A099837 -id:A099837 - cp's OEIS frontend

A004483 Tersum n + 2.

2, 0, 1, 5, 3, 4, 8, 6, 7, 11, 9, 10, 14, 12, 13, 17, 15, 16, 20, 18, 19, 23, 21, 22, 26, 24, 25, 29, 27, 28, 32, 30, 31, 35, 33, 34, 38, 36, 37, 41, 39, 40, 44, 42, 43, 47, 45, 46, 50, 48, 49, 53, 51, 52, 56, 54, 55, 59, 57, 58, 62, 60, 61, 65, 63, 64, 68, 66, 67

Offset: 0

N. J. A. Sloane

Tersum m + n: write m and n in base 3 and add mod 3 with no carries; e.g., 5 + 8 = "21" + "22" = "10" = 1.

Also Sprague-Grundy values for game of Wyt Queens.

E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 76.

Andreas Dress, Achim Flammenkamp, and Norbert Pink, Additive periodicity of the Sprague-Grundy function of certain Nim games, Adv. Appl. Math., 22, p. 249-270 (1999).
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

This sequence is row 2 of table A004481.

Second column of triangle in A296339.

a[n_] := If[Divisible[n, 3], n+2, n-1]; Table[a[n], {n, 0, 70}] (* Jean-François Alcover, Oct 25 2013 *)
LinearRecurrence[{1,0,1,-1},{2,0,1,5},70] (* Harvey P. Dale, Feb 07 2018 *)

Periodic with period and saltus 3: a(n) = 3*floor(n/3) + ((n+2) mod 3).
a(n) = n + 2*cos(2*n*Pi/3). - Wesley Ivan Hurt, Sep 27 2017
From R. J. Mathar, Dec 14 2017: (Start)
G.f.: ( 2+x^2+2*x^3-2*x ) / ( (1+x+x^2)*(x-1)^2 ).
a(n) = n + A099837(n) if n > 0. (End)
Sum_{n>=2} (-1)^n/a(n) = 2*Pi/(3*sqrt(3)) + log(2)/3 - 1/2. - Amiram Eldar, Aug 21 2023

Edited by N. J. A. Sloane at the suggestion of Philippe Deléham, Nov 20 2007

A110307 Expansion of (1+2x)/((1+x+x^2)(1+5*x+x^2)).

Original entry on oeis.org

1, -4, 17, -80, 384, -1842, 8827, -42292, 202631, -970862, 4651680, -22287540, 106786021, -511642564, 2451426797, -11745491420, 56276030304, -269634660102, 1291897270207, -6189851690932, 29657361184451, -142096954231322, 680827409972160, -3262040095629480

Offset: 0

Creighton Dement, Jul 19 2005

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-6,-7,-6,-1).

Cf. A002320, A004253, A049347, A099837, A110308, A110309, A110310, A110311.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( (1+2*x)/((1+x+x^2)*(1+5*x+x^2)) )); // G. C. Greubel, Jan 03 2023

seriestolist(series((1+2*x)/((x^2+x+1)*(x^2+5*x+1)), x=0,25));

LinearRecurrence[{-6,-7,-6,-1}, {1,-4,17,-80}, 41] (* G. C. Greubel, Jan 03 2023 *)

Vec((1+2*x)/((1+x+x^2)*(1+5*x+x^2)) + O(x^25)) \\ Colin Barker, Apr 30 2019

def U(n,x): return chebyshev_U(n, x)
def A110307(n): return (1/4)*(3*U(n,-5/2) +U(n-1,-5/2) +U(n,-1/2) -U(n-1,-1/2))
[A110307(n) for n in range(41)] # G. C. Greubel, Jan 03 2023

a(n+2) = - 5*a(n+1) - a(n) - A099837(n+1).
a(n) + a(n+1) + a(n+2) = A002320(n).
a(n) = -6*a(n-1) - 7*a(n-2) - 6*a(n-3) - a(n-4) for n>3. - Colin Barker, Apr 30 2019
a(n) = (1/4)*(3*U(n,-5/2) + U(n-1,-5/2) + U(n,-1/2) - U(n-1,-1/2)), where U(n, x) = ChebyshevU(n, x). - G. C. Greubel, Jan 03 2023

A110308 Expansion of -x(2+x)/((1+x+x^2)(1+5*x+x^2)).

