A054451 -id:A054451 - cp's OEIS frontend

A054450 Triangle of partial row sums of unsigned triangle A049310(n,m), n >= m >= 0 (Chebyshev S-polynomials).

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 4, 4, 1, 1, 8, 8, 5, 5, 1, 1, 13, 12, 12, 6, 6, 1, 1, 21, 21, 17, 17, 7, 7, 1, 1, 34, 33, 33, 23, 23, 8, 8, 1, 1, 55, 55, 50, 50, 30, 30, 9, 9, 1, 1, 89, 88, 88, 73, 73, 38, 38, 10, 10, 1, 1, 144, 144, 138, 138, 103, 103, 47, 47, 11, 11, 1, 1

Offset: 0

Wolfdieter Lang, Apr 27 2000 and May 08 2000

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is Fib(z)/(1-x*z/(1-z^2)) where Fib(x)=1/(1-x-x^2) = g.f. for A000045(n+1) (Fibonacci numbers without 0).

This is the first member of the family of Riordan-type matrices obtained from the unsigned convolution matrix A049310 by repeated application of the partial row sums procedure.

			Triangle begins as:
   1;
   1,  1;
   2,  1,  1;
   3,  3,  1,  1;
   5,  4,  4,  1,  1;
   8,  8,  5,  5,  1,  1;
  13, 12, 12,  6,  6,  1,  1;
  21, 21, 17, 17,  7,  7,  1,  1;
  34, 33, 33, 23, 23,  8,  8,  1,  1;
  55, 55, 50, 50, 30, 30,  9,  9,  1, 1;
  89, 88, 88, 73, 73, 38, 38, 10, 10, 1, 1;
  ...
Fourth row polynomial (n=3): p(3,x) = 3 + 3*x + x^2 + x^3.

Cf. A000027, A000045, A029907 (row sums), A038793, A038794, A038795, A038796.

Cf. A049310, A052952, A053121, A054451, A054453, A099571, A099572, A152948.

A049310:= func< n,k | ((n+k) mod 2) eq 0 select (-1)^(Floor((n+k)/2)+k)*Binomial(Floor((n+k)/2), k) else 0 >;
A054450:= func< n,k | (&+[Abs(A049310(n,j)): j in [k..n]]) >;
[A054450(n,k): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

A049310[n_, k_]:= A049310[n, k]= If[n<0, 0, If[k==n, 1, A049310[n-1, k-1] - A049310[n-2, k] ]];
A054450[n_, k_]:= A054450[n, k]= Sum[Abs[A049310[n,j]], {j,k,n}];
Table[A054450[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

@CachedFunction
def A049310(n, k):
    if (n<0): return 0
    elif (k==n): return 1
    else: return A049310(n-1, k-1) - A049310(n-2, k)
def A054450(n,k): return sum( abs(A049310(n,j)) for j in (k..n) )
flatten([[A054450(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

T(n, m) = Sum_{k=m..n} |A049310(n, k)| (sequence of partial row sums in column m).

Column m recursion: T(n, m) = Sum_{j=m..n} T(j-1, m)*|A049310(n-j, 0)| + |A049310(n, m)|, n >= m >= 0, a(n, m) := 0 if n

T(n, 0) = A000045(n+1).

T(n, 1) = A052952(n-1).

T(n, 2) = A054451(n-2).

Sum_{k=0..n} T(n, k) = A029907(n) = A054453(n, 0).

G.f. for column m: Fib(x)*(x/(1-x^2))^m, m >= 0, with Fib(x) = g.f. A000045(n+1).

The corresponding square array has T(n, k) = Sum_{j=0..floor(k/2)} binomial(n+k-j, j). - Paul Barry, Oct 23 2004

From G. C. Greubel, Jul 25 2022: (Start)

T(n, 3) = A099571(n-3).

T(n, 4) = A099572(n-4).

T(n, n) = T(n, n-1) = A000012(n).

T(n, n-2) = A000027(n), n >= 2.

T(n, n-3) = A000027(n), n >= 3.

T(n, n-4) = A152948(n), n >= 4.

T(n, n-5) = A152948(n), n >= 5.

T(n, n-6) = A038793(n), n >= 6.

T(n, n-8) = A038794(n), n >= 8.

T(n, n-10) = A038795(n), n >= 10.

T(n, n-12) = A038796(n), n >= 12. (End)

A054452 Partial sums of A027941(n-1) with a(-1) = 0.

