 7, 11, 17, 21, 27, 31, 37, 41, 47, 51, 57, 61, 67, 71, 77, 81, 87, 91, 97, 101, 107, 111, 117, 121, 127, 131, 137, 141, 147, 151, 157, 161, 167, 171, 177, 181, 187, 191, 197, 201, 207, 211, 217, 221, 227, 231, 237, 241, 247, 251, 257, 261, 267, 271, 277, 281

Offset: 1

Gary W. Adamson, Jun 20 2007

Row sums of triangle A131228.
Take the sum of the squares of the first n triangular numbers and divide it by the sum of these n triangular numbers.  The sum evenly divides the sum of the squares for the n in this sequence. - J. M. Bergot, May 09 2012
a(n) = the difference between the sum of the terms in antidiagonal(n) and antidiagonal(n-1) in A204008. - J. M. Bergot, Jul 15 2013

			11 = a(3) = 1 mod 10.
11 = a(3) = row 4 sums, triangle A131228: (1 + 3 + 7).

David Lovler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001622, A128174, A051340, A063097, A131228.

Select[Range[300],MemberQ[{1,7},Mod[#,10]]&] (* or *) LinearRecurrence[ {1,1,-1},{1,7,11},60] (* Harvey P. Dale, May 20 2012 *)

is(n)=n%10==1 || n%10==7 \\ Charles R Greathouse IV, Jul 16 2013

From R. J. Mathar, Apr 17 2008: (Start)
a(n) = a(n-2) + 10 = 5*n + ((-1)^n - 7)/2.
G.f.: x*(1 + 6*x + 3*x^2)/((-1+x)^2*(1+x)). (End)
a(n) = 10*n - a(n-1) - 12 with n > 1, a(1)=1. - Vincenzo Librandi, Nov 23 2010
a(n) = a(n-1) + a(n-2) - a(n-3) for n > 3; a(1)=1, a(2)=7, a(3)=11. - Harvey P. Dale, May 20 2012
a(n) = A063097(n) for n >= 2. - Georg Fischer, Nov 02 2018
E.g.f.: 3 + ((10*x - 7)*exp(x) + exp(-x))/2. - David Lovler, Sep 07 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt((5+sqrt(5))/2)*Pi/10 + 3*log(phi)/(2*sqrt(5)), where phi is the golden ratio (A001622). - Amiram Eldar, Apr 15 2023

More terms from R. J. Mathar, Apr 17 2008

A130301 Triangle read by rows: A130296 * A007318, as infinite lower triangular matrices.

Original entry on oeis.org

1, 3, 1, 5, 3, 1, 7, 6, 4, 1, 9, 10, 10, 5, 1, 11, 15, 20, 15, 6, 1, 13, 21, 35, 35, 21, 7, 1, 15, 28, 56, 70, 56, 28, 8, 1, 17, 36, 84, 126, 126, 84, 36, 9, 1, 19, 45, 120, 210, 252, 210, 120, 45, 10, 1, 21, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1

Offset: 1

Gary W. Adamson, May 20 2007

Row sums = A083706: (1, 4, 9, 18, 35, 68, ...).
A130300 = A007318 * A130296.
The lower triangular matrix A130296 is equal to the restriction of the square array A051340 to its lower left triangular part. So this is also equal to (A051340) * A007318, where (A051340) is the lower triangular part of A051340, i.e., A051340[i,j] replaced by zero for j > i: see  Mathar's Maple code. - M. F. Hasler, Aug 15 2015

			First few rows of the triangle:
   1;
   3,  1;
   5,  3,  1;
   7,  6,  4,  1;
   9, 10, 10,  5,  1;
  11, 15, 20, 15,  6,  1;
  13, 21, 35, 35, 21,  7,  1;
  ...

Cf. A083706, A130296, A130300.

A130301 := proc(n, k)
    add( A051340(n, i)*binomial(i, k), i=k..n);
end proc: # R. J. Mathar, Jul 16 2015

A130301[m,n] = A121775[m,n] for n >= m/2. A130301[m,1] = 2m-1, A130301[m,2] = A000217[m-1], A130301[m,m] = 1, A130301[m,m-1] = m for m>2. - M. F. Hasler, Aug 15 2015

Corrected (missing a(15)=1 inserted) by M. F. Hasler, Aug 15 2015

a(26) = 27 corrected and more terms from Georg Fischer, May 29 2023

A130295 Erroneous duplicate of A125026.

Original entry on oeis.org

1, 3, 2, 5, 7, 3, 7, 15, 13, 4, 9, 26, 34, 21, 5, 11, 40, 70, 65, 31, 6, 13, 57, 125, 155, 111, 43, 7, 15, 77, 203, 315, 301, 175, 57, 8, 17, 100, 308, 574, 686, 532, 260, 73, 9, 19, 126, 444, 966, 1386, 1344, 876369, 91, 10

Offset: 1

Gary W. Adamson, May 20 2007

dead

This sequence was initially defined as "A051340 * A007318". However, the matrix product A051340 * A007318 is not well defined, because all elements of A051340 are strictly positive integers, as are all elements of the lower left of A007318. Therefore the matrix A051340 must be truncated to its lower left (setting A[i,j]=0 if j>i), which actually equals A130296. Then the product yields this sequence, which is identical to A125026.