Original entry on oeis.org

0, -2, 11, -52, 247, -1182, 5664, -27140, 130037, -623044, 2985181, -14302860, 68529120, -328342742, 1573184591, -7537580212, 36114716467, -173036002122, 829065294144, -3972290468600, 19032387048857, -91189644775684, 436915836829561, -2093389539372120

Offset: 0

Creighton Dement, Jul 19 2005

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-6,-7,-6,-1).

Cf. A004253, A049347, A099837, A110307, A110309, A110310, A110311.

R:=PowerSeriesRing(Integers(), 40); [0] cat Coefficients(R!( -x*(2+x)/((1+x+x^2)*(1+5*x+x^2)) )); // G. C. Greubel, Jan 03 2023

seriestolist(series(-x*(2+x)/((x^2+x+1)*(x^2+5*x+1)), x=0,25));

LinearRecurrence[{-6,-7,-6,-1}, {0,-2,11,-52}, 40] (* G. C. Greubel, Jan 03 2023 *)

concat(0, Vec(-x*(2+x)/((1+x+x^2)*(1+5*x+x^2)) + O(x^25))) \\ Colin Barker, Apr 30 2019

def U(n, x): return chebyshev_U(n,x)
def A110308(n): return (1/4)*(2*U(n, -5/2) +U(n-1, -5/2) -2*U(n, -1/2) -U(n-1, -1/2))
[A110308(n) for n in range(41)] # G. C. Greubel, Jan 03 2023

a(n+2) = - 5*a(n+1) - a(n) - A099837(n+2).
a(n) = -6*a(n-1) - 7*a(n-2) - 6*a(n-3) - a(n-4) for n>3. - Colin Barker, Apr 30 2019
a(n) = (1/4)*(2*U(n, -5/2) + U(n-1, -5/2) - 2*U(n, -1/2) - U(n-1, -1/2)), where U(n, x) = ChebyshevU(n, x). - G. C. Greubel, Jan 03 2023

A219233 Alternating row sums of Riordan triangle A110162.

Original entry on oeis.org

1, -3, 7, -18, 47, -123, 322, -843, 2207, -5778, 15127, -39603, 103682, -271443, 710647, -1860498, 4870847, -12752043, 33385282, -87403803, 228826127, -599074578, 1568397607, -4106118243, 10749957122, -28143753123, 73681302247, -192900153618, 505019158607

Offset: 0

Wolfdieter Lang, Nov 16 2012

If a(0) is put to 2 instead of 1 this becomes a(n) = (-1)^n*A005248(n), n >= 0. These are then the alternating row sums of triangle A127677.

Also abs(a(n)) is the number of rounded area of pentagon or pentagram in series arrangement. - Kival Ngaokrajang, Mar 27 2013

Colin Barker, Table of n, a(n) for n = 0..1000
Richard M. Low and Ardak Kapbasov, Non-Attacking Bishop and King Positions on Regular and Cylindrical Chessboards, Journal of Integer Sequences, Vol. 20 (2017), Article 17.6.1, Table 8.
Kival Ngaokrajang, Pentagram for n = 1..6
Eric Weisstein's World of Mathematics, Pentagram
Index entries for linear recurrences with constant coefficients, signature (-3,-1).

Cf. A099837 (row sums of A110162).

Cf. A000032, A000045, A001254, A005248, A075150, A127677.

