A094727 -id:A094727 - cp's OEIS frontend

A002260 Triangle read by rows: T(n,k) = k for n >= 1, k = 1..n.

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

Offset: 1

Angele Hamel (amh(AT)maths.soton.ac.uk)

Old name: integers 1 to k followed by integers 1 to k+1 etc. (a fractal sequence).
Start counting again and again.
This is a "doubly fractal sequence" - see the Franklin T. Adams-Watters link.
The PARI functions t1, t2 can be used to read a square array T(n,k) (n >= 1, k >= 1) by antidiagonals downwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002
Reading this sequence as the antidiagonals of a rectangular array, row n is (n,n,n,...); this is the weight array (Cf. A144112) of the array A127779 (rectangular). - Clark Kimberling, Sep 16 2008
The upper trim of an arbitrary fractal sequence s is s, but the lower trim of s, although a fractal sequence, need not be s itself. However, the lower trim of A002260 is A002260. (The upper trim of s is what remains after the first occurrence of each term is deleted; the lower trim of s is what remains after all 0's are deleted from the sequence s-1.) - Clark Kimberling, Nov 02 2009
Eigensequence of the triangle = A001710 starting (1, 3, 12, 60, 360, ...). - Gary W. Adamson, Aug 02 2010
The triangle sums, see A180662 for their definitions, link this triangle of natural numbers with twenty-three different sequences, see the crossrefs. The mirror image of this triangle is A004736. - Johannes W. Meijer, Sep 22 2010
A002260 is the self-fission of the polynomial sequence (q(n,x)), where q(n,x) = x^n + x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. Sequence A002260 is reluctant sequence of sequence 1,2,3,... (A000027). - Boris Putievskiy, Dec 12 2012
This is the maximal sequence of positive integers, such that once an integer k has occurred, the number of k's always exceeds the number of (k+1)'s for the remainder of the sequence, with the first occurrence of the integers being in order. - Franklin T. Adams-Watters, Oct 23 2013
A002260 are the k antidiagonal numerators of rationals in Cantor's proof of 1-to-1 correspondence between rationals and naturals; the denominators are k-numerator+1. - Adriano Caroli, Mar 24 2015
T(n,k) gives the distance to the largest triangular number < n. - Ctibor O. Zizka, Apr 09 2020

			First six rows:
  1
  1   2
  1   2   3
  1   2   3   4
  1   2   3   4   5
  1   2   3   4   5   6

Clark Kimberling, "Fractal sequences and interspersions," Ars Combinatoria 45 (1997) 157-168. (Introduces upper trimming, lower trimming, and signature sequences.)
M. Myers, Smarandache Crescendo Subsequences, R. H. Wilde, An Anthology in Memoriam, Bristol Banner Books, Bristol, 1998, p. 19.
F. Smarandache, Sequences of Numbers Involved in Unsolved Problems, Hexis, Phoenix, 2006.

N. J. A. Sloane, Table of n, a(n) for n = 1..11325
Franklin T. Adams-Watters, Doubly Fractal Sequences
Matin Amini and Majid Jahangiri, A Novel Proof for Kimberling's Conjecture on Doubly Fractal Sequences, arXiv:1612.09481 [math.NT], 2017.
Bruno Berselli, Illustration of the initial terms
Jerry Brown et al., Problem 4619, School Science and Mathematics (USA), Vol. 97(4), 1997, pp. 221-222.
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida, and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
Clark Kimberling, Fractal sequences
Clark Kimberling, Numeration systems and fractal sequences, Acta Arithmetica 73 (1995) 103-117.
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
F. Smarandache, Sequences of Numbers Involved in Unsolved Problems.
Aaron Snook, Augmented Integer Linear Recurrences, 2012. - _N. J. A. Sloane_, Dec 19 2012
Michael Somos, Sequences used for indexing triangular or square arrays
Eric Weisstein's World of Mathematics, Smarandache Sequences
Eric Weisstein's World of Mathematics, Unit Fraction