Original entry on oeis.org

0, 0, 1, 5, 17, 50, 138, 370, 979, 2575, 6755, 17700, 46356, 121380, 317797, 832025, 2178293, 5702870, 14930334, 39088150, 102334135, 267914275, 701408711, 1836311880, 4807526952, 12586269000, 32951280073, 86267571245, 225851433689, 591286729850

Offset: 0

Wolfdieter Lang, Apr 27 2000

Colin Barker, Table of n, a(n) for n = 0..1000
László Németh, Pascal pyramid in the space H^2 x R, arXiv:1701.06022 [math.CO], 2017 (5th line of Table 1 is a(n-2)).
A. Shriki and O. Liba, Polygons with Fibonacci Number Coordinates: Problem B-1167, Fib. Quart. 54,2 May 2016, p. 180-181.
Index entries for linear recurrences with constant coefficients, signature (5,-8,5,-1).

Cf. A027941, A054451, A001906, A052952.

I:=[0,0,1,5]; [n le 4 select I[n] else 5*Self(n-1)-8*Self(n-2)+5*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Mar 26 2015

a[0]:=0: a[1]:=1: for n from 2 to 50 do a[n] := 3*a[n-1]-a[n-2] od: seq(a[n]-n, n=0..27); # Zerinvary Lajos, Mar 20 2008
with(combinat): seq(fibonacci(2*n)-n, n=0..27); # Zerinvary Lajos, Jun 19 2008
g:=z/(1-3*z+z^2): gser:=series(g, z=0, 43): seq(abs(coeff(gser, z, n)-n), n=0..27); # Zerinvary Lajos, Mar 22 2009

CoefficientList[Series[x^2 / ((1 - x)^2 (1 - 3 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Mar 26 2015 *)

makelist(sum(fib(k)*binomial(n+1,k+2),k,0,n),n,0,20); /* Vladimir Kruchinin, Oct 21 2016 */

concat(vector(2), Vec(x^2/((1-x)^2*(1-3*x+x^2)) + O(x^40))) \\ Colin Barker, Jan 28 2017

[(lucas_number1(n, 3, 1)-lucas_number1(n, 2, 1)) for n in range(1, 28)]# Zerinvary Lajos, Mar 13 2009

a(n) = +5*a(n-1) -8*a(n-2) +5*a(n-3) -1*a(n-4).
G.f.: x^2/((1-x)^2*(1-3*x+x^2)).
a(n) = Sum_{k=0..n} A027941(k-1) = F(2*n)-n = A054450(2*n-1, 2) = A054451(2*n-3).
G.f.: x^2*Fibe(x)/(1-x)^2, with Fibe(x) := 1/(1-3*x+x^2) = g.f. A001906(n+1) (Fibonacci numbers F(2(n+1))).
Fourth diagonal of array defined by T(i, 1) = T(1, j) = 1, T(i, j) = Max(T(i-1, j) + T(i-1, j-1); T(i-1, j-1) + T(i, j-1)). - Benoit Cloitre, Aug 05 2003
a(n) = Sum_{k=0..n-2} binomial(2*n-k-1, k). - Johannes W. Meijer, Aug 12 2013
a(n) = Sum_{i=1..n-1} Sum_{j=1..n-1} binomial(i+j, i-j). - Wesley Ivan Hurt, Mar 25 2015
a(n) = Sum_{k=0..n} (binomial(n+1,k+2)*Fibonacci(k)). - Vladimir Kruchinin, Oct 21 2016
a(n) = (-((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n)/sqrt(5) - n. - Colin Barker, Jan 28 2017

More terms from James Sellers, Apr 28 2000

a(0) added by Arkadiusz Wesolowski, Jun 07 2011

A099572 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+4, k).

Original entry on oeis.org

1, 1, 6, 7, 23, 30, 73, 103, 211, 314, 581, 895, 1560, 2455, 4135, 6590, 10890, 17480, 28590, 46070, 74946, 121016, 196326, 317342, 514123, 831465, 1346148, 2177613, 3524441, 5702054, 9227311, 14929365, 24157645, 39087010, 63245795, 102332805

Offset: 0

Paul Barry, Oct 23 2004

Fifth column of triangle A054450. In general Sum_{k=0..floor(n/2)} binomial(n-k+r, k), r>=0, will have g.f. 1/((1-x^2)^r*(1-x-x^2)) and for r>0, a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+r-1, r-1)*(1+(-1)^k)/2.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,5,-4,-10,6,10,-4,-5,1,1).