A219233:= func< n | n eq 0 select 1 else (-1)^n*Lucas(2*n) >; // G. C. Greubel, Jun 13 2025

A219233[n_]:= (-1)^n*LucasL[2*n] - Boole[n==0]; (* G. C. Greubel, Jun 13 2025 *)

Vec((1-x^2)/(1+3*x+x^2) + O(x^40)) \\ Colin Barker, Oct 14 2015

def A219233(n): return (-1)**n*lucas_number2(2*n,1,-1) - int(n==0) # G. C. Greubel, Jun 13 2025

a(0) = 1 and a(n) = (-1)^n*(F(2*(n+1)) - F(2*(n-1))) = (-1)^n*L(2*n), n>=1, with F=A000045 (Fibonacci) and L=A000032 (Lucas).
O.g.f.: (1-x^2)/(1+3*x+x^2).
G.f.: (W(0) -6)/(5*x) -1 , where W(k) = 5*x*k + x + 6 - 6*x*(5*k-9)/W(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Aug 19 2013
From Colin Barker, Oct 14 2015: (Start)
a(n) = -3*a(n-1) - a(n-2) for n>2.
a(n) = (1/2*(-3-sqrt(5)))^n + (1/2*(-3+sqrt(5)))^n for n>0. (End)
E.g.f.: 2*exp(-3*x/2)*cosh(sqrt(5)*x/2) - 1. - Stefano Spezia, Dec 26 2021
From G. C. Greubel, Jun 13 2025: (Start)
a(-n) = a(n).
a(n) = (-1)^n*A001254(n) - 2 - [n=0] = A075150(n) - 2 - [n=0]. (End)

A234044 Period 7: repeat [2, -2, 1, 0, 0, 1, -2].

Original entry on oeis.org

2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2, 1, 0, 0, 1, -2, 2, -2

Offset: 0

Wolfdieter Lang, Feb 27 2014

This is a member of the six sequences which appear for the instance N=7 of the general formula 2*exp(2*Pi*n*I/N) = R(n, x^2-2) + x*S(n-1, x^2-2)*s(N)*I, for n >= 0, with I = sqrt(-1), s(N) = sqrt(2-x)*sqrt(2+x), x = rho(N) := 2*cos(Pi/N) and R and S are the monic Chebyshev polynomials whose coefficient tables are given in A127672 and A049310. If powers x^k with k >= delta(N) = A055034(N) enter in R or x*S then C(N, x), the minimal polynomial of x = rho(N) (see A187360) is used for a reduction. If delta(N) = 2 it may happen that sqrt(2+x) or sqrt(2-x) is an integer in the number field Q(rho(N)). See the N=5 case comment on A164116.

For N=7 with delta(7) = 3, and C(7, x) = x^3 - x^2 - 2*x + 1 the final result becomes 2*exp(2*Pi*n*I/7) = (a(n) + b(n)*x + c(n)*x^2) + (A(n) + B(n)*x + C(n)*x^2)*s(7)*I, with x = rho(7) = 2*cos(Pi/7), a(n) the present sequence, b(n) = A234045(n), c(n) = A234046(n), A(n) = A238468(n), B(n) = A238469(n) and C(n) = A238470(n). The a, b, c and A, B, C brackets are integers in Q(rho(7)).

			n = 4: 2*exp(8*Pi*I/7) = (2-16*x^2+20*x^4-8*x^6+x^8) + (4*x+10*x^3-6*x^5+x^7)*s(7)*I, reduced with C(7, x) = x^3 - x^2 - 2*x + 1 = 0 this becomes = (-x) + (-1)*s(7)*I with x= 2*cos(Pi/7) and s(7) = 2*sin(Pi/7).The power basis coefficients are thus (a(4), b(4), c(4)) = (0, -1, 0) and (A(4), B(4), C(4)) = (-1, 0, 0).

Index entries for linear recurrences with constant coefficients, signature (-1,-1,-1,-1,-1,-1).

Cf. A234045, A234046, A238468, A238469, A238470, A099837 (N=3), A056594 (N=4), A164116 (N=5), A057079 (N=6).

