A048739 -id:A048739 - cp's OEIS frontend

A069341 Number of 19 X n binary arrays with path of adjacent 1's from upper right corner to lower left corner.

1, 11309768, 22779900825195, 24471786016855404727, 19494870504116920632360116, 13160595194449992172457391361101, 8025991217666841006376543207614110979, 4574735745210954097724726033484092793843149, 2485328828679106654042733426185318122603041421600

Offset: 1

R. H. Hardin, Mar 16 2002

nonn

Cf. 2 X n A048739, 3 X n A069325, 4 X n A069326, 5 X n A069327, 6 X n A069328, 7 X n A069329, 8 X n A069330, 9 X n A069331, 10 X n A069332, 11 X n A069333, 12 X n A069334, 13 X n A069335, 14 X n A069336, 15 X n A069337, 16 X n A069338, 17 X n A069339, 18 X n A069340, 20 X n A069342, n X n A069343, n X n symmetric A069344.

a(5)-a(9) from Sean A. Irvine, Apr 28 2024

A049652 a(n) = (F(3*n+2) - 1)/4, where F=A000045 (the Fibonacci sequence).

Original entry on oeis.org

0, 1, 5, 22, 94, 399, 1691, 7164, 30348, 128557, 544577, 2306866, 9772042, 41395035, 175352183, 742803768, 3146567256, 13329072793, 56462858429, 239180506510, 1013184884470, 4291920044391, 18180865062035, 77015380292532, 326242386232164, 1381984925221189, 5854182087116921

Offset: 0

Clark Kimberling

From Anant Godbole, Apr 27 2006: (Start)
"a(n) equals the number of 2 by 2 determinants that need to be computed while finding the determinant of an n X n matrix using the method discovered by C. L. Dodgson (Lewis Carroll) in the 19th century.
"To evaluate the determinant of an n X n matrix A, set up a 2 by 2 determinant with entries that equal the determinants of the "northwest, northeast, southwest and southeast" (n-1) by (n-1) submatrices of A. Divide this by determinant of the "central" (n-2) by (n-2) submatrix of A. If the latter is zero, the problem can be fixed by row interchanges.
"The Dodgson method does better than the standard method of using cofactors and an expansion in terms of a row/column if n is 6 or larger. By this we mean that a fewer number of 2 by 2 determinants need to be calculated. Of course, the method of choice is diagonalization which can be achieved in polynomial time. Dodgson's method runs in exponential time, whereas the "standard" method requires one to evaluate n!/2 two by two determinants.
"A beautiful combinatorial proof of Dodgson's result was recently given by Zeilberger and an application is presented by Amdeberhan and Ekhad, where a conjecture of Kuperberg and Propp is proved using Dodgson's formula." (End)
This is the sequence A(0,1;4,1;1)of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000
T. Amdeberhan and S. Ekhad, A condensed condensation proof of a determinant evaluation conjectured by Greg Kuperberg and Jim Propp.
C. L. Dodgson, Condensation of determinants, Proceedings of the Royal Society of London, 15 (1866), 150-155.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
Doron Zeilberger, Dodgson's Determinant-Evaluation Rule Proved by Two-Timing Men and Women, Electron. J. Combin. 4 (no. 2, "The Wilf Festschrift") (1997), #R22, 2 pp. (p. 12).
Index entries for linear recurrences with constant coefficients, signature (5,-3,-1).

Cf. A000045, A000071, A048739, A099919.

[(Fibonacci(3*n+2) - 1)/4: n in [0..30]]; // G. C. Greubel, Dec 05 2017

a:= n-> add(fibonacci(i,4), i=0..n): seq(a(n), n=0..22); # Zerinvary Lajos, Mar 20 2008

s = 0; lst = {s}; Do[s += Fibonacci[n, 4]; AppendTo[lst, s], {n, 1, 22, 1}]; lst (* Zerinvary Lajos, Jul 14 2009 *)
LinearRecurrence[{5, -3, -1}, {0, 1, 5}, 30] (* or *) Table[(Fibonacci[ 3*n+2] - 1)/4, {n,0,30}] (* G. C. Greubel, Dec 05 2017 *)