Cf. A000045, A054450, A054451, A052952, A099571.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( 1/((1-x^2)^4*(1-x-x^2)) )); // G. C. Greubel, Jul 25 2022

Table[Fibonacci(n+5) +(-1)^n*(n^3+9*n^2+35*n+33)/96 -(n^3+21*n^2+155*n+417)/96, {n,0,40}] (* G. C. Greubel, Jul 25 2022 *)

[fibonacci(n+5) + (-1)^n*(n^3+9*n^2+35*n+33)/96 - (n^3+21*n^2+155*n + 417)/96 for n in (0..40)] # G. C. Greubel, Jul 25 2022

G.f.: 1/((1-x^2)^4*(1-x-x^2)). - corrected by R. J. Mathar, Feb 20 2011
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2+3, 3)*((1+(-1)^k)/2).
a(n) = Fibonacci(n+5) + (-1)^n*(n^3 + 9*n^2 + 35*n + 33)/96 - (n^3 + 21*n^2 + 155*n + 417)/96. - G. C. Greubel, Jul 25 2022

A173284 Triangle by columns, Fibonacci numbers in every column shifted down twice, for k > 0.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 5, 2, 1, 8, 3, 1, 13, 5, 2, 21, 8, 3, 1, 34, 13, 5, 2, 1, 55, 21, 8, 3, 1, 89, 34, 13, 5, 2, 1, 144, 55, 21, 8, 3, 1, 233, 89, 34, 13, 5, 2, 1, 377, 144, 55, 21, 8, 3, 1, 610, 233, 89, 34, 13, 5, 2, 1

Offset: 0

Gary W. Adamson, Feb 14 2010

The row sums equal A052952.
Let the triangle = M. Then lim_{n->infinity} M^n = A173285 as a left-shifted vector.
A173284 * [1, 2, 3, ...] = A054451: (1, 1, 4, 5, 12, 17, 33, ...). - Gary W. Adamson, Mar 03 2010
From Johannes W. Meijer, Sep 05 2013: (Start)
Triangle read by rows formed from antidiagonals of triangle A104762.
The diagonal sums lead to A004695. (End)

			First few rows of the triangle:
    1;
    1;
    2,   1;
    3,   1;
    5,   2,  1;
    8,   3,  1;
   13,   5,  2,  1;
   21,   8,  3,  1;
   34,  13,  5,  2,  1;
   55,  21,  8,  3,  1;
   89,  34, 13,  5,  2, 1;
  144,  55, 21,  8,  3, 1;
  233,  89, 34, 13,  5, 2, 1;
  377, 144, 55, 21,  8, 3, 1;
  610, 233, 89, 34, 13, 5, 2, 1;
  ...

Cf. A000045, A004695, A052952, A054451, A173285.

Cf. (Similar triangles) A008315 (Catalan), A011973 (Pascal), A102541 (Losanitsch), A122196 (Fractal), A122197 (Fractal), A128099 (Pell-Jacobsthal), A152198, A152204, A207538, A209634.

T := proc(n, k): if n<0 then return(0) elif k < 0 or k > floor(n/2) then return(0) else combinat[fibonacci](n-2*k+1) fi: end: seq(seq(T(n, k), k=0..floor(n/2)), n=0..14); # Johannes W. Meijer, Sep 05 2013

Triangle by columns, Fibonacci numbers in every column shifted down twice, for k > 0.
From Johannes W. Meijer, Sep 05 2013: (Start)
T(n,k) = A000045(n-2*k+1), n >= 0 and 0 <= k <= floor(n/2).
T(n,k) = A104762(n-k, k). (End)

Term a(15) corrected by Johannes W. Meijer, Sep 05 2013

A058255 Distinct values of lcm_{i=1..n} (p(i)-1), where p() are the primes.

Original entry on oeis.org

1, 2, 4, 12, 60, 240, 720, 7920, 55440, 1275120, 16576560, 480720240, 19709529840, 39419059680, 197095298400, 3350620072800, 177582863858400, 532748591575200, 19711697888282400, 59135093664847200

Offset: 1

Labos Elemer, Dec 06 2000

nonn

The prime A095365(n) is the least prime yielding an LCM of a(n). This sequence and A095365 are related to A095366. - T. D. Noe, Jun 04 2004

			For p = 29, 31, 37, 41, 43 these LCMs are equal to 55440 = 11*7! = lcm[1, 2, 4, 6, 10, 12, 16, 22, 28, 30, 36, 40, 42] = lcm[1, 2, 4, 6, 10, 12, 16, 22, 28]. The values was put on the stage only once. Repetitions skipped.