&cat [[2, -2, 1, 0, 0, 1, -2]^^20]; // Wesley Ivan Hurt, Jul 16 2016

seq(op([2, -2, 1, 0, 0, 1, -2]), n=0..20); # Wesley Ivan Hurt, Jul 16 2016

PadRight[{}, 100, {2, -2, 1, 0, 0, 1, -2}] (* Wesley Ivan Hurt, Jul 16 2016 *)

a(n)=[2, -2, 1, 0, 0, 1, -2][n%7+1] \\ Charles R Greathouse IV, Jul 17 2016

G.f.: (2 - 2*x + x^2 + x^5 - 2*x^6)/(1 - x^7).
a(n+7) = a(n) for n>=0, with a(0) = -a(1) = -a(6) = 2, a(3) = a(4) =0 and a(2) = a(5) = 1.
From Wesley Ivan Hurt, Jul 16 2016: (Start)
a(n) + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) + a(n-6) = 0 for n>5.
a(n) = (1/7) * Sum_{k=1..6} 2*cos((2k)*n*Pi/7) - 2*cos((2k)*(1+n)*Pi/7) + cos((2k)*(2+n)*Pi/7) + cos((2k)*(5+n)*Pi/7) - 2*cos((2k)*(6+n)*Pi/7).
a(n) = 2 + 4*floor(n/7) - 3*floor((1+n)/7) + floor((2+n)/7) - floor((4+n)/7) + 3*floor((5+n)/7) - 4*floor((6+n)/7). (End)

A008577 Crystal ball sequence for planar net 4.8.8.

Original entry on oeis.org

1, 4, 9, 17, 28, 41, 57, 76, 97, 121, 148, 177, 209, 244, 281, 321, 364, 409, 457, 508, 561, 617, 676, 737, 801, 868, 937, 1009, 1084, 1161, 1241, 1324, 1409, 1497, 1588, 1681, 1777, 1876, 1977, 2081, 2188, 2297, 2409, 2524, 2641, 2761, 2884, 3009, 3137, 3268

Offset: 0

N. J. A. Sloane

			G.f. = 1 + 4*x + 9*x^2 + 17*x^3 + 28*x^4 + 41*x^5 + 67*x^6 + ... - _Michael Somos_, May 02 2020

T. D. Noe, Table of n, a(n) for n = 0..1000
Brian Galebach, Uniform Tiling 2 of 11
W. M. Meier and H. J. Moeck, Topology of 3-D 4-connected nets ..., J. Solid State Chem 27 1979 349-355, esp. p. 351.
Index entries for crystal ball sequences
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Partial sums of A008576.

Cf. A099837, A164359.

A099837[0] = 1; A099837[n_] := Mod[n+2, 3] - Mod[n, 3]; a[n_] := 4*n/3*(n+1) + 10/9 + A099837[n+2]/9; Table[a[n], {n, 0, 39}] (* Jean-François Alcover, Feb 15 2012, after R. J. Mathar *)
CoefficientList[Series[((1 + x)^2 (1 + x^2))/((1 - x)^3 (1 + x + x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Dec 31 2015 *)
LinearRecurrence[{2,-1,1,-2,1},{1,4,9,17,28},40] (* Harvey P. Dale, Dec 17 2017 *)
a[ n_] := Quotient[(n^2 + n + 1)*4, 3]; (* Michael Somos, May 02 2020 *)

a(n) = (n^2 + n + 1)*4\3; /* Michael Somos, May 02 2020 */

G.f.: ((1+x)^2*(1+x^2)) / ((1-x)^3*(1+x+x^2)). - Ralf Stephan, Apr 24 2004
a(n) = 4*(n/3)*(n+1)+10/9+A099837(n+2)/9. - R. J. Mathar, Nov 20 2010
The above g.f. and formula were originally stated as conjectures, but I now have a proof. This also justifies the b-file. Details will be added later. - N. J. A. Sloane, Dec 29 2015
From Michael Somos, May 02 2020: (Start)
Euler transform of length 3 sequence [4, -1, 1, -1].
a(n) = a(-1-n) = floor((n^2+n+1)*4/3) for all n in Z.
a(n) - 2*a(n+1) + a(n+2) = A164359(n) unless n=0.
(End)

A100050 A Chebyshev transform of n.