a(n)=fibonacci(3*n+2)\4 \\ Charles R Greathouse IV, Jun 11 2015

a(n) = A099919(n)/2.
From Benoit Cloitre, May 06 2003: (Start)
a(0) = 0, a(1) = 1; a(n) = ceiling(r(4)*a(n-1)) where r(4) = 2+sqrt(5) is the positive root of X^2 = 4*X+1. More generally the sequence a(1) = 1, a(n) = ceiling(r(z)*a(n-1)) where r(z) = (1/2)*(z+sqrt(z^2+4)) is the positive root of X^2 = z*X+1 satisfies the linear recurrence: n > 3, a(n) = (z+1)*a(n-1) - (z-1)*a(n-2) - a(n-3) and the closed form formula: a(n) = floor(t(z)*r(z)^n) where t(z) = (1/(2*z))*(1+(z+2)/sqrt(z^2+4)) is the positive root of z*(z^2+4)*X^2 = (z^2+4)*X+1.
a(0) = 0, a(1) = 1, a(2) = 5, a(3) = 22, a(n) = 5*a(n-1) - 3*a(n-2) - a(n-3); a(n) = floor(t(4)*r(4)^n) where t(4) = (1/8)*(1+3/sqrt(5)) is the positive root of 80*X^2 = 20*X+1. (End)
a(n+2) = 4*a(n+1) + a(n) + 1. - Anant Godbole, Apr 27 2006
G.f.: x/((x-1)*(x^2+4*x-1)). - R. J. Mathar, Nov 23 2007
E.g.f.: exp(x)*(exp(x)*(5*cosh(sqrt(5)*x) + 3*sqrt(5)*sinh(sqrt(5)*x)) - 5)/20. - Stefano Spezia, May 24 2024

A104698 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(k, j)*binomial(n-j+1, k+1).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 6, 1, 5, 16, 19, 8, 1, 6, 25, 44, 33, 10, 1, 7, 36, 85, 96, 51, 12, 1, 8, 49, 146, 225, 180, 73, 14, 1, 9, 64, 231, 456, 501, 304, 99, 16, 1, 10, 81, 344, 833, 1182, 985, 476, 129, 18, 1, 11, 100, 489, 1408, 2471, 2668, 1765, 704, 163, 20, 1, 12

Offset: 0

Gary W. Adamson, Mar 19 2005

The n-th column of the triangle is the binomial transform of the n-th row of A081277, followed by zeros. Example: column 3, (1, 6, 19, 44, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). A104698 = reversal by rows of A142978. - Gary W. Adamson, Jul 17 2008
This sequence is jointly generated with A210222 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n > 1, u(n,x) = x*u(n-1,x) + v(n-1) + 1 and v(n,x) = 2x*u(n-1,x) + v(n-1,x) + 1. See the Mathematica section at A210222. - Clark Kimberling, Mar 19 2012
This Riordan triangle T appears in a formula for A001100(n, 0) = A002464(n), for n >= 1. - Wolfdieter Lang, May 13 2025

			The Riordan triangle T begins:
  n\k  0   1   2    3    4    5    6   7   8  9 10 ...
  ----------------------------------------------------
  0:   1
  1:   2   1
  2:   3   4   1
  3:   4   9   6    1
  4:   5  16  19    8    1
  5:   6  25  44   33   10    1
  6:   7  36  85   96   51   12    1
  7:   8  49 146  225  180   73   14   1
  8:   9  64 231  456  501  304   99  16   1
  9:  10  81 344  833 1182  985  476 129  18  1
  10: 11 100 489 1408 2471 2668 1765 704 163 20  1
  ... reformatted and extended by _Wolfdieter Lang_, May 13 2025
From _Wolfdieter Lang_, May 13 2025: (Start)
Zumkeller recurrence (adapted for offset [0,0]): 19 = T(4, 2) = T(2, 1) + T(3, 1) + T(3,3) = 4 + 9 + 6 = 19.
A-sequence recurrence: 19 = T(4, 2) = 1*T(3. 1) + 2*T(3. 2) - 2*T(3, 3) = 9 + 12 - 2 = 19.
Z-sequence recurrence: 5 = T(4, 0) = 2*T(3, 0) - 1*T(3, 1) + 2*T(3, 2) - 6*T(3, 3) = 8 - 9 + 12 + 6 = 5.
Boas-Buck recurrence: 19 = T(4, 2) = (1/2)*((2 + 0)*T(2, 2) + (2 + 2*2)*T(3, 2)) = (1/2)*(2 + 36) = 19. (End)