Cf. A000217, A001710, A002262, A003056, A004736 (ordinal transform), A025581, A056534, A094727, A127779.
Cf. A140756 (alternating signs).
Triangle sums (see the comments): A000217 (Row1, Kn11); A004526 (Row2); A000096 (Kn12); A055998 (Kn13); A055999 (Kn14); A056000 (Kn15); A056115 (Kn16); A056119 (Kn17); A056121 (Kn18); A056126 (Kn19); A051942 (Kn110); A101859 (Kn111); A132754 (Kn112); A132755 (Kn113); A132756 (Kn114); A132757 (Kn115); A132758 (Kn116); A002620 (Kn21); A000290 (Kn3); A001840 (Ca2); A000326 (Ca3); A001972 (Gi2); A000384 (Gi3).
Cf. A108872.

a002260 n k = k
a002260_row n = [1..n]
a002260_tabl = iterate (\row -> map (+ 1) (0 : row)) [1]
-- Reinhard Zumkeller, Aug 04 2014, Jul 03 2012

at:=0; for n from 1 to 150 do for i from 1 to n do at:=at+1; lprint(at,i); od: od: # N. J. A. Sloane, Nov 01 2006
seq(seq(i,i=1..k),k=1..13); # Peter Luschny, Jul 06 2009

FoldList[{#1, #2} &, 1, Range[2, 13]] // Flatten (* Robert G. Wilson v, May 10 2011 *)
Flatten[Table[Range[n],{n,20}]] (* Harvey P. Dale, Jun 20 2013 *)

T(n,k):=sum((i+k)*binomial(i+k-1,i)*binomial(k,n-i-k+1)*(-1)^(n-i-k+1),i,max(0,n+1-2*k),n-k+1); /* Vladimir Kruchinin, Oct 18 2013 */

t1(n)=n-binomial(floor(1/2+sqrt(2*n)),2) /* this sequence */

A002260(n)=n-binomial((sqrtint(8*n)+1)\2,2) \\ M. F. Hasler, Mar 10 2014

from math import isqrt, comb
def A002260(n): return n-comb((m:=isqrt(k:=n<<1))+(k>m*(m+1)),2) # Chai Wah Wu, Nov 08 2024

a(n) = 1 + A002262(n).
n-th term is n - m*(m+1)/2 + 1, where m = floor((sqrt(8*n+1) - 1) / 2).
The above formula is for offset 0; for offset 1, use a(n) = n-m*(m+1)/2 where m = floor((-1+sqrt(8*n-7))/2). - Clark Kimberling, Jun 14 2011
a(k * (k + 1) / 2 + i) = i for k >= 0 and 0 < i <= k + 1. - Reinhard Zumkeller, Aug 14 2001
a(n) = (2*n + round(sqrt(2*n)) - round(sqrt(2*n))^2)/2. - Brian Tenneson, Oct 11 2003
a(n) = n - binomial(floor((1+sqrt(8*n))/2), 2). - Paul Barry, May 25 2004
T(n,k) = A001511(A118413(n,k)); T(n,k) = A003602(A118416(n,k)). - Reinhard Zumkeller, Apr 27 2006
a(A000217(n)) = A000217(n) - A000217(n-1), a(A000217(n-1) + 1) = 1, a(A000217(n) - 1) = A000217(n) - A000217(n-1) - 1. - Alexander R. Povolotsky, May 28 2008
a(A169581(n)) = A038566(n). - Reinhard Zumkeller, Dec 02 2009
T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n-k,n-i) (regarded as triangle, see the example). - Mircea Merca, Apr 11 2012
T(n,k) = Sum_{i=max(0,n+1-2*k)..n-k+1} (i+k)*binomial(i+k-1,i)*binomial(k,n-i-k+1)*(-1)^(n-i-k+1). - Vladimir Kruchinin, Oct 18 2013
G.f.: x*y / ((1 - x) * (1 - x*y)^2) = Sum_{n,k>0} T(n,k) * x^n * y^k. - Michael Somos, Sep 17 2014
a(n) = n - S(n) where S(n) = sum of distinct terms in {a(1), a(2), ..., a(n-1)}. - David James Sycamore, Mar 10 2025