Michael De Vlieger, Table of n, a(n) for n = 1..360

A058254 with duplicates removed.
Cf. A002110, A005867, A003418, A054451, A000142, A000010.
Cf. A095366 (least k > 1 such that k divides 1^n + 2^n + ... + (k-1)^n).

Union@ Table[LCM @@ (Prime@ Range[1, n] - 1), {n, 38}] (* Michael De Vlieger, Dec 06 2018 *)

A099571 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+3, k).

Original entry on oeis.org

1, 1, 5, 6, 17, 23, 50, 73, 138, 211, 370, 581, 979, 1560, 2575, 4135, 6755, 10890, 17700, 28590, 46356, 74946, 121380, 196326, 317797, 514123, 832025, 1346148, 2178293, 3524441, 5702870, 9227311, 14930334, 24157645, 39088150, 63245795

Offset: 0

Paul Barry, Oct 23 2004

Fourth column of triangle A054450.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,4,-3,-6,3,4,-1,-1).

Cf. A000045, A052952, A054450, A054451, A099572.

[Fibonacci(n+4) +(-1)^n*(n^2+4*n+7)/16 -(n^2+12*n+39)/16: n in [0..40]]; // G. C. Greubel, Jul 25 2022

Table[Sum[Binomial[n-k+3,k],{k,0,Floor[n/2]}],{n,0,40}] (* or *) LinearRecurrence[{1,4,-3,-6,3,4,-1,-1},{1,1,5,6,17,23,50,73},40] (* Harvey P. Dale, Jun 04 2021 *)

[fibonacci(n+4) +(-1)^n*(n^2+4*n+7)/16 -(n^2+12*n+39)/16 for n in (0..40)] # G. C. Greubel, Jul 25 2022

G.f.: 1/((1-x^2)^3*(1-x-x^2)).
a(n) = a(n-1) + 4*a(n-2) - 3*a(n-3) - 6*a(n-4) + 3*a(n-5) + 4*a(n-6) - a(n-7) - a(n-8);
a(n) = Sum_{k=0..n} Fibonacci(n-k+1)*binomial(k/2 +2, 2)*(1+(-1)^k)/2.
a(n) = Fibonacci(n+4) + (-1)^n*(n^2 + 4*n + 7)/16 - (n^2 + 12*n + 39)/16. - G. C. Greubel, Jul 25 2022

A099573 Reverse of number triangle A054450.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 4, 5, 1, 1, 5, 5, 8, 8, 1, 1, 6, 6, 12, 12, 13, 1, 1, 7, 7, 17, 17, 21, 21, 1, 1, 8, 8, 23, 23, 33, 33, 34, 1, 1, 9, 9, 30, 30, 50, 50, 55, 55, 1, 1, 10, 10, 38, 38, 73, 73, 88, 88, 89, 1, 1, 11, 11, 47, 47, 103, 103, 138, 138, 144, 144, 1, 1, 12, 12, 57, 57, 141, 141, 211, 211, 232, 232, 233

Offset: 0

Paul Barry, Oct 23 2004

			First few rows of the array:
  1, 1, 2, 3,  5,  8, ... (A000045)
  1, 1, 3, 4,  8, 12, ... (A052952)
  1, 1, 4, 5, 12, 17, ... (A054451)
  1, 1, 5, 6, 17, 23, ... (A099571)
  1, 1, 6, 7, 23, 30, ... (A099572)
  ...
Triangle begins as:
  1;
  1, 1;
  1, 1, 2;
  1, 1, 3, 3;
  1, 1, 4, 4,  5;
  1, 1, 5, 5,  8,  8;
  1, 1, 6, 6, 12, 12, 13;
  1, 1, 7, 7, 17, 17, 21, 21;
  1, 1, 8, 8, 23, 23, 33, 33, 34;
  1, 1, 9, 9, 30, 30, 50, 50, 55, 55;

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

Cf. A000045, A029907 (row sums), A052951, A052952, A054450, A054451, A052952.

Cf. A099571, A099572, A099574 (diagonal sums), A099575.