Reinhard Zumkeller, First 100 rows of triangle, flattened

Diagonal sums are A008937(n+1).
Cf. A048739 (row sums), A008288, A005900 (column 3), A014820 (column 4)
Cf. A081277, A142978 by antidiagonals, A119328, A110271 (matrix inverse).
Cf. A001100, A002464, A010673, A040000, A112478, A133872, A383857.

a104698 n k = a104698_tabl !! (n-1) !! (k-1)
a104698_row n = a104698_tabl !! (n-1)
a104698_tabl = [1] : [2,1] : f [1] [2,1] where
   f us vs = ws : f vs ws where
     ws = zipWith (+) ([0] ++ us ++ [0]) $
          zipWith (+) ([1] ++ vs) (vs ++ [0])
-- Reinhard Zumkeller, Jul 17 2015

A104698 := proc(n, k) add(binomial(k, j)*binomial(n-j+1, n-k-j), j=0..n-k) ; end proc:
seq(seq(A104698(n, k), k=0..n), n=0..15); # R. J. Mathar, Sep 04 2011
T := (n, k) -> binomial(n + 1, k + 1)*hypergeom([-k, k - n], [-n - 1], -1):
for n from 0 to 9 do seq(simplify(T(n, k)), k = 0..n) od;
T := proc(n, k) option remember; if k = 0 then n + 1 elif k = n then 1 else T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) fi end: # Peter Luschny, May 13 2025

u[1, ] = 1; v[1, ] = 1;
u[n_, x_] := u[n, x] = x u[n-1, x] + v[n-1, x] + 1;
v[n_, x_] := v[n, x] = 2 x u[n-1, x] + v[n-1, x] + 1;
Table[CoefficientList[u[n, x], x], {n, 1, 11}] // Flatten (* Jean-François Alcover, Mar 10 2019, after Clark Kimberling *)

T(n,k)=sum(j=0,n-k,binomial(k,j)*binomial(n-j+1,k+1)) \\ Charles R Greathouse IV, Jan 16 2012

The triangle is extracted from the product A * B; A = [1; 1, 1; 1, 1, 1; ...], B = [1; 1, 1; 1, 3, 1; 1, 5, 5, 1; ...] both infinite lower triangular matrices (rest of the terms are zeros). The triangle of matrix B by rows = A008288, Delannoy numbers.
From Paul Barry, Jul 18 2005: (Start)
Riordan array (1/(1-x)^2, x(1+x)/(1-x)) = (1/(1-x), x)*(1/(1-x), x(1+x)/(1-x)).
T(n, k) = Sum_{j=0..n} Sum_{i=0..j-k} C(j-k, i)*C(k, i)*2^i.
T(n, k) = Sum_{j=0..k} Sum_{i=0..n-k-j} (n-k-j-i+1)*C(k, j)*C(k+i-1, i). (End)
T(n, k) = binomial(n+1, k+1)*2F1([-k, k-n], [-n-1], -1) where 2F1 is a Gaussian hypergeometric function. - R. J. Mathar, Sep 04 2011
T(n, k) = T(n-2, k-1) + T(n-1, k-1) + T(n-1, k) for 1 < k < n; T(n, 0) = n + 1; T(n, n) = 1. - Reinhard Zumkeller, Jul 17 2015
From Wolfdieter Lang, May 13 2025: (Start)
The Riordan triangle T = (1/(1 - x)^2, x*(1 + x)/(1 - x)) has the o.g.f. G(x, y) = 1/((1 - x)*(1 - x - y*x*(1+x))) for the row polynomials R(n, y) = Sum_{k=0..n} T(n, k)*y^k.
The o.g.f. for column k is G(k, x) = (1/(1 - x)^2)*(x*(1 + x)/(1 - x))^k, for k >= 0.
The o.g.f. for the diagonal m is D(m, x) = N(m, x)/(1 - x)^(m+1), with the numerator polynomial N(m, x) = Sum_{k=0..floor(m/2)} A034867(m, k)*x^(2*k) for m >= 0.
The row sums with o.g.f. R(x) = 1/((1 -x)*(1 - 2*x -x^2) give A048739.
The alternating row sums with o.g.f. 1/((1 - x)(1 + x^2)) give A133872.
The A-sequence for this Riordan triangle has o.g.f. A(x) = 1 + x + sqrt(1 + 6*x + x^2))/2 giving A112478(n). Hence T(n, k) = Sum_{j=0..n-k} A112478(j)*T(n-1, k-1+j), for n >= 1, k >= 1, T(n, k) = 0 for n < k, and T(0, 0) = 1.
The Z-sequence has o.g.f. (5 + x - sqrt(1 + 6*x + x^2))/2 = 3 + x - A(x) giving Z(n) = {2, -1, -A112478(n >= 2)}. Hence T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), for n >= 1. For A- and Z-sequences of Riordan triangles see a W. Lang link at A006232 with references.
The Boas-Buck sequences alpha and beta for the Riordan triangle T (see A046521 for the Aug 10 2017 comment and reference) are alpha(n) = A040000(n+1) = repeat{2} and beta(n) = A010673(n+1) = repeat{2,0}. Hence the recurrence for column T(n, k){n>=k}, with input T(k, k) = 1, for k >= 0, is T(n, k) = (1/(n-k)) * Sum{j=k..n-1} (2 + k*(1 + (-1)^(n-1-j))) *T(j,k), for n >= k+1. (End)