Row sums = A099035 (not A083706 as stated initially): (1, 5, 15, 39, 95, 223, 511, ...).

			First few rows of the triangle A125026:
   1;
   3,  2;
   5,  7,   3;
   7, 15,  13,   4;
   9, 26,  34,  21,   5;
  11, 40,  70,  65,  31,  6;
  13, 57, 125, 155, 111, 43, 7;
  ...

Cf. A051340, A099035.

(A051340) * A007318 as infinite lower triangular matrices. [Here (A051340) is that matrix with the upper right triangle set to zero, which is actually A130296. - M. F. Hasler, Aug 15 2015]

Restored and edited by M. F. Hasler, Aug 15 2015

A263646 Coefficients for an expansion of the Schwarzian derivative of a power series.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 1, 4, 1, 1, 5, 1, 1, 1, 6, 1, 1, 1, 7, 1, 1, 1, 1, 8, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 10, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 12, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 14, 1, 1, 1, 1, 1, 1, 1, 15, 1, 1, 1, 1, 1, 1, 1, 1, 16, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Tom Copeland, Oct 31 2015

Coefficients for an expansion of the Schwarzian derivative of a power series f(x), with f'(0) = 1, expressed in terms of an expansion of the natural logarithm of the derivative of the function G(x) = log(D f(x)) = Sum_{n >= 1} -F(n) x^n/n.

			Partitions by powers of x^n:
n=0: -(F2 + F1^2/2)
n=1: -(2 F3 + F1 F2)
n=2: -(3 F4 + F1 F3 + F2^2/2) =  -[3 F4 + (F1 F3 + F2 F2 + F3 F1) / 2]
n=3: -(4 F5 + F1 F4 + F2 F3)  =  -[4 F5 + (F1 F4 + F2 F3 + F3 F2 + F4 F1) / 2]
n=4: -(5 F6 + F1 F5 + F2 F4 + F3^2/2)
--------------------
Example series:
f(x)= (1/2) / (1-x)^2 = 1/2 + x + (3/2) x^2 + 2x^3 + (5/2)x^4 + ... .
log(f'(x)) = log(1 + 3x + 6x^2 + 10x^3 + ...) = 3x + 3 x^2/2 + 3 x^3/3  + ... .
Then F(n) = -3 for n>=1, and the Schwarzian derivative series is
S{f(x)} = - [(-3 + 3^2/2) + (-2*3 + 3^2) x + (-3*3 + 3^2 + 3^2/2) x^2 + ...] = -3/2 - 3x - (9/2)x^2 - 6x^3 - (15/2)x^4 - ... .
--------------------
The Schwarzian vanishes if and only if acting on a Moebius, or linear fractional, transformation. This corresponds to F(n) = (-1)^(n+1) 2 * d^n, where d is an arbitrary constant.
--------------------
Example polynomial:
f(x) = (x-x1)(x-x2)/-(x1+x2)
log(f'(x)) = log[1 - 2x/(x1+x2)] = Sum_{n>= 1} -(2/(x1+x2))^n x^n/n.
Then F(n) = (2/(x1+x2))^n, and the Schwarzian derivative series is
S{f(x)} = (Sum_{n>= 0} -6(n+1) 2^n (x/(x1+x2))^n) / (x1+x2)^2 = -6 / (x1+x2-2*x)^2 (cf. A001787 and A085750).

Cf. A263916, A001787, A085750, A051340.

print(sum(([n]+[1]*((n+1)//2) for n in range(1, 18)), [])) # Andrey Zabolotskiy, Mar 07 2024

Schwarzian{f(x)} = S{f(x)} = (D^3 f(x)) / (D f(x)) - (3/2) [(D^2 f(x)) / D f(x)]^2 = D [(D^2 f(x)) / D f(x)] - (1/2) [(D^2 f(x)) / D f(x)]^2 = D^2 log[D f(x)] - (1/2) [D log[D f(x)]]^2.
Then, with G(x) = log[D f(x)], S{f(x)} = D^2 G(x) - (1/2) [D G(x)]^2.
With f'(0) = 1, G(x) = log[D f(x)] = sum[n >= 1, -F(n) * x^n/n], and F(n) as Fn,
S{f(x)} = -[(F2 + F1^2/2) + (2 F3 + F1 F2) x + (3 F4 + F1 F3 + F2^2/2) x^2 + (4 F5 + F1 F4 + F2 F3) x^3 + (5 F6 + F1 F5 + F2 F4 + F3^2/2) x^4 + (6 F7 + F1 F6 + F2 F5 + F3 F4) x^5 + (7 F8 + F1 F7 + F2 F6 + F3 F5 + F4^2/2) x^6 + ...] .
This entry's a(m) are the numerators of the coefficients of the binary partitions in the brackets. For the singular partition of the integer n, the coefficient is (n-1); for the symmetric partition, 1/2; and for the rest, 1.
More symmetrically, x^2 S{f(x)}= - sum{n>=2, x^n [(n-1)F(n) + (1/2) sum(k=1 to n-1, F(n-k) F(k))]}.
With f(x)= c(0) + x + c(2) x^2 + ... , F(n) are given by the Faber polynomials of A263916: F(n) = Faber(n,2c(2),3c(3),..,(n+1)c(n+1)).

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