Original entry on oeis.org

0, 1, 2, 0, -4, -5, 0, 7, 8, 0, -10, -11, 0, 13, 14, 0, -16, -17, 0, 19, 20, 0, -22, -23, 0, 25, 26, 0, -28, -29, 0, 31, 32, 0, -34, -35, 0, 37, 38, 0, -40, -41, 0, 43, 44, 0, -46, -47, 0, 49, 50, 0, -52, -53, 0, 55, 56, 0, -58, -59, 0, 61, 62, 0, -64, -65, 0, 67, 68, 0, -70, -71, 0, 73, 74, 0, -76, -77, 0, 79, 80, 0, -82, -83, 0, 85, 86, 0

Offset: 0

Paul Barry, Oct 31 2004

A Chebyshev transform of x/(1-x)^2: if A(x) is the g.f. of a sequence, map it to ((1-x^2)/(1+x^2))A(x/(1+x^2)).

			x + 2*x^2 - 4*x^4 - 5*x^5 + 7*x^7 + 8*x^8 - 10*x^10 - 11*x^11 + 13*x^13 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Michael Somos, Rational Function Multiplicative Coefficients
Index entries for linear recurrences with constant coefficients, signature (2,-3,2,-1).

Cf. A165202 (partial sums).

Cf. A099837, A099443, A011655, A100047, A100048, A100051, A091684.

LinearRecurrence[{2, -3, 2, -1}, {0, 1, 2, 0},50] (* G. C. Greubel, Aug 08 2017 *)

{a(n) = n * (-1)^(n\3) * sign( n%3)} /* Michael Somos, Mar 19 2011 */

{a(n) = local(A, p, e); if( abs(n)<1, 0, A = factor(abs(n)); prod( k=1, matsize(A)[1], if( p=A[k,1], e=A[k,2]; if( p==2, -(-2)^e, (kronecker( -12, p) * p)^e))))} /* Michael Somos, Mar 19 2011 */

[lucas_number1(n,2,1)*lucas_number1(n,1,1) for n in range(0,88)] # Zerinvary Lajos, Jul 06 2008

Euler transform of length 6 sequence [ 2, -3, -2, 0, 0, 2]. - Michael Somos, Mar 19 2011
a(n) is multiplicative with a(2^e) = -(-2)^e if e>0, a(3^e) = 0^e, a(p^e) = p^e if p == 1 (mod 6), a(p^e) = (-p)^e if p == 5 (mod 6). - Michael Somos, Mar 19 2011
G.f.: x*(1 - x^2)^3 *(1 - x^3)^2 / ((1 - x)^2 *(1 - x^6)^2) = x *(1 + x)^2 *(1 - x^2) / (1 + x^3)^2. - Michael Somos, Mar 19 2011
a(3*n) = 0, a(3*n + 1) = (-1)^n * (3*n + 1), a(3*n + 2) = (-1)^n * (3*n + 2). a(-n) = a(n). - Michael Somos, Mar 19 2011
G.f.: x(1-x^2)/(1-x+x^2)^2.
a(n) = 2*a(n-1) -3*a(n-2) +2*a(n-3) -a(n-4).
a(n) = n*Sum_{k=0..floor(n/2)} (-1)^k*binomial(n-k,k)*(n-2k)/(n-k).

A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

Original entry on oeis.org

1, 3, 19, 144, 1171, 9878, 85216, 746371, 6609043, 59008563, 530279894, 4790262348, 43458522976, 395683988547, 3613641184739, 33088666355144, 303670285138067, 2792497004892302, 25724693177503987, 237350917999324431, 2193027397174233046, 20288470364637624223

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) defined as the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases A099837(m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333095 (m = 3), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(2*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^2 = 1 + 2*x + O(x^2)
  n = 2: c(x)^4 = 1 + 4*x + 14*x^2 + O(x^3)
  n = 3: c(x)^6 = 1 + 6*x + 27*x^2 + 110*x^3 + O(x^4)
  n = 4: c(x)^8 = 1 + 8*x + 44*x^2 + 208*x^3 + 910*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 2 = 3, a(2) = 1 + 4 + 14 = 19, a(3) = 1 + 6 + 27 + 110 = 144 and a(4) = 1 + 8 + 44 + 208 + 910 = 1171.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(2*n), n >= 0, in descending powers of x begins
                                          row sums
  n = 0 |   1                                 1
  n = 1 |   2    1                            3
  n = 2 |  14    4    1                      19
  n = 3 | 110   27    6   1                 144
  n = 4 | 910  208   44   8   1            1171
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 395683988547 - 3 = (2^6)*(3^2)*(13^3)*312677 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 20288470364637624223 - 144 = (7^3)*17*269*12934629208861 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 150194008594715226556753 - 9878 = (5^6)*2593*5471* 677584325533 == 0 ( mod 5^6 ).