A359573 Array read by antidiagonals: T(m,n) is the number of m X n binary arrays with all 1's connected and a path of 1's from upper left corner to lower right corner.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 8, 8, 1, 1, 20, 45, 20, 1, 1, 49, 234, 234, 49, 1, 1, 119, 1193, 2423, 1193, 119, 1, 1, 288, 6049, 24455, 24455, 6049, 288, 1, 1, 696, 30616, 245972, 482443, 245972, 30616, 696, 1, 1, 1681, 154861, 2473317, 9469361, 9469361, 2473317, 154861, 1681, 1

Offset: 1

Andrew Howroyd, Jan 06 2023

			Array begins:
================================================================
m\n| 1   2     3       4         5           6             7
---+------------------------------------------------------------
1  | 1   1     1       1         1           1             1 ...
2  | 1   3     8      20        49         119           288 ...
3  | 1   8    45     234      1193        6049         30616 ...
4  | 1  20   234    2423     24455      245972       2473317 ...
5  | 1  49  1193   24455    482443     9469361     185899132 ...
6  | 1 119  6049  245972   9469361   360923899   13742823032 ...
7  | 1 288 30616 2473317 185899132 13742823032 1012326365581 ...
  ...

Andrew Howroyd, Table of n, a(n) for n = 1..435

Main diagonal is A163002.
Rows 1..10 are A000012, A048739(n-1), A163003, A163004, A163005, A163006, A163007, A163008, A163009, A163010.
Cf. A287151, A359574, A359575, A359576.

T(m,n) = T(n,m).

A083087 Square table read by antidiagonals which forms a permutation of the natural numbers: T(n,0) = floor(nx/(x-1))+1, T(n,k+1) = ceiling(xT(n,k)), for n>=0, k>=0, where x = 1 + sqrt(2).

Original entry on oeis.org

1, 3, 2, 8, 5, 4, 20, 13, 10, 6, 49, 32, 25, 15, 7, 119, 78, 61, 37, 17, 9, 288, 189, 148, 90, 42, 22, 11, 696, 457, 358, 218, 102, 54, 27, 12, 1681, 1104, 865, 527, 247, 131, 66, 29, 14, 4059, 2666, 2089, 1273, 597, 317, 160, 71, 34, 16, 9800, 6437, 5044, 3074

Offset: 0

Paul D. Hanna, Apr 21 2003

The array in A083087 is the dispersion of the sequence given floor(n+1+n*sqrt(2)). The Mathematica program at A191438 generates A083087 using f[n_]:=Floor[n*x+n+1] instead of f[n_]:=Floor[n*x+n]. - Clark Kimberling, Jun 04 2011

			Table begins:
   1  3   8  20  49  119  288 ...
   2  5  13  32  78  189  457 ...
   4 10  25  61 148  358  865 ...
   6 15  37  90 218  527 1273 ...
   7 17  42 102 247  597 1442 ...
   9 22  54 131 317  766 1850 ...
  11 27  66 160 387  935 2258 ...
  12 29  71 172 416 1005 2427 ...
  14 34  83 201 486 1174 2835 ...
  16 39  95 230 556 1343 3243 ...
  18 44 107 259 626 1512 3651 ...
  19 46 112 271 655 1582 3820 ...
  21 51 124 300 725 1751 4228 ...
  23 56 136 329 795 1920 4636 ...
  24 58 141 341 824 1990 4805 ...