[(&+[Binomial(n-j,j): j in [0..Floor(k/2)]]): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 25 2022

T[n_, k_]:= Sum[Binomial[n-j,j], {j,0,Floor[k/2]}];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 25 2022 *)

def A099573(n,k): return sum(binomial(n-j, j) for j in (0..(k//2)))
flatten([[A099573(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 25 2022

Number triangle T(n, k) = Sum_{j=0..floor(k/2)} binomial(n-j, j) if k <= n, 0 otherwise.
T(n, n) = A000045(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A099574(n).
Sum_{k=0..n} T(n, k) = A029907(n+1).
Antidiagonals of the following array: the first row equals the Fibonacci numbers, (1, 1, 2, 3, 5, ...), and the (n+1)-st row is obtained by the matrix-vector product A128174 * n-th row. - Gary W. Adamson, Jan 19 2011
From G. C. Greubel, Jul 25 2022: (Start)
T(n, n-1) = A052952(n-1), n >= 1.
T(n, n-2) = A054451(n-2), n >= 2.
T(n, n-3) = A099571(n-3), n >= 3.
T(n, n-4) = A099572(n-4), n >= 4. (End)

A007382 Number of strict (-1)st-order maximal independent sets in path graph.

Original entry on oeis.org

0, 0, 3, 4, 11, 16, 32, 49, 87, 137, 231, 369, 608, 978, 1595, 2574, 4179, 6754, 10944, 17699, 28655, 46355, 75023, 121379, 196416, 317796, 514227, 832024, 1346267

Offset: 1

N. J. A. Sloane, Mira Bernstein

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. Yanco and A. Bagchi, K-th order maximal independent sets in path and cycle graphs, J. Graph Theory, submitted, 1994.

R. Yanco, Letter and Email to N. J. A. Sloane, 1994

Equals A054451(n+1) - 1.

Table[Sum[Binomial[n - i + 1, i], {i, Floor[(n - 1)/2]}], {n, 30}] (* or *)
Rest@ Abs@ CoefficientList[Series[x^3*(x^3 + 2 x^2 - x - 3)/((1 - x - x^2) (1 - x^2)^2), {x, 0, 30}], x] (* Michael De Vlieger, Sep 19 2017 *)

John W. Layman observes that if b(n) = 1+A007382(n) then b(n) = b(n-1) + 3b(n-2) - 2b(n-3) - 3b(n-4) + b(n-5) + b(n-6) for all 27 terms shown.
G.f.: x^3*(x^3+2x^2-x-3)/((1-x-x^2)*(1-x^2)^2).
a(n) = Sum_{i=1..floor((n-1)/2)} C(n-i+1, i). - Wesley Ivan Hurt, Sep 19 2017

A058256 a(n) = A058254(n+1)/A058254(n).

Original entry on oeis.org

2, 2, 3, 5, 1, 4, 3, 11, 7, 1, 1, 1, 1, 23, 13, 29, 1, 1, 1, 1, 1, 41, 1, 2, 5, 17, 53, 3, 1, 1, 1, 1, 1, 37, 1, 1, 3, 83, 43, 89, 1, 19, 2, 7, 1, 1, 1, 113, 1, 1, 1, 1, 5, 4, 131, 67, 1, 1, 1, 47, 73, 1, 31, 1, 79, 1, 1, 173, 1, 1, 179, 61, 1, 1, 191, 97, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Labos Elemer, Dec 06 2000

nonn

a(n) = 1 if in prime(n+1)-1 no new prime divisor or new power of a prime appear, like LCM[{1, 2, 4, 6, 10, 12, 16, 22}]= LCM[{1, 2, 4, 6, 10, 12, 16, 22, 28}].

a(n) > 1 if in prime(n+1)-1 new prime divisor(s) or new power(s) of a prime arise, like in A058254(15) compared with A058254(14), where the new prime divisor is 23 only, so a(14)=23. Such sites of increase do not correspond to the natural order of primes and prime-powers like in A054451.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A058254, A002110, A005867, A003418, A054451, A000142, A000010, A003418, A000961.

f(n) = lcm(apply(p->p-1, primes(n))); \\ A058254
a(n) = f(n+1)/f(n); \\ Michel Marcus, Mar 22 2020

a(n) = lcm{i=1..n+1} (prime(i)-1) / lcm{i=1..n} (prime(i)-1).

Offset corrected by Amiram Eldar, Sep 24 2019

Showing 1-9 of 9 results.

cp's OEIS Frontend

A054450 Triangle of partial row sums of unsigned triangle A049310(n,m), n >= m >= 0 (Chebyshev S-polynomials).

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A054452 Partial sums of A027941(n-1) with a(-1) = 0.

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A099572 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+4, k).

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A173284 Triangle by columns, Fibonacci numbers in every column shifted down twice, for k > 0.

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A058255 Distinct values of lcm_{i=1..n} (p(i)-1), where p() are the primes.

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A099571 a(n) = Sum_{k=0..floor(n/2)} binomial(n-k+3, k).

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A099573 Reverse of number triangle A054450.

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