More terms from Reinhard Zumkeller, Apr 27 2006
Incorrect program removed by Franklin T. Adams-Watters, Mar 19 2010
New name from Omar E. Pol, Jul 15 2012

A000267 Integer part of square root of 4n+1.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

1^1, 2^1, 3^2, 4^2, 5^3, 6^3, 7^4, 8^4, 9^5, 10^5, ...
Start with n, repeatedly subtract the square root of the previous term; a(n) gives number of steps to reach 0. - Robert G. Wilson v, Jul 22 2002
Triangle A094727 read by diagonals. - Philippe Deléham, Mar 21 2014
Partial sums of A240025; a(n) = number of quarter squares <= n. - Reinhard Zumkeller, Jul 05 2014
Every number k is present consecutively (floor((2*k+3)/4)) times. - Bernard Schott, Jun 08 2019

			From _Philippe Deléham_, Mar 21 2014: (Start)
Triangle A094727 begins:
  1;
  2,  3;
  3,  4,  5;
  4,  5,  6,  7;
  5,  6,  7,  8,  9;
  6,  7,  8,  9, 10, 11; ...
Read by diagonals:
   1;
   2;
   3,  3;
   4,  4;
   5,  5,  5;
   6,  6,  6;
   7,  7,  7,  7;
   8,  8,  8,  8;
   9,  9,  9,  9,  9;
  10, 10, 10, 10, 10; (End)
G.f. = 1 + 2*x + 3*x^2 + 3*x^3 + 4*x^4 + 4*x^5 + 5*x^6 + 5*x^7 + 5*x^8 + 6*x^9 + ...

Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 73, problem 20.
Bruce C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, 1994, see p. 77, Entry 23.

T. D. Noe, Table of n, a(n) for n = 0..10000
Gal Cohensius, Urban Larsson, Reshef Meir, and David Wahlstedt, Cumulative subtraction games, arXiv:1805.09368 [math.CO], 2018-2020.
S. Ramanujan, Question 723, J. Ind. Math. Soc., Vol. 7 (1915), p. 240, Vol. 10 (1918), pp. 357-358.

Cf. A055086, A080037, A227368.

Cf. A000196, A002620, A016813, A070939, A094727, A240025.

a000267 = a000196 . a016813  -- Reinhard Zumkeller, Dec 13 2012

[Floor(Sqrt(4*n+1)): n in [0..100]]; // Vincenzo Librandi, Jun 08 2019

A000267:=seq(floor(sqrt(4*n+1)), n=0..100); // Bernard Schott, Jun 08 2019

Table[Floor[Sqrt[4*n + 1]], {n, 0, 100}] (* T. D. Noe, Jun 19 2012 *)

{a(n) = if( n<0, 0, sqrtint(4*n + 1))};

from math import isqrt
def A000267(n): return isqrt((n<<2)|1) # Chai Wah Wu, Nov 23 2024

floor(a(n)/2) = A000196(n).
a(n) = 1 + a(n - floor(n^(1/2))), if n>0. - Michael Somos, Jul 22 2002
a(n) = floor( 1 / ( sqrt(n + 1) - sqrt(n) ) ). - Robert A. Stump (bob_ess107(AT)yahoo.com), Apr 07 2003
a(n) = |{floor(n/k): k in Z+}|. - David W. Wilson, May 26 2005
a(n) = ceiling(2*sqrt(n+1) - 1). - Mircea Merca, Feb 03 2012
a(n) = A000196(A016813(n)). - Reinhard Zumkeller, Dec 13 2012
a(n) = A070939(A227368(n+1)), conjectured. - Antti Karttunen, Dec 28 2013
a(n) = floor( sqrt(n) + sqrt(n+2) ). [Bruno Berselli, Jan 08 2015]
a(n) = floor( sqrt(4*n + k) ) where k = 1, 2, or 3. - Michael Somos, Mar 11 2015
G.f.: (Sum_{k>0} x^floor(k^2 / 4)) / (1 - x). - Michael Somos, Mar 11 2015
a(n) = 1 + A055086(n). - Michael Somos, Sep 02 2017
a(n) = floor(sqrt(n+1)+1/2) + floor(sqrt(n)). - Ridouane Oudra, Jun 07 2019
Sum_{k>=0} (-1)^k/a(k) = Pi/8 + log(2)/4. - Amiram Eldar, Jan 26 2024

More terms from Michael Somos, Jun 13 2000

A193091 Augmentation of the triangular array A158405. See Comments.