Cf. A083088 (first column), A048739 (first row), A083090 (diagonal), A083091 (antidiagonal sums), A083044, A083047, A083050.

z:=10; x:=1+Sqrt(2); S:=[]; for n in [0..z] do for k in [0..n] do if n-k eq 0 then Append(~S, Floor(n*x/(x-1))+1); else Append(~S, Ceiling(x*S[k+1+(n*(n-1) div 2)])); end if; end for; end for; S; // Klaus Brockhaus, Jan 04 2011

Mathematica
```
(See Comments.)
```

T(n,k+1) = 2*T(n,k) + T(n,k-1) + 1 for n>=0, k>=1.

A098790 a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.

Original entry on oeis.org

1, 2, 6, 15, 37, 90, 218, 527, 1273, 3074, 7422, 17919, 43261, 104442, 252146, 608735, 1469617, 3547970, 8565558, 20679087, 49923733, 120526554, 290976842, 702480239, 1695937321, 4094354882, 9884647086, 23863649055, 57611945197

Offset: 0

Creighton Dement, Oct 30 2004

nonn

Previous name was: a(n) = A048739(n) - A000129(n).

Partial sums of Pell numbers A000129 except omit next-to-last Pell number. E.g., 37 = 0+1+2+5+12+29 - 12.

M. Bicknell-Johnson and G. E. Bergum, The Generalized Fibonacci Numbers {C(n)}, C(n)=C(n-1)+C(n-2)+K, Applications of Fibonacci Numbers, 1986, pp. 193-205.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
M. Bicknell, A Primer on the Pell Sequence and related sequences, Fibonacci Quarterly, Vol. 13, No. 4, 1975, pp. 345-349.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434.
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (3, -1, -1).

Cf. A006451, A000129, A048739, A124124, A024537, A074323, A052542.

a[0] = 1; a[1] = 2; a[n_] := a[n] = 2a[n - 1] + a[n - 2] + 1; Table[ a[n], {n, 0, 28}] (* Robert G. Wilson v, Nov 04 2004 *)
LinearRecurrence[{3,-1,-1},{1,2,6},31] (* Harvey P. Dale, Oct 15 2011 *)
CoefficientList[Series[(x^2 - x + 1)/((1 - x) (1 - 2 x - x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Aug 14 2014 *)

a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.
G.f.: (x^2-x+1)/((1-x)(1-2x-x^2)).
a(n+1) = - A024537(n+1) + 2*A048739(n+1) - 2*A048739(n).
a(n) = - A024537(n) + A052542(n+1).
Partial sums of A074323. - Paul Barry, Mar 11 2007
a(n) = (sqrt(2)+1)^n*(3/4+sqrt(2)/4)+(sqrt(2)-1)^n*(3/4-sqrt(2)/4)*(-1)^n-1/2; - Paul Barry, Mar 11 2007
a(0)=1, a(1)=2, a(2)=6, a(n)=3*a(n-1)-a(n-2)-a(n-3). [Harvey P. Dale, Oct 15 2011]
a(2*n) = A124124(2*n+1). - Hermann Stamm-Wilbrandt, Aug 03 2014
a(2*n+1) = A006451(2*n+1). - Hermann Stamm-Wilbrandt, Aug 26 2014
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>5. - Hermann Stamm-Wilbrandt, Aug 26 2014
2*a(n) = A135532(n+1)-1. - R. J. Mathar, Jan 13 2023

More terms from Robert G. Wilson v, Nov 04 2004
Definition edited by N. J. A. Sloane, Aug 03 2014
New name from existing formula by Joerg Arndt, Aug 13 2014

A352319 Numbers whose minimal (or greedy) Pell representation (A317204) is palindromic.