Cf. A000108, A069271, A333090 through A333097.

seq(add(2*n/(2*n+k)*binomial(2*n+2*k-1, k), k = 0..n), n = 1..25);
#alternative program
c:= x -> (1/2)*(1-sqrt(1-4*x))/x:
G := (x, n) -> series(c(x)^(2*n), x, 76):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Table[SeriesCoefficient[((1 + x)^2 * (1 - x - Sqrt[(1 - 3*x)*(1 + x)]) / (2*x^2))^n, {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 2*n/(2*n+k)*binomial(2*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^2(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + ... = (1/x)*Revert( x/c^2(x) ) is the o.g.f. of A069271.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), x*f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 2^(8*n + 7/2) / (13 * sqrt(Pi*n) * 3^(3*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} n/(2*n+2*k)*binomial(2*n+2*k, k) for n >= 1. - Peter Bala, Apr 19 2024

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 4, 34, 337, 3554, 38754, 431521, 4874377, 55639010, 640177033, 7412165034, 86256322816, 1007980394849, 11820510331777, 139032549536551, 1639506780365337, 19376785465043938, 229458302589724067, 2721958273545613513, 32339465512495259708, 384758834631081248554

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences: a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333096 (m = 4), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(3*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^3 = 1 + 3*x + O(x^2)
  n = 2: c(x)^6 = 1 + 6*x + 27*x^2 + O(x^3)
  n = 3: c(x)^9 = 1 + 9*x + 54*x^2 + 273*x^3 + O(x^4)
  n = 4: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + 2907*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 3 = 4, a(2) = 1 + 6 + 27 = 34, a(3) = 1 + 9 + 54 + 273 = 337 and a(4) = 1 + 12 + 90 + 544 + 2907 = 3554.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^n, n >= 0, in descending powers of x begins
                                              row sums
  n = 0 |    1                                    1
  n = 1 |    3       1                            4
  n = 2 |   27       6    1                      34
  n = 3 |  273      54    9   1                 337
  n = 4 | 2907     544   90  12   1            3554
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of supercongruences:
a(13) - a(1) = 11820510331777 - 4 = 3*11*(13^3)*(43^2)*88177 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 4583419703934987639046 - 337 = (3^2)*(7^4)*2441* 86893477573061 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 93266278848727959965820004 - 38754 = 2*(5^7)*19* 31416009717466260199 == 0 ( mod 5^6 ).

Cf. A000108, A118970, A333090 through A333097.

seq(add(3*n/(3*n+k)*binomial(3*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(3*n), x, 101):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[3*Binomial[5*n-1, n] * HypergeometricPFQ[{1, -4*n, -n}, {1/2 - 5*n/2, 1 - 5*n/2}, 1/4]/4, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 3*n/(3*n+k)*binomial(3*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^3(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 3*x + 18*x^2 + 136*x^3 + 1155*x^4 + ... = (1/x)*Revert( x/c^3(x) ) is the o.g.f. of A118970.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 5^(5*n + 3/2) / (7 * 2^(8*n + 3/2) * sqrt(Pi*n)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 3*n/(3*n+2*k)*binomial(3*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

Original entry on oeis.org

1, 5, 53, 647, 8373, 111880, 1525511, 21093476, 294663349, 4148593604, 58770091928, 836722722951, 11961868391175, 171601856667701, 2469036254872996, 35615467194043147, 514888180699419829, 7458193213805231529, 108219144962546395364, 1572690742149983040857