Original entry on oeis.org

1, 1, 3, 1, 6, 14, 1, 9, 37, 79, 1, 12, 69, 242, 494, 1, 15, 110, 516, 1658, 3294, 1, 18, 160, 928, 3870, 11764, 22952, 1, 21, 219, 1505, 7589, 29307, 85741, 165127, 1, 24, 287, 2274, 13355, 61332, 224357, 638250, 1217270, 1, 27, 364, 3262, 21789, 115003

Offset: 0

Clark Kimberling, Jul 30 2011

Suppose that P is an infinite triangular array of numbers:
p(0,0)
p(1,0)...p(1,1)
p(2,0)...p(2,1)...p(2,2)
p(3,0)...p(3,1)...p(3,2)...p(3,3)...
...
Let w(0,0)=1, w(1,0)=p(1,0), w(1,1)=p(1,1), and define
    W(n)=(w(n,0), w(n,1), w(n,2),...w(n,n-1), w(n,n)) recursively by W(n)=W(n-1)*PP(n), where PP(n) is the n X (n+1) matrix given by
...
row 0 ... p(n,0) ... p(n,1) ...... p(n,n-1) ... p(n,n)
row 1 ... 0 ..... p(n-1,0) ..... p(n-1,n-2) .. p(n-1,n-1)
row 2 ... 0 ..... 0 ............ p(n-2,n-3) .. p(n-2,n-2)
...
row n-1 . 0 ..... 0 ............. p(2,1) ..... p(2,2)
row n ... 0 ..... 0 ............. p(1,0) ..... p(1,1)
...
The augmentation of P is here introduced as the triangular array whose n-th row is W(n), for n>=0. The array P may be represented as a sequence of polynomials; viz., row n is then the vector of coefficients:  p(n,0), p(n,1),...,p(n,n), from p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n). For example, (C(n,k)) is represented by ((x+1)^n); using this choice of P (that is, Pascal's triangle), the augmentation of P is calculated one row at a time, either by the above matrix products or by polynomial substitutions in the following manner:
...
row 0 of W:  1, by decree
row 1 of W:  1 augments to 1,1
...polynomial version:  1 -> x+1
row 2 of W:  1,1 augments to 1,3,2
...polynomial version:  x+1 -> (x^2+2x+1)+(x+1)=x^2+3x+2
row 3 to W:  1,3,2 augments to 1,6,11,6
...polynomial version:
    x^2+3x+2 -> (x+1)^3+3(x+1)^2+2(x+1)=(x+1)(x+2)(x+3)
...
Examples of augmented triangular arrays:
(p(n,k)=1) augments to A009766, Catalan triangle.
Catalan triangle augments to A193560.
Pascal triangle augments to A094638, Stirling triangle.
A002260=((k+1)) augments to A023531.
A154325 augments to A033878.
A158405 augments to A193091.
((k!)) augments to A193092.
A094727 augments to A193093.
A130296 augments to A193094.
A004736 augments to A193561.
...
Regarding the specific augmentation W=A193091: w(n,n)=A003169.
From Peter Bala, Aug 02 2012: (Start)
This is the table of g(n,k) in the notation of Carlitz (p. 124). The triangle enumerates two-line arrays of positive integers
............a_1 a_2 ... a_n..........
............b_1 b_2 ... b_n..........
such that
1) max(a_i, b_i) <= min(a_(i+1), b_(i+1)) for 1 <= i <= n-1
2) max(a_i, b_i) <= i for 1 <= i <= n
3) max(a_n, b_n) = k.
See A071948 and A211788 for other two-line array enumerations.
(End)