Original entry on oeis.org

0, 1, 3, 6, 8, 13, 20, 30, 35, 40, 44, 49, 71, 88, 102, 119, 170, 182, 194, 204, 216, 238, 242, 254, 266, 276, 288, 409, 450, 484, 525, 559, 580, 621, 655, 696, 986, 1015, 1044, 1068, 1097, 1150, 1160, 1189, 1218, 1242, 1271, 1334, 1363, 1392, 1396, 1425, 1454

Offset: 1

Amiram Eldar, Mar 12 2022

A052937(n) = A000129(n+1)+1 is a term for n>0, since its minimal Pell representation is 10...01 with n-1 0's between two 1's.
A048739 is a subsequence since these are repunit numbers in the minimal Pell representation.
A001109 is a subsequence. The minimal Pell representation of A001109(n), for n>1, is 1010...01, with n-1 0's interleaved with n 1's.

			The first 10 terms are:
   n  a(n)  A317204(a(n))
  --  ----  -------------
   1     0              0
   2     1              1
   3     3             11
   4     6            101
   5     8            111
   6    13           1001
   7    20           1111
   8    30          10001
   9    35          10101
  10    40          10201

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

Cf. A000129, A317204.
Subsequences: A001109, A048739, A052937 \ {2}.
Similar sequences: A002113, A006995, A014190, A094202, A331191, A351712, A351717, A352087.

pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; q[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; PalindromeQ[IntegerDigits[Total[3^(s - 1)], 3]]]; Select[Range[0, 1500], q]

A352341 Numbers whose maximal Pell representation (A352339) is palindromic.

Original entry on oeis.org

0, 1, 3, 6, 8, 10, 20, 27, 40, 49, 54, 58, 63, 68, 88, 93, 119, 136, 150, 167, 221, 238, 288, 300, 310, 322, 334, 338, 360, 372, 382, 394, 406, 508, 530, 542, 696, 737, 771, 812, 833, 867, 908, 942, 983, 1242, 1276, 1317, 1392, 1681, 1710, 1734, 1763, 1792, 1802

Offset: 1

Amiram Eldar, Mar 12 2022

A000129(n) - 2 is a term for n > 1. The maximal Pell representations of these numbers are 0, 11, 121, 1221, 12221, ... (0 and A132583).
A048739 is a subsequence since these are the repunit numbers in the maximal Pell representation.
A065113 is a subsequence since the maximal Pell representation of A065113(n) is 2*n 2's.

			The first 10 terms are:
   n  a(n)  A352339(a(n))
  --  ----  -------------
   1    0               0
   2    1               1
   3    3              11
   4    6              22
   5    8             111
   6   10             121
   7   20            1111
   8   27            1221
   9   40            2222
  10   49           11111

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

Cf. A000129, A048739, A065113, A132583, A352339.

Similar sequences: A002113, A006995, A014190, A094202, A331191, A351712, A351717, A352087, A352105, A352319.

pell[1] = 1; pell[2] = 2; pell[n_] := pell[n] = 2*pell[n - 1] + pell[n - 2]; pellp[n_] := Module[{s = {}, m = n, k}, While[m > 0, k = 1; While[pell[k] <= m, k++]; k--; AppendTo[s, k]; m -= pell[k]; k = 1]; IntegerDigits[Total[3^(s - 1)], 3]]; lazy[n_] := Module[{v = pellp[n]}, nv = Length[v]; i = 1; While[i <= nv - 2, If[v[[i]] > 0 && v[[i + 1]] == 0 && v[[i + 2]] < 2, v[[i ;; i + 2]] += {-1, 2, 1}; If[i > 2, i -= 3]]; i++]; i = Position[v, _?(# > 0 &)]; If[i == {}, 0, FromDigits[v[[i[[1, 1]] ;; -1]]]]]; Select[Range[0, 2000], PalindromeQ[lazy[#]] &]

A270342 Positive integers n such that the sum of the Pell numbers A000129(0) + ... + A000129(n-1) is divisible by n.