Offset: 0

Peter Bala, Mar 15 2020

The sequence satisfies the Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k.
We conjecture that the sequence satisfies the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. Examples of these congruences are given below.
More generally, for each integer m, we conjecture that the sequence a_m(n) := the n-th order Taylor polynomial of c(x)^(m*n) evaluated at x = 1 satisfies the same supercongruences. For cases see A099837 (m = -2), A100219 (m = -1), A000012 (m = 0), A333093 (m = 1), A333094 (m = 2), A333095 (m = 3), A333097 (m = 5).

			n-th order Taylor polynomial of c(x)^(4*n):
  n = 0: c(x)^0 = 1 + O(x)
  n = 1: c(x)^4 = 1 + 4*x + O(x^2)
  n = 2: c(x)^8 = 1 + 8*x + 44*x^2 + O(x^3)
  n = 3: c(x)^12 = 1 + 12*x + 90*x^2 + 544*x^3 + O(x^4)
  n = 4: c(x)^16 = 1 + 16*x + 152*x^2 + 1120*x^3 + 7084*x^4 + O(x^5)
Setting x = 1 gives a(0) = 1, a(1) = 1 + 4 = 5, a(2) = 1 + 8 + 44 = 53, a(3) = 1 + 12 + 90 + 544 = 647 and a(4) = 1 + 16 + 152 + 1120 + 7084 = 8373.
The triangle of coefficients of the n-th order Taylor polynomial of c(x)^(4*n), n >= 0, in descending powers of x begins
                                         row sums
  n = 0 |    1                               1
  n = 1 |    4     1                         5
  n = 2 |   44     8     1                  53
  n = 3 |  544    90    12     1           647
  n = 4 | 7084  1120   152    16   1      8373
   ...
This is a Riordan array belonging to the Hitting time subgroup of the Riordan group.
Examples of congruences:
a(13) - a(1) = 171601856667701 - 5 = (2^4)*3*(7^2)*(13^3)*33208909 == 0 ( mod 13^3 ).
a(3*7) - a(3) = 333475516822140871773101 - 647 = 2*(3^2)*(7^3)* 54012879303877692221 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 15187725485911657497382846255 - 111880 = (3^3)*(5^7)*29* 248279548173268475053 == 0 ( mod 5^6 ).

Cf. A000108, A212073, A333090 through A333097.

seq(add(4*n/(4*n+k)*binomial(4*n+2*k-1,k), k = 0..n), n = 1..25);
#alternative program
c:= x → (1/2)*(1-sqrt(1-4*x))/x:
G := (x,n) → series(c(x)^(4*n), x, 126):
seq(add(coeff(G(x, n), x, n-k), k = 0..n), n = 0..25);

Join[{1}, Table[4*Binomial[6*n-1, n] * HypergeometricPFQ[{1, -5*n, -n}, {1/2 - 3*n, 1 - 3*n}, 1/4]/5, {n, 1, 20}]] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = Sum_{k = 0..n} 4*n/(4*n+k)*binomial(4*n+2*k-1, k) for n >= 1.
a(n) = [x^n] ( (1 + x)*c^4(x/(1 + x)) )^n.
O.g.f.: ( 1 + x*f'(x)/f(x) )/( 1 - x*f(x) ), where f(x) = 1 + 4*x + 30*x^2 + 280*x^3 + 2925*x^4 + ... = (1/x)*Revert( x/c^4(x) ) is the o.g.f. of A212073.
Row sums of the Riordan array ( 1 + x*f'(x)/f(x), f(x) ) belonging to the Hitting time subgroup of the Riordan group.
a(n) ~ 2^(6*n + 3) * 3^(6*n + 3/2) / (31 * sqrt(Pi*n) * 5^(5*n + 1/2)). - Vaclav Kotesovec, Mar 28 2020
a(n) = Sum_{k = 0..n} 4*n/(4*n+2*k)*binomial(4*n+2*k, k) for n >= 1. - Peter Bala, May 03 2024

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A333094 a(n) is the n-th order Taylor polynomial (centered at 0) of c(x)^(2n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the Catalan numbers A000108.

A333095 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(3n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.

A333096 a(n) = the n-th order Taylor polynomial (centered at 0) of c(x)^(4n) evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4x))/(2*x) is the o.g.f. of the sequence of Catalan numbers A000108.