			The triangle P, at A158405, is given by rows
  1
  1...3
  1...3...5
  1...3...5...7
  1...3...5...7...9...
The augmentation of P is the array W starts with w(0,0)=1, by definition of W.
Successive polynomials (rows of W) arise from P as shown here:
  ...
  1->x+3, so that W has (row 1)=(1,3);
  ...
  x+3->(x^2+3x+5)+3*(x+3), so that W has (row 2)=(1,6,14);
  ...
  x^2+6x+14->(x^3+3x^2+5x+7)+6(x^2+3x+5)+14(x+3), so that (row 3)=(1,9,37,79).
  ...
First 7 rows of W:
  1
  1    3
  1    6    14
  1    9    37    79
  1   12    69   242    494
  1   15   110   516   1658    3294
  1   18   160   928   3870   11764   22952

L. Carlitz, Enumeration of two-line arrays, Fib. Quart., Vol. 11 Number 2 (1973), 113-130.

Cf. A003169, A193092, A193093, A193094.

Cf. A071948, A211788.

p[n_, k_] := 2 k + 1
Table[p[n, k], {n, 0, 5}, {k, 0, n}] (* A158405 *)
m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}]
TableForm[m[4]]
w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1];
v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]};
v[n_] := v[n - 1].m[n]
TableForm[Table[v[n], {n, 0, 6}]] (* A193091 *)
Flatten[Table[v[n], {n, 0, 9}]]

From Peter Bala, Aug 02 2012: (Start)
T(n,k) = (n-k+1)/n*Sum_{i=0..k} C(n+1,n-k+i+1)*C(2*n+i+1,i) for 0 <= k <= n.
Recurrence equation: T(n,k) = Sum_{i=0..k} (2*k-2*i+1)*T(n-1,i).
(End)

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

Original entry on oeis.org

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511

Offset: 0

Clark Kimberling, Aug 02 2011

nonn

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).

Since degree(D(p))

Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.

D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14

D(D(p))=2(x+1)+7(1)+14=2x+23

D(D(D(p)))=2(1)+23=25;

the Q-residue of p is 25.

We may regard the sequence Q of polynomials as the triangular array formed by coefficients:

t(0,0)

t(1,0)....t(1,1)

t(2,0)....t(2,1)....t(2,2)

t(3,0)....t(3,1)....t(3,2)....t(3,3)

and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.

Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:

Q.....P...................Q-residue of P

1.....1...................A000079, 2^n

1....(x+1)^n..............A007051, (1+3^n)/2

1....(x+2)^n..............A034478, (1+5^n)/2

1....(x+3)^n..............A034494, (1+7^n)/2

1....(2x+1)^n.............A007582

1....(3x+1)^n.............A081186

1....(2x+3)^n.............A081342

1....(3x+2)^n.............A081336

1.....A040310.............A193649

1....(x+1)^n+(x-1)^n)/2...A122983

1....(x+2)(x+1)^(n-1).....A057198

1....(1,2,3,4,...,n)......A002064

1....(1,1,2,3,4,...,n)....A048495

1....(n,n+1,...,2n).......A087323

1....(n+1,n+2,...,2n+1)...A099035

1....p(n,k)=(2^(n-k))*3^k.A085350

1....p(n,k)=(3^(n-k))*2^k.A090040

1....A008288 (Delannoy)...A193653

1....A054142..............A101265

1....cyclotomic...........A193650

1....(x+1)(x+2)...(x+n)...A193651

1....A114525..............A193662

More examples:

Q...........P.............Q-residue of P

(x+1)^n...(x+1)^n.........A000110, Bell numbers

(x+1)^n...(x+2)^n.........A126390

(x+2)^n...(x+1)^n.........A028361

(x+2)^n...(x+2)^n.........A126443

(x+1)^n.....1.............A005001

(x+2)^n.....1.............A193660

A094727.....1.............A193657

(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)

(k+1).....(x+1)^n.........A112091

(x+1)^n...(k+1)...........A029761

(k+1)......A049310........A193663

(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)

A051162...(x+1)^n.........A193658

A094727...(x+1)^n.........A193659

A049310...(x+1)^n.........A193664

A075362....A075362........A193665

Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

q[n_, k_] := 1;
r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
f[n_, x_] := Fibonacci[n + 1, x];
p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
Table[v[n], {n, 0, 24}]    (* A193649 *)
TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
Table[r[k], {k, 0, 8}]  (* 2^k *)
TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

A094728 Triangle read by rows: T(n,k) = n^2 - k^2, 0 <= k < n.