Original entry on oeis.org

3, 4, 5, 7, 8, 11, 13, 16, 17, 19, 23, 24, 29, 31, 32, 37, 41, 43, 47, 48, 53, 59, 61, 64, 67, 71, 72, 73, 79, 83, 89, 96, 97, 101, 103, 107, 109, 113, 120, 127, 128, 131, 137, 139, 144, 149, 151, 157, 163, 167, 168, 169, 173, 179, 181, 191, 192, 193, 197, 199, 211, 216, 223, 227, 229

Offset: 1

Altug Alkan, Mar 15 2016

nonn

Sequence contains all odd primes because of the fact that ((1-sqrt(2))^p + (1+sqrt(2))^p - 2) is divisible by p where p is an odd prime.

			3 is a term because 0 + 1 + 2 = 3 is divisible by 3.
4 is a term because 0 + 1 + 2 + 5 = 8 is divisible by 4.
5 is a term because 0 + 1 + 2 + 5 + 12 = 20 is divisible by 5.
7 is a term because 0 + 1 + 2 + 5 + 12 + 20 + 79 = 119 is divisible by 7.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000040, A000129, A048739.

Module[{nn=250,pell},pell=LinearRecurrence[{2,1},{0,1},nn];Position[ Table[ Total[Take[pell,n]]/n,{n,nn}],?(IntegerQ[#]&)]]//Flatten (* _Harvey P. Dale, Nov 11 2021 *)

a048739(n) = local(w=quadgen(8)); -1/2+(3/4+1/2*w)*(1+w)^n+(3/4-1/2*w)*(1-w)^n;
for(n=1, 1e3, if(a048739(n-1) % (n+1) == 0, print1(n+1, ", ")));

A359575 Array read by antidiagonals: T(m,n) is the number of m X n binary arrays with a path of adjacent 1's from upper right corner to lower left corner.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 8, 8, 1, 1, 20, 51, 20, 1, 1, 49, 295, 295, 49, 1, 1, 119, 1632, 3828, 1632, 119, 1, 1, 288, 8830, 46557, 46557, 8830, 288, 1, 1, 696, 47239, 546286, 1225194, 546286, 47239, 696, 1, 1, 1681, 251261, 6279393, 30754544, 30754544, 6279393, 251261, 1681, 1

Offset: 1

Andrew Howroyd, Jan 06 2023

			Array begins:
================================================================
m\n| 1   2     3       4         5           6             7
---+------------------------------------------------------------
1  | 1   1     1       1         1           1             1 ...
2  | 1   3     8      20        49         119           288 ...
3  | 1   8    51     295      1632        8830         47239 ...
4  | 1  20   295    3828     46557      546286       6279393 ...
5  | 1  49  1632   46557   1225194    30754544     749866185 ...
6  | 1 119  8830  546286  30754544  1636193228   83949041929 ...
7  | 1 288 47239 6279393 749866185 83949041929 9009490924794 ...
  ...

Main diagonal is A069343.
Rows 1..20 are A000012, A048739(n-1), A069325, A069326, A069327-A069342.
Cf. A359573, A359576.

T(m,n) = T(n,m).

cp's OEIS Frontend

A069341 Number of 19 X n binary arrays with path of adjacent 1's from upper right corner to lower left corner.

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A049652 a(n) = (F(3*n+2) - 1)/4, where F=A000045 (the Fibonacci sequence).

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A104698 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(k, j)*binomial(n-j+1, k+1).

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A359573 Array read by antidiagonals: T(m,n) is the number of m X n binary arrays with all 1's connected and a path of 1's from upper left corner to lower right corner.

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A083087 Square table read by antidiagonals which forms a permutation of the natural numbers: T(n,0) = floor(n*x/(x-1))+1, T(n,k+1) = ceiling(x*T(n,k)), for n>=0, k>=0, where x = 1 + sqrt(2).

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A098790 a(n) = 2*a(n-1) + a(n-2) + 1, a(0) = 1, a(1) = 2.

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A352319 Numbers whose minimal (or greedy) Pell representation (A317204) is palindromic.

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A352341 Numbers whose maximal Pell representation (A352339) is palindromic.

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A083087 Square table read by antidiagonals which forms a permutation of the natural numbers: T(n,0) = floor(nx/(x-1))+1, T(n,k+1) = ceiling(xT(n,k)), for n>=0, k>=0, where x = 1 + sqrt(2).