Original entry on oeis.org

1, 4, 3, 9, 8, 5, 16, 15, 12, 7, 25, 24, 21, 16, 9, 36, 35, 32, 27, 20, 11, 49, 48, 45, 40, 33, 24, 13, 64, 63, 60, 55, 48, 39, 28, 15, 81, 80, 77, 72, 65, 56, 45, 32, 17, 100, 99, 96, 91, 84, 75, 64, 51, 36, 19, 121, 120, 117, 112, 105, 96, 85, 72, 57, 40, 21

Offset: 1

Reinhard Zumkeller, May 24 2004

(T(n,k) mod 4) <> 2, see A042965, A016825.
All numbers m occur A034178(m) times.
The row polynomials T(n,x) appear in the calculation of the column g.f.s of triangle A120070 (used to find the frequencies of the spectral lines of the hydrogen atom).

			n=3: T(3,x) = 9+8*x+5*x^2.
Triangle begins:
   1;
   4,  3;
   9,  8,  5;
  16, 15, 12,  7;
  25, 24, 21, 16,  9;
  36, 35, 32, 27, 20, 11;
  49, 48, 45, 40, 33, 24, 13;
  64, 63, 60, 55, 48, 39, 28, 15;
  81, 80, 77, 72, 65, 56, 45, 32, 17;
  ... etc. - _Philippe Deléham_, Mar 07 2013

G. C. Greubel, Table of n, a(n) for n = 1..5050

Cf. A000290, A000567, A004736, A005408, A005563, A008586, A008590.
Cf. A008594, A008598, A016825, A016945, A017329, A028347, A028560.
Cf. A028566, A034178, A042965, A094727, A120070, A128624.
Cf. A000384 (alternating row sums), A002412 (row sums).

[n^2-k^2: k in [0..n-1], n in [1..15]]; // G. C. Greubel, Mar 12 2024

Table[n^2 - k^2, {n,12}, {k,0,n-1}]//Flatten (* Michael De Vlieger, Nov 25 2015 *)

flatten([[n^2-k^2 for k in range(n)] for n in range(1,16)]) # G. C. Greubel, Mar 12 2024

Row polynomials: T(n,x) = n^2*Sum_{m=0..n} x^m - Sum_{m=0..n} m^2*x^m = Sum_{k=0..n-1} T(n,k)*x^k, n >= 1.
T(n, k) = A004736(n,k)*A094727(n,k).
T(n, 0) = A000290(n).
T(n, 1) = A005563(n-1) for n>1.
T(n, 2) = A028347(n) for n>2.
T(n, 3) = A028560(n-3) for n>3.
T(n, 4) = A028566(n-4) for n>4.
T(n, n-1) = A005408(n).
T(n, n-2) = A008586(n-1) for n>1.
T(n, n-3) = A016945(n-2) for n>2.
T(n, n-4) = A008590(n-2) for n>3.
T(n, n-5) = A017329(n-3) for n>4.
T(n, n-6) = A008594(n-3) for n>5.
T(n, n-8) = A008598(n-2) for n>7.
T(A005408(k), k) = A000567(k).
Sum_{k=0..n} T(n, k) = A002412(n) (row sums).
From G. C. Greubel, Mar 12 2024: (Start)
Sum_{k=0..n-1} (-1)^k * T(n, k) = A000384(floor((n+1)/2)).
Sum_{k=0..floor((n-1)/2)} T(n-k, k) = A128624(n).
Sum_{k=0..floor((n-1)/2)} (-1)^k*T(n-k, k) = (1/2)*n*(n+1 - (-1)^n*cos(n*Pi/2)). (End)
G.f.: x*(1 - 3*x^2*y + x*(1 + y))/((1 - x)^3*(1 - x*y)^2). - Stefano Spezia, Aug 04 2025

A128076 Triangle T(n,k) = 2*n-k, read by rows.

Original entry on oeis.org

1, 3, 2, 5, 4, 3, 7, 6, 5, 4, 9, 8, 7, 6, 5, 11, 10, 9, 8, 7, 6, 13, 12, 11, 10, 9, 8, 7, 15, 14, 13, 12, 11, 10, 9, 8, 17, 16, 15, 14, 13, 12, 11, 10, 9, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11

Offset: 1

Gary W. Adamson, Feb 14 2007

From Boris Putievskiy, Jan 24 2013: (Start)
Table T(n,k) = n+2*k-2 n, k > 0, read by antidiagonals.
General case A209304. Let m be natural number. The first column of the table T(n,1) is the sequence of the natural numbers A000027. Every next column is formed from previous shifted by m elements.
For m=0 the result is A002260,
for m=1 the result is A002024,
for m=2 the result is A128076,
for m=3 the result is A131914,
for m=4 the result is A209304. (End)

			First few rows of the triangle are:
1;
3, 2;
5, 4, 3;
7, 6, 5, 4;
9, 8, 7, 6, 5;
...

Boris Putievskiy, Transformations [Of] Integer Sequences And Pairing Functions, arXiv preprint arXiv:1212.2732 [math.CO], 2012.

Cf. A128064, A004736, A000326 (row sums), A003056, A002260, A002024, A131914, A209304, A094727 (rows reversed).

A128076 := proc(n,k)
    2*n-k ;
end proc:
seq(seq( A128076(n,k),k=1..n),n=1..12) ;# R. J. Mathar, Sep 27 2021

Table[(Round[Sqrt[2 n]]^2 + 3 Round[Sqrt[2 n]] - 2 n)/2, {n, 100}] (* Wesley Ivan Hurt, Sep 19 2021 *)

Matrix product A128064 * A004736 as infinite lower triangular matrices.
From Boris Putievskiy, Jan 24 2013: (Start)
For the general case:
a(n) = m*A003056 -(m-1)*A002260.
a(n) = m*(t+1) + (m-1)*(t*(t+1)/2-n), where t=floor((-1+sqrt(8*n-7))/2).
For m = 2:
a(n) = 2*A003056 -A002260.
a(n) = 2*(t+1)+(t*(t+1)/2-n), where t=floor((-1+sqrt(8*n-7))/2). (End)
a(n) = (r^2 + 3*r - 2*n)/2, where r = round(sqrt(2*n)). - Wesley Ivan Hurt, Sep 19 2021
a(n) = A105020(n-1)/A002260(n). - Wesley Ivan Hurt, Sep 22 2021

NAME simplified. - R. J. Mathar, Sep 27 2021

A258993 Triangle read by rows: T(n,k) = binomial(n+k,n-k), k = 0..n-1.

Original entry on oeis.org

1, 1, 3, 1, 6, 5, 1, 10, 15, 7, 1, 15, 35, 28, 9, 1, 21, 70, 84, 45, 11, 1, 28, 126, 210, 165, 66, 13, 1, 36, 210, 462, 495, 286, 91, 15, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1, 66, 715, 3003, 6435, 8008, 6188, 3060, 969, 190, 21

Offset: 1

Reinhard Zumkeller, Jun 22 2015

T(n,k) = A085478(n,k) = A007318(A094727(n),A004736(k)), k = 0..n-1;
rounded(T(n,k)/(2*k+1)) = A258708(n,k);
rounded(sum(T(n,k)/(2*k+1)): k = 0..n-1) = A000967(n).

			.  n\k |  0  1    2    3     4     5     6     7    8    9  10 11
. -----+-----------------------------------------------------------
.   1  |  1
.   2  |  1  3
.   3  |  1  6    5
.   4  |  1 10   15    7
.   5  |  1 15   35   28     9
.   6  |  1 21   70   84    45    11
.   7  |  1 28  126  210   165    66    13
.   8  |  1 36  210  462   495   286    91    15
.   9  |  1 45  330  924  1287  1001   455   120   17
.  10  |  1 55  495 1716  3003  3003  1820   680  153   19
.  11  |  1 66  715 3003  6435  8008  6188  3060  969  190  21
.  12  |  1 78 1001 5005 12870 19448 18564 11628 4845 1330 231 23  .

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

If a diagonal of 1's is added on the right, this becomes A085478.
Essentially the same as A143858.
Cf. A007318, A004736, A094727.
Cf. A027941 (row sums), A117671 (central terms), A143858, A000967, A258708.
T(n,k): A000217 (k=1), A000332 (k=2), A000579 (k=3), A000581 (k=4), A001287 (k=5), A010965 (k=6), A010967 (k=7), A010969 (k=8), A010971 (k=9), A010973 (k=10), A010975 (k=11), A010977 (k=12), A010979 (k=13), A010981 (k=14), A010983 (k=15), A010985 (k=16), A010987 (k=17), A010989 (k=18), A010991 (k=19), A010993 (k=20), A010995 (k=21), A010997 (k=22), A010999 (k=23), A011001 (k=24), A017714 (k=25), A017716 (k=26), A017718 (k=27), A017720 (k=28), A017722 (k=29), A017724 (k=30), A017726 (k=31), A017728 (k=32), A017730 (k=33), A017732 (k=34), A017734 (k=35), A017736 (k=36), A017738 (k=37), A017740 (k=38), A017742 (k=39), A017744 (k=40), A017746 (k=41), A017748 (k=42), A017750 (k=43), A017752 (k=44), A017754 (k=45), A017756 (k=46), A017758 (k=47), A017760 (k=48), A017762 (k=49), A017764 (k=50).
T(n+k,n): A005408 (k=1), A000384 (k=2), A000447 (k=3), A053134 (k=4), A002299 (k=5), A053135 (k=6), A053136 (k=7), A053137 (k=8), A053138 (k=9), A196789 (k=10).
Cf. A165253.

Flat(List([1..12], n-> List([0..n-1], k-> Binomial(n+k,n-k) ))); # G. C. Greubel, Aug 01 2019

a258993 n k = a258993_tabl !! (n-1) !! k
a258993_row n = a258993_tabl !! (n-1)
a258993_tabl = zipWith (zipWith a007318) a094727_tabl a004736_tabl

[Binomial(n+k,n-k): k in [0..n-1], n in [1..12]]; // G. C. Greubel, Aug 01 2019

Table[Binomial[n+k,n-k], {n,1,12}, {k,0,n-1}]//Flatten (* G. C. Greubel, Aug 01 2019 *)

T(n,k) = binomial(n+k,n-k);
for(n=1, 12, for(k=0,n-1, print1(T(n,k), ", "))) \\ G. C. Greubel, Aug 01 2019

[[binomial(n+k,n-k) for k in (0..n-1)] for n in (1..12)] # G. C. Greubel, Aug 01 2019

T(n,k) = A085478(n,k) = A007318(A094727(n),A004736(k)), k = 0..n-1;
rounded(T(n,k)/(2*k+1)) = A258708(n,k);
rounded(sum(T(n,k)/(2*k+1)): k = 0..n-1) = A000967(n).

Showing 1-10 of 21 results. Next

cp's OEIS Frontend

A193093 Augmentation of the triangular array P=A094727 given by p(n,k)=n+k+1 for 0<=k<=n. See Comments.

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A193659 Q-residue of the triangle A094727, where Q=Pascal's triangle. (See Comments.)

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A104587 Triangle read by rows, given by the matrix product A * B where A (A094727) = [1; 2, 3; 3, 4, 5; 4, 5, 6, 7; ...] and B = [1; 1, 1; 1, 1, 1; ...] (both are infinite lower triangular matrices with the other terms zero).

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A002260 Triangle read by rows: T(n,k) = k for n >= 1, k = 1..n.

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A000267 Integer part of square root of 4n+1.

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A193091 Augmentation of the triangular array A158405. See Comments.

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